M.Sc_.-Mathematics-PAPER-I-Algebra-II-SUBJECT-CODE-PSMTPAMT201-Inside-Pages-munotes

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20 ALGEBRA II
For
()
() .
() , ( 1 )
(2)dd n
d
d
dkaa e
oa d
If o a k k d
Then a e
nd ki e d d d k
dk 
 
      

      
a
a
a
a
=b
b 

aa a=b b
=b
From (1) and (2), k = d.
() ( ) .ddoa o a d   aa=   
Thus, we have a subgroup. H = /ndad   Da   of order d.
We show H is a unique subgroup of G of order d.
Let K be any subgroup of G of order d.
K < G and G is cyclic
= K is cyclic.
Let K = . o(k) = d. () .oa d=A 
= d is the least positive integer such that () .dae A       
By division algorithm,  q, r ‰ z such that
1
.
()dq d d r dq n r dq n r d n q r d q r dr dqd r r d
dq d dr d q nr d
aa a a a a ae aaaaa b b 
a =  
=  ¸  AA
A
        
         
          
(*)rd daae b y=A    
But 0rd d d n    ab  and n is the least positive integer such that nae
rd  = which means r = 0 (d is positive)
.
.qd dqd a a a K H
But K H d
KH        
  
aaa= =       = ‡

=AA

= G has unique subgroup of order d for each positive divisor d of n. munotes.in

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23Chapter 1: Introduction to Groups
Proposition 2: Let G be an infinite cyclic group generated by `a’ and a1 are the
only generators of G.
Proof:- Suppose ,,mGam   ‰ ]
1and ( ) , .
and 1 ( distinct powers of `a' are distinct)
1
Generators of are andm
mm k
mkGa a
aa a ak
aa m k
m
Ga a  
=‰    ‰
= 
=v
=]
' 
    
    

    
Note : ] is an infinite cyclic group and 1 and – 1 are the only generation of ]
1.6 Permutation Groups
We have defined symmetric group {/ :nn n Sf f I Il   s.t. f is bijective} where
nI = {1, …., n}. (, )nS D is a group of order n! where ‘ D’ is composition of maps.
Reason: T‰Sn is determined by specifying T(1),.…, T(n). T(1) can be choosen
in n ways. After choosing T(1), T(2) can be choose in n - 1 ways and finally T(n)
can be choose in 1 way.
= No of elements in Sn= n (n – 1) = n!.
We have already seen T‰Sn is denoted by
T = 123 . . . . .
(1) (2) (3) ..... ( )n
n¬­ ž ­ ž ­ ž ­ TTT TŸ®
It is non-abelian if n b3
12
12 211234. . . 1234. . .Consider ,2314. . . 2134. . .
1234. . . 1234. . .,3214. . . 1324. . .nn
nn
nn
nn¬ ¬­­ žžT T ­­ žž ­­ žž ­­ Ÿ® Ÿ®
¬ ¬­­ žž TT TT  ­­ žž ­­ žž ­­ Ÿ® Ÿ®DD 
                       
Cycle: nST‰ i s s a i d t o b e a cycle o f l e n g t h r i f t h e r e e x i s t s
121. . . . .r ii inb b        s u c h t h a t T(i1) = i2, 23()iiT… 1() ,rriiT
1 ()riiTand ()kkTfor k v 1 } ,,{r iiy. It is denoted by 1(. . . .)r ii munotes.in

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24 ALGEBRA II
Note : 12 23 1 1 1() ( ) ( . . . . )rr r r ii i ii ii ii i             
A cycle of length 2 is called a transposition . A cycle of length 1 is denoted by ( k),
which means k is fixed.
Cycles x a n d y ‰Sn a r e s a i d t o b e d i s j o i n t s i f 12()rxii i    , 1()syjj    
11 and { }{ }rsii j j       .
We note that if x and y are disjoint cycles then xy = yx
Reason :- 11 1 1(. . .) ( ) , { } { }rs r s xi i y j j i i j j ‚            
1
1Then ( ) ( ) 1... 1xyi x yi i i f r
ii f r   
AA A A
A    
     
1
1
1() ( ) 1 1kk
kxy j x j if k
j
j if k s
 b 

   
 
      
11
21 1
11
1
1
1() { } { }
() ( ) ( 1 1 )
() ()
() () 1 1
() . . . } {rs
r
kr k
rixy k k if k i i j jyxi y i i f r i
yx i y i i
yx j y j j if k s
ji f k s
yx k k for k i i j            
       
 
    
      
    ^    
ƒ
b b  

 b b 

ŠƒAA A
}sj
xy yx
=
Theorem:
1. Every T‰Sn can be written as a product of disjoint cycles (unique upto order)
2. Every T‰Sn can be written as product of transpositions.
Proof : proof of (1) easy .
(2) Let T= 1T…. T2 where Tis are disjoint cycles.
Let 11(. . . .)r ii T
11 1 1 3 1 2 Then ( )( )....( ) ( )ir rii ii ii ii    T
=Each cycle is a product of transpositions
= T is a product of transposition.
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25Chapter 1: Introduction to Groups
Order of a cycle of length r in group nS:
1(. . . )r Let i i    T
21
12 13 1 11
11Then we note ( ) , ( ) , ( ) , ( )
() 1rr
r ii ii k ii ii
ii f o rk rTT  T T 
=T v b b             
  
Similarly for 2,..r ii 
() 1
() 1k
mm k
k
mm k rii f o r r k
ii f o r r k m r
T b 
T   b b   
    
() 1k
mmii f o r k k r =T  b        
()
()r
mm
r
mmand i i
ii f o r i m rT
=T  b b 
       
,rI =T  and r is the least positive integer such that Tr = I
() . r =TT  
2. Order of nST‰ w h e r e T1,. . .kT  a r e d i s j o i n t c y c l e s o f l e n g t h 1.... .k rr
11 1(. . . . ) [ () . . . . () ] [ . . . . .]kk k Ol c m O o l c m r rTT  T T      
Definition : (Even and Odd Permutation) A permutation that can be expressed as
a product of an even number of 2-cycles is called an even permutation . A
permutation that can be expressed as a product of an odd number of 2-cycles is
called an odd permutation .
Definition: (Sign of a permutation) The sign of a permutation Bis said to be 1 if
B is even and – 1 if B is odd.
1i f i s e v e n)1i f s(i o d dsign£ B ¦¦B¤¦B¦¥
Lemma: For 12 nS TT‰ , 12 1 2 )() ( )( )is i g n s i g n s i g n TT  T T   
1)( ) ( )ii sign signT T  
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26 ALGEBRA II
Proof: ( i ) I f 1Tis a product of k 1 transpositions and 2Tis a product of k 2
transpositions then 12.TTis a product of 12kktranspositions.
Thus, 1212.) 1(( )kksignTT 
1212.) ( 1 )(. ( 1 )kksign TT   º
12 1 2.) () . ()( sign sig insg n TT  T Tº
Let An = (nST‰ : Sign( T) = 1) ( This group is called Alternating group i.e. it is
the group of all even permutation of S n )
We shall prove later that A n < G and .2n
nSA
Generators and relations : A group generated by a finite set S, stashing a set of
relations, which give all possible finite products of elements in the group can be
described by the generators and relations.
e.g. Dn = {e, a, …a1n, b, ab,…. a1nb} = , 21 nnae b b a a b   
1.7 Lagrange Theorem
This is one of the most importaut theorems for finite groups
Definition: Cosets: Let G be a group and H be a subgroup of G for a ‰G, we define,
Ha = { :}aHMM ‰ to be the right coset of H in G containing a and
aH = {a M‰H} to be the left coset of H in G containing a
For example, if G = S3= {I, (123), (132), (12), (13), (23)} and H = {I, (1,2)}
Then the right cosets of H in G are,
HI = H = {I, (12)}
H (123) = {(123), (12) (123)} = {(123), (23)}
H (132) = {(132), (12) (132)} = {(132), (13)}
We now prove Lagrange theorem
Theorem: If G is a finite group and H is a subgroup of G then O (H) \ O(G). munotes.in

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27Chapter 1: Introduction to Groups
Proof : We define a relation  in G as follows : For a,b, ‰G, a  b if and only if
1. ab H‰ We show  is an equivalence relation in G we note a  a for each a
‰G since 1. aa e H‰ is is reflexive=
For 1,,ab Ga b a b H   ‰º ‰
11 1()ab H ba H b a      º‰ º ‰ º 
. is symmetric=
11,,,, , , ,For a b c G a b b c ab H bc H        ‰º ‰ ‰
11 1ab bc H ac H c a     º‰ º ‰ º .
is transitive=
Thus  is an equivalence relation an G.
1,[ ] { / } { / }( )
{/ , } {/ }aG a xG a x xG x a i s s y m m e t r i cxGx a H x Gx H H a        
          ‰ ‰  ‰
‰  M M ‰ ‰  M ‰  ' 

2) Thus, we have the following,
For a ‰G, a ‰ [a] = Ha
aGaG
‰ˆ* .
Moreover, as two equivalence classes [a], [b] are either disjoint or equal,
Ha‚Hb = Gor Ha = Hb for a, b ‰G.
aGGH a
‰=*
3) Let a ‰G
We now define a map :( ) .HH a s . t . a      l  M  M
The map is well-defined and for
12 1 2 1 2 1 2,, ( ) ( ) Ha a          MM‰  M  Mº M M º M M =The map is 1-1
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28 ALGEBRA II
The map is clearly onto.
= the map is bijective.
12 1.
....... , ,k
kHH a
GH a H a H a w h e r e H a
Ha are distinct right c sets of H in G 
     
   R   =
=   
() ( ) ( ) ()Gk H
ie o G k o H and o H \o G  
    =

Definition : The number of right cosets of H in G is called index of H in G and is
denoted by (G : H) or :.GH
1:,, Note For a b G Ha Hb ab H          ‰ ” ‰
We also note that the number of left gets of H in G is same as the number of right
cosets of H in G.
Left coset of H contain `a’ = aH = {ah : h ‰H}.Further, aH = bH iff 1. abH‰
Proof:
1
1,
,.
,,
,,aH bH b be b H and hence b a H b a for some
ab H
Conversely a b H b ah b a H
xb H x b H x a a H
bH a H
º  ‰ ‰ º  M M ‰
º M M ‰
M M‰ º  ‰
=‰ ºM M ‰ ºM M ‰
=ˆ         +
     
          
      
  
1
1,
.xaH x a H x b bH
aH bH
aH bH iff a b H
aa a ‰º  M M ‰ º  M M ‰
=ˆ
= ‰          
 
   
There is 1-1 correspondences between the set of right cosets of H in G and the set
of left cosets of H in G
1. Ha a H l 11 1 1 1 1() Ha Hb ab H a b H H b       D   +    ” ‰ ” ‰ ” 

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31Chapter 1: Introduction to Groups
1.9 Unit and Exercises
1) If R is an equivalence relation on a non-empty set A, then show that –
a) <>aa‰ (by reflexivity)
b) <><><> ba a b‰º  
c) For ,ab A‰, either <><><><>ab o r a b G          
Thus, an equivalence relation divides the set into disjoint equivalence classes.
2) Determine if f: l' defined by mfmn¬­ž­ž­žŸ® is well defined.
12Is f f ?12¬¬ ¬­ ž­­žž ­ ­­žžž ­ ­­žž ­ žŸ® Ÿ®Ÿ® 
Determine if f: l' defined by mfmn¬­ž­ž­žŸ® is well defined.
12Is f f ?12¬¬ ¬­ ž­­žž ­ ­­žžž ­ ­­žž ­ žŸ® Ÿ®Ÿ® 

3) Determine whether the following functions are injective, surjective, and
bijective.

i) f: l'' defined by
fx 3 x 2
ii) f:l defined by
1fxx
iii) f: l defined by
fz z
iv) f: l defined by
2fx x 
4) Let f: A B l be a surjective map. Show that the relation R defined by aRb
if and only if

fa fb is an equivalence relation. Find the equivalence
classes.
5) Consider a relation R on ' defined by aRb if and only if ab 0p. Is R an
equivalence relation?
6) Find

200 , 350GG 
7) List elements of U(24). munotes.in

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32 ALGEBRA II
8) Which elements in ] have an inverse w.r.t. multiplication.
9) Find inverse of 5 in 6]
10) Is multiplication a binary operation in the set of odd integer?
11) List elements in 10]which are invertible (we will prove a general result later)
12) Show that `0’ is a binary operation in z where aob = a+b+5, for a, b ‰].
Show `o’ is commutative and associative.
13) Find identity of 10] and inverse of 2.
(14) Find generators of Z6, Z20 and subgroups of Z10
Ans:
(i) 66(6, ) 1, 1 5 ( ) ri f f r r Z i =         bb  
6 5, 6 (2) (3) 1 2 2 generaton of Z are              = G G  q 
(ii) 20 (20, ) 1,1 20 rZ i f f r r r        bb º
137 , 9 , 1 1 , 1 3 , 1 7 , 1 9      
22(( 2 0 ) ) ( 5 ) ( 2 2 )4 8 )     G GG   <
(iii) Subgroups of 10Z, order 1, 2, 5, 10
10 0{ 0 } , 5 { 0 , 5 } , 2 { 0 , 2 4 , 6 , 8 , 1 0 } 1Z                  
(15) List the elements of subgroup <20> in Z30.
(16) List the elements of subgroups <3>, <7> in Z20.
(Ans :- 3{ 1 3 9 7 }
7{ 1 7 9 3 } 
  

(17) Let a be a group and a ‰G. If o(a) = 15 state orders of 35 4,,.aa D
(18) Let G be a group and a ‰G. If o(a) = 24. Find a generator for
21 10 6: aa A n s a ‚     
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34 ALGEBRA II UNIT 1
2
ISOMORPHISM OF GROUPS
Unit Structure
2.0 Objectives
2.1 Homomorphism And Isomorphism
2.2 Cayley’s Theorem
2.3 Automorphisms and Inner Automorphisms
2.4 External Direct Product of groups.
2.5 Normal sub groups and Quotient (Factor groups)
2.6 Isomorphism Theorems
2.7 Classification of groups of order b7 upto isomorphism
2.8 Fundamental Theorem Of Finite Abelian Groups
2.9 Summary
2.10 Unit and Exercises
2.0 Objectives
After going through this unit you shall come to know about
x Relation between the groups and the existence of isomorphisms between
groups
x Special kind of subgroups called normal subgroups and its impotance in the
construction of factor groups
x Classification of groups upto order 7 using various results of isomorphism
theorem


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35Chapter 2: Isomorphism of Groups
2.1 Homomorphism and Isomorphism
Definition: 1] Let G, G be groups. A map f : G Gl is called a homomorphism
of group G to group G if f(ab) = f(a) f(b) for each a, b ‰G (i.e. f preserves group
operation)
Definition: 2] Let G, G be groups. A map f : G Gl is called an is isomorphism
of group G to group G if i) f (ab) = f(a) f(b) for for each a,b ‰G (ie f is a group
homomorphism) and ii) f is bijective.
This is denoted by G x G
Properties of group homomorphisms and isomorphisms.
Let G, G be groups and f : G l Gbe a group homomorphism and a ‰G
Then
(i) f(e) = eawhere e, ea are identity elements of G, G respectively.
(ii) f (a1) = (f (a))1
(iii) f(ak) = (f(a))k for each k ‰ ]
Proof : (i) ea= f(e) = f(e.e) = f(e) . f(e)
= By right cancellation law, f(e) = ea
(ii) f (aa1) = f(e) = ea
11 1
11() ( ) () ( () )
() ( ( ) )fa fa e fa fa
fa fa
= 
=    
 
(iii) We prove inductively, f (an) = (f(a))n “nN‰
The result is true for n = 1
Suppose the result is true for k, k ‰N
f (a)( ( ) )kkfa 
Then, 11 1()( )( ) ( ) ( ( ) ) ( ) ( ( ) )kk k k kfa fa a fa fa fa fa fa        
,( ) ( ( ) )nnInductively f a f a   “
01 0
11 1() ( ) ( )) ( ( ) ) ( ) ( ( ( ) ) )() ( ) ( ( ) ) , ,
,{ , , },
nm m m mfa fa fa fa b yi i fanNfaf e e f a t h e r e s u l t i s t r u e f o r
for n n n m m N
     
       
        
    ‰
 
‰  ‰]
(( ) ) (( ) )mnfaf a 
Thus, the result is true for each n ‰]. munotes.in

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36 ALGEBRA II
Properties : Let G, G be groups and f : G Gl be an onto group homomorphism
(i) If G is abelian, so is G
(ii) If G is cyclic, so is G.
(iii) O(f(a)) () .oa If f is an is isomorphism, then o(f(a)) = o(a).
Proof : (i) Let ,,ab Gaa‰ As f is onto, for a,b, ‰G such that
() , ()faa f bb  aa
Then, () () ( ) ( ) () () ab f a fb f a b fb a fb f a b a       aa aa  
Thus, G is abelian.
(ii) Let G = < a > = {: }nan  ‰]
() ( { : } ) {( ) : } { (( ) : } ( )
.nn nGfGfanfanfanfa
Thus G is cyclic          
 ‰  ‰  ‰   ]]]
(iii) Suppose o(a) = n, then nae
(( ) ) ( )
(( ) )( )nnfaf a e
fa o a  
a = 
=
If f : G Gl is an isomorphism, then
(( ) ) (( ) ) ( ) (( ) ) ( )mm mofa m fa e fa fa e fe     aa º º  
() /maef i s o n e o n e n m    º  º '
() / .By iii m n m n          =
Result : If G, G, are groups, and f : G Gl i s a n i s o m o r p h i s m , t h e n
1:fGGl    is an isomorphism.
Examples of homomorphism and isomorphism .
(1) For any group G, I : G l G (I = identity) and the trivial may f : G lG defined
by f(x) = e are group homomorphism.
(2) The map :RR l   defined by 3()xx   is a group homomorphism.
(3) Consider the map f : ]n l] defined by f(m) = m is a group homomorphism. munotes.in

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37Chapter 2: Isomorphism of Groups
Theorem 1: Any finite cyclic group of order n is isomorphic to Z ,nthe group of
integer residue classes modulo under addition.
Proof : Let G be a cyclic group of order n generated by `a’.
Then, G = {e, a, … a1n} and for m ,mraa‰]
01
:, m o d
() 1 ( m o d )
,, 1
,.
\mr
nn
r
rs
rsrn
Ga a b y w h e r e m r n
we define a r for o r n r r n
is well defined for a a r,s n
ae o r s n
nr s a n d r s
           
             
      R   
  
      bb
l ‰  w
 b b 
  b b 
º b  
º   º]
rs rs º
 is a group homomorphism, since
mod
() ()rsand r s t n
trsrs a a 
      w
ºw   
.
{, 1 , 1 } , ( ) .
.r
nnis clearly onto
on a n d f o r r r a
is a group isomorphism 
                  
 
=  ‰  
=]]
() () , , 1
,1rs
rsaa r s n
rs rs rsn n
aa
is one one          
     R  

   b b 
ºº bb
º
='
Corollary: Any two finite cyclic groups of same order are isomorphic
(Consider 1)rrab b y a b o r n l  l bb      



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38 ALGEBRA II
Theorem 2: An infinite cyclic group is isomorphic to the group of integers under
addition.
Proof : Let G be an infinite cyclic group generated by `a’, {}nGa n z‰ 
Consider the map :( ) .rGd e f i n e d b y a r f o r r         l   ‰]]
For ,( , )rsaa Gr s    ‰‰ ]
(, ) ( ) () ()rs r s r saa a r s a a              
is a group homomorphism= 
() ()rs r saa r s a a º  º   
is one one= 
,( )rrFor r a G such that a r‰‰  ]       . is onto=
Thus,  is a group isomorphism
Corollary : Any two infinite cyclic groups are isomorphic
(If < a >, < b > are infinite cyclic groups consider ,) .rrab r  l‰ ]

2.2 Cayley’s Theorem
Cayley’s Theorem : Every group is isomorphic to a group of permutations
Proof : Let G be a group
For a G‰, we define f a : G l G by f a (x) = ax (f a is multiplication by a on left)
We show f a is bijective map, ie f a is a permutations on G.
fa (x) = f a (y) for x, y Ga x a yx y‰º º 
= fa is one one
For y ‰G, 1xayG ‰  such that

1
.
{: }a
a
afx a x a y y
fi s o n t o
let G f a G
 
    
=
‰ munotes.in

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39Chapter 2: Isomorphism of Groups
Then G i s a g r o u p u n d e r c o m p o s i t i o n o f m a p s f o r a , b f o r


,, ( ) ( ) ( ) ( ) ( )ab a b a a b ab G f of x f f x f b x ab x a b x f x‰
















,ab ab a bfof G f of f =


‰






ef is the identity element of G.
()efx e x x V




' ()eaxGfx

‰



 ax
() () () ()ae a e a eafof x f x f x f of x










1
af is the inverse of af
111 1
aa a a e a a aafof f f f f of  













'
Gis a group and :GG


l
defined by
a af

is an group isomorphism.
' For


,,ab a b ab G a b f f of a o b‰
















() ()ab a b ab f f f e f e a e b e a b


º



º



º



º


 is clearly onto.
= G is isomorphic to ,Ga group of permutations.
Note : The group G consummated above is called the left regular representation
of G
2.3 Automorphisms and Inner Automorphisms
Definition: An isomorphism of a group G onto itself is called an automorphism
of G
Examples
1) Identity map is an automorphism of any group to itself
2) f : G lG defined by f(a + ib) = a – ib is automorphism of the group (G, +).
Let Aut G = {f : f : G lG, f automorphism of G }
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40 ALGEBRA II
Definition : Let G be a group and a .G‰ T h e m a p :aiG G l d e f i n e d b y
1()aix a x a f o r xG‰     is called the inner automorphism of G induced by a.
Let Inn (G) =1{: ( ) , }aaiG G ix a x a aG         l‰
Theorem: Aut G is a group and Inn G < Aut G
Proof : If f, g ‰ aut G, f D g ‰ Aut G, D is associative
i ‰ Aut G, For f ‰ Aut G. then map f1 ‰ Aut G,
= Aut G is a group
11 1.
,, ,
.ed
ab
ab ab a biiI n n GFor i i Inn G where a b G
ii ii i I n n G
Inn G Ant G ‰
‰‰
 ‰
= 
      
     
  
2.4 External Direct Product of Groups
Definition : Let, 12,, . . . .n GG G   be finite collection of groups. The external direct
product of 1,...n GG denoted by 12 ...n GG G€€ €     is defined by
12 1 {( , .... ) / 1 }nn i i GG G g g gG i n€€ € ‰ b b              where
12 1 2 1(, , ) ( , . . . . , ) ( , . . . . . . . . . , )nn n n gg g g g g g g g g           aa a a a <
It can be easily verified that 12 .... .n GG G i s a g r o u p   €€€
11 2 1 1
111 1 1 1
111(, . . . ) , ( , . . . . , ) , (" . . " ) . . . . . .
(( , ..... ) ( .. )) ( ",..... ") ( , ...., ) ( ,".. ")
(( ", ... ( ") "nn n n
nn n n n n
nn nFor g g g g g g g G Ggg g g g g g g g g g gggg g g g            
                      
  aa a ‰€€
aa
a
11 1 1
11 1(( " ) ( " ) ) ( , . . )
(, . . . . , ) ( ( ) (" . . . . . " ) ) .nn n n
nn ngg g g gg B y a s s o c i a t i v i t y i n G G
gg g g g g                    
               aa
aa 
Thus, associatively holds in 1 n GG        €€ munotes.in

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41Chapter 2: Isomorphism of Groups
Let 11,......., .......nn ee b e i d e n t i t y e l e m e n t s g r o u p s o f G G r e s p e c t i v e l y     
11
11 1 1 1
11 1,......., .......
,( , . . . . . , ) ( , ) ( , . . . . . . , ) ( , . . . . . , )
(, . . . . . . . . . . . . . , ) ( ) (nn
nn n n n
nn nee b e i d e n t i t y e l e m e n t s g r o u p s o f G G r e s p e c t i v e l y
Then g g e e g e g e g g
eg eg e e    
        
     ¸ 
1 )n gg
1(, . . . )n ee=  is identity element of 1 .n GG€€   
Let
1
11 1 1
11 1
1
1111 .
1.
,( ,. . )( ,. . . . , ) ( )
(, . . . . , ) ( ,ii i
ii i
nn n n
ngG f o r i n a n d G i s a g r o u p f o r i n
Let g be the inverse of g in G for i n
Then g g g g g g g g
ee g g        
            
              
           
  
‰b b b b
bb

11 1
11 )( , . . . . . . , ) ( , ) .nn n ngg gg g      
Thus, 11
11(, . . . . . . . . . . , ) ( , . . . . . . . . . . , )nn gg h a s a n i n v e r s e g g    
11......... . .........nn in G G G G is a group     €€ = €€
This is called the external direct product of 1.........n GG €€
Examples :
1) 23 {(0, 0), (1, 0), (0,1), (0, 2), (1,1), (1, 2)} ZZ       €
This is an abelian group.

Properties of external direct product.
Let 1,n GG be finite groups and 1 n GG        €€ be their external direct
product.
(1) 11 2 1 2() ( ) ( ) . . . . ( ) .nn n OG G O G O G O G o r G G G              €€  ¸ ¸¸
(2) For 11,.., ) .... ,nn gg G G   ‰€€ then 11(, . . , ) ) ( () , . . , ( ) ) .nn Og g l c mo g og   


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42 ALGEBRA II
Proof : (1) is clear by property of cardinality of Cartesian product
(2) Let () 1iiOg n f o r i n    b b
1
1111 1
111
1[, . . . , ]
1, 1 .
(, . . . . , ) ( ) ( , )
(, . . . , )( , . . . . . , )
(k
ii i
nr nr nnnnn
r r nnnLet cm n n
Then n for i n let n r for i n
Then g g g g g g
ee ee
Og 
        
              
   

bb  bb


=AA AAA
AA
,) \ngA
Then 1(, . . . ) \nLet O g g k ie k      A
11
1
2(, ) ( , ) ( * )
(, )
1
.kk k
nn
n
k
igg g g
ee
ge f o r i n
k

= b b
A    
  
      
 
Note, by induction, the result is true “m‰
Theorem : Let 1,n GG be finite cyclic groups of orders 1,n nn respectively,
Then, 1....n GG  €€ is cyclic if and only if g.c.d (, ) 1 1 ,ijnn f o r ij n       b b
Corollary: Let 1. k mn n     Then , mn n k   !€ €'' '
,.ij if and only if n n are relatively prime for i j        v
Note : If 12,GG are abelian groups, so is 12GG€
Theorem: Let m and n be positive integers of g.c.d (m,n) = 1, then U(mn) is
isomorphic to U(m) € U(n).
Proof :Consider the map f : U(mn) l U(m) € U(n) defined by,
f(x) = (x mod m, x mod m)
The map f is well-defined,
For if x, y ‰ U(mn), f(xy) = (xy mod m, xy mod m)
= x mod m, y mod m, x modm, y mod m)
= (x mod m, y mod m) (y modm, y mod m)
= f(x) f(y)
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43Chapter 2: Isomorphism of Groups
Let f(x) = f(y) º(x mod m , x mod m) = (y mod m, y mod m)
ºx mod m = y mod m x mod m = y mod m
ºx w y mod mn ( = m, n are relatively prime)
= f is one – one
f is onto by Chinese Remainder Theorem.
Therefore, f is bijective
Clearly f is a homomorphism.
Thus, U(mn) is isomorphic to U(m) € U(n).
Corollary : Let 12 1,. . , . . ( , ) 1j mn n g c d n n f o r i j          v
12 () () ( ) . . . ( )k ThenU m U n U n U n  !€€ €
2.5 Normal Sub Groups and Quotient (Factor Groups)
Normal Subgroup
Definition: Let G be a group. A subgroup H of G is called a normal subgroup of
G if aHa1 ˆ H for each aG‰
1(, )or aha H for each a G each h H        ‰‰ ‰
This is denoted by H  G
Theorem 1 : Let H be a subgroup of a group G.
Then, the following statements are equivalent.
11) (HG i e a H a H  ˆ“ 
1)
2)
3)
4) ,a
ab a baG
aHa H for each a G
aH H for each a G
HH H f o r e a c h a b G
      
    
      ‰
‰
‰
‰
(or aHbH = abH for each a, b ‰G)

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44 ALGEBRA II
Proof :- () ( )ii i
º

Suppose HG
1
11
11 1
1,
,( * )
(( * )
.Let a G then aHa H
Also a G a Ha H
Ha a a a a H a b y
aHa H   
        
  +    
 


‰ˆ
‰= ˆ
= ˆ
=
1() ( )
.ii iii
Ha aHa a a He aHº
 
  
() ( )
()abiii iv
HH H H a b H a b H i s c l o s e d u n d e r m u l t i p l i c a t i o n
    º
ˆ =
()
(Hab Heab HHab HaHb Ha aH
Hab HaHb
HaHb Hab or aHbH abHH ab H     

      = ˆ 
=ˆ
= ˆ
.
()He HH H
aHbH abHH abH 
 ˆ
=
(iv)º(i)
11 1 1,
.For a H aHa eaHa HaHa Haa H
HG      
  ‰ ˆ ˆ 
=
Note : If G is abelian, and H < G, then H G.
Theorem : Let H G, Let /{ :}GH H a a G    ‰
Then, the operation HaHb = Hab on G/H is a well defined binary operation in G/H
and G/H is a group under this binary operation.



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45Chapter 2: Isomorphism of Groups
Proof : We first show that this binary operation is well defined
Let
11
12 1 2
11 1 1 1 1 1 1
22 1 2
11
13 3 2,, .
()
()Let aa h bb h h h H
we show Hab Ha b
ab a b ab b a ah a aa a h a h a h a
hh w h e r eh ab a H H G      
  
     
      
    
a a ‰
aa
aa a a a a aaa  
a  ‰ = 
1
13 () .
.
.ab a b h h H
Hab Ha b
ie HaHb a Hbaa= ‰
aa =
aa    
 
 +
and the operation is well defined.
For
() (),/ ,
(abc
ab c a b c a b c a b c a b c a bcHH G H
H‰  +   
+  +   +   +  +   +  + +    + ++ 
= The binary operation is associative:
/e
ae a e a e a eaHH G H
HH H H H HH‰
  
   
=He = H is, the identity element of G/H
For
1
11 1 1/, /aa
aa a a e a a aHG H H G H s u c h t h a t
HH H H a H H
 ‰ ‰
     
  +  
1
aH= is the inverse of aH
Thus, G/H is a group.
It is called the quotient groups of G by H (or the factor group of G/H).
Note : If G is finite o(G/H) = o(G)/o(H) = [G : H].
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46 ALGEBRA II
Example: Let us consider ]/n]
Let us find the cosets ]/n]
For a ‰ Z, we know ,qr ‰] such that ,1 aq nr rn   b b 
., 1 .
,, 1arn an rn rn
Moreover for r s n     
        =‰ =  bb
bb ]]]

,rn sn rs n
rs o r rs n rs rs    
        º  ‰
º ºº]] ]

= The distinct cosets of Z/n ] are o + ] , 1 + ],……, (n – 1 ) + ].
Propositions 1 : Let G be a group. If G/Z(G) is cyclic, then G is abelian.
Proof : Suppose G/Z(G) = {(Z(G)x)n : n‰]} =
,( ) , ( ) ,( ) , , , ,nmFor a b Z G let a Z G x b Z G m n ‰‰ ‰ ‰ ]      
12 1 2
12 1 2 2 1 2 1,, , ( )
.nm
nm n m m n n naz x bz xz z Z G
ab z x z x z z x x z z x x z x z x ba
Gi sa b e l i a n=  ‰
   
=     
     

Proposition 2 : G/Z(G) x Inn G
Proof : C onsider f : G/Z(G) l Inn G defined by f(Z(G)x) = ix where
1(( ) )xia x a x 
11
11 1 1 1 1 1() () () , , ()
,( ) () ( ) ( )x yZGx Gy x y G L e t x y zz ZG
For a G i a xax zya zy zyay z yay zz yay i a  =   =      
         
     =º ‰  ‰
‰    
.xyiifis well defined=
=
 
f (Z(G)x. Z(G)y) = f (Z(G) = y) = ixy munotes.in

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47Chapter 2: Isomorphism of Groups
11 1,( )( ) ( ) ( ) ( )
(() ()) (() ) (())
hom .xy x y
xy x y
xy x yFor a G i a xy a xy x yay x i i a
ii ifZG x ZGy i ii f ZGx f ZGy
fi s a g r o u p o m o r p h i s m       
 
      
  ‰  
=
=  
=
,, ( ( ) ) ( ( ) ) ( ) ( )xy x y For x y G f z G x f z G y i i i a i a     ‰ º  º  “
11aG
xax yay
 ‰
º“11 yaG yx a a y x     ‰º  “1(
() ()aG yx G
zG y zGx
fi s o n e o n e   =  
‰º ‰
º
=
Clearly for ix‰ Z(G), f(Z(G)x) = xi and f is onto
=f is a group isomorphism.
Definition: Internal Direct Product :- Let 12,nHHH      be normal
subgroups of G we say G is the internal direct product of 12HH if
(i) G = 12 1 2,{ / , 1 }nn i iHHHh hh h Hi n                    ‰ b b
(ii) 12 1,{ } 1 , 1 .ni HH H H ef o r i n                 ‚  
Note : {} )ijHH e i f ij‚ v  
In particular, if H, K are normal subgroups of G, then G is the internal direct product
of H, K if.
(i) G = HK = {/ , }hk h H k k ‰‰
(ii) . Hk‚ 
Theorem : If a group G is the internal direct product of subgroups H1,..,Hn then
12 ..n GH H H€ €    
Proof: For 11 1 1,, ( )ii j j i j i j i j i j jhH hH h h h h h h h h H             ‰‰  ‰
11 1 1
1()
{}ij i j i j i j i
ij i i j
ij j ihh h h h h h h H
hh h H H e
hh h h    
   
 
‰
=‰ 
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48 ALGEBRA II
1
11
12 12
11 1 1
12 1 1 2 2 1
11 1 1 1
11 2 2 1 1,, 1
() ( ) ( )ii l
nn
nn n n n
nnFor h h H i n
hh h hh h
hh h h h h h h h
hh h h h h


 
=‰ b b

º
       
  
    
    
11
1
12 1 1nn
nn
ni nhh e
hh
hh h h h
º
º
= 

    
By similar argument, 1
11nnhh
Proceeding in this manner, 11iihh f o r i nb b   
We define 1 :. . . . . . .n GH H b y      l€ €
11() () ( )nn hh hhi s w e l l d e f i n e d                       €
11 1 1 1 1
11 1 1 1 1
11 1 1
11 1 1(( ,..., ) ( )) ( ,.., ) ( ,..., )
(, . . , ) ( ) (, . . . , ) ( )nn n n n n
nn n nhh h h h hh h h h h h
hh h h h h h h           
                  
  
= is a group homomorphism
11 11
11
11
1() ( ) () ( )
1.
.ni n n i n
ii ni nhh h h hh h h
hh f o r i n h h h h
is one one
is clearly onto                        
            
  
   º
º b b º 
= 

= is a group isomorphism.
2.6 Isomorphism Theorems
We recall that a homomorphism from a group G to group G is a map :fGGl  
such that f(ab) = f(a) f(b) for each a, b ‰ G.
Let G, G be groups and f : G Gl be a group homomorphism we define kernel
of f (denoted by ker f) asker f {x G/f x =e } c  is the identity element of G.
Proposition 1: Let G, G be groups and f : G Gl be a group homomorphism.
Then, Kerf  G. munotes.in

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49Chapter 2: Isomorphism of Groups
Proof :- We know, ()fe e e K e r f K e r f   a= ‰ º v 
11 1 1 1 1 1
1,, ( ) ( ) ( ) ( ) ( ( ) )
.For a b Kerffabfafbfafbe e e
ab Ker f and Ker f G        
       
‰   
=‰ 
11 1,k e r , ( )( ) ( ) ( )( ) ( ( ) )For a G h f f aha f a f h f a f a e f a e          aa ‰‰   
1ker , ker
keraha f for a G h f
fG         
 =‰ ‰ ‰
=
Proposition 2: Let G, G be groups and f : G Gl be a group homomorphism.
Then, f is one – one if and only if ker f = {e}.
Proof :- Suppose f is one-one
Then x ‰ kerf
() () () k e r { }fxe f x f ee x e f e       aa º º º  =
Conversely, suppose ker f = {e}
11
11 1,, , ( ) ( ) ( ) ( ) ( ) ()
() k e r
.Then for x y G f x f y f x f y e f x f y e
fx y e x y f x y e x y
fi s o n e o n e               
           
 
 aa ‰ º  º 
a º º ‰ º  º 
=
Note: If G, G are groups and f : G Gl is a group homomorphism, then
Imf = {f(x) : x G‰} is a subgorup of G.
Proposition: Let G be a group and H ,GThen p: G lG/H is surjective group
homomorphism, where p(x) = Hx. For x ‰ G and ker p = H
Proof: For x, y ‰ G, (p(xy) = () ()xy x yHH H p x p y  
=p is a group homomorphism.
p is clearly onto. (For /, ( ) ( )xHGHp x H x x G s u c h t h a t    ‰  ‰
kerxxpH H ‰”     ( identity element of G/H is H)
kerxH
pH”‰
=
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50 ALGEBRA II
Note:- Thus, if H G, where G is a group, then H is a kernel of a group
homomorphism from G to a suitable group.
We next prove the Isomorphism Theorems.
First Isomorphism Theorem : Let f : G Gl be a homomomorphism of groups.
If f is onto, G/kerf Gx (or in general G/kerf x Im f).

Proof : Consider the mapping f: G/ker f lG defined by,
(ker ) ( ) ( ( ) ( )) kerffa f a o rf K a f a w h e r e f          .  
This map is well defined
For Ka = Kb, a, b ‰ G
11 1(k e r) ( ) ( )( ) ab K f f ab e f a f b e         aa º‰  º  º 
1() ( () ) () ()faf b e f a f b   a º º  f(Ka.Kb) = f(Kab) = f(ab) = f(a) f(b) = f(Ka)f(Kb)
=f is a group homomorphism. f(Ka) = f(Kb) º f(a) = f(b)
=f is one – one.
If f is onto, for aGc a ‰ G such that f(a) = aa
() ( )
.fKa f a a
fi s o n t o  
a = 
=
Note :- 1. If f is not onto, we consider f : G/ker f l Imf and G/kerf x Imf.
2. The above theorem, is also called. The Fundamental Theorem Of
Homomorphism Of Groups.
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51Chapter 2: Isomorphism of Groups
Second Isomorphism Theorem: Let H and K be subgroups of a group G and
K  G. Then, HHKHKK ˆ
Proof : We note H K < G and K < HK,
:/ ( )KG KH K
we define a map f H HK K by f h hK for h H
            º
l ‰

Then, for 12 1 2 1 2 1 2 1 2,, ( ) ( ) ( ) ( ) ( ) ( )hh H fh h h h h Kh K fhfh     .  ‰ 

= f is a group homomorphism.
f is onto for any element in HK / K is of the form hkk(hk)
And hk = f(h)
Ker f = {h ‰H : hK = K} hKŸŸ h = H ..‚
= By First Isomorphism Theorem of groups HHKHKK ˆ
Third Isomorphism Theorem of groups:
Let G be a group and H, K be normal subgroups of G If K H, then
(G/K) / (H/K) x G/H
Proof : Consider the mapping f: G/K lG/H given by f(Kx) = Hx
This map is well defined, for
11
() ( ) ( ) ( )xyK Ȁxy k xy H Hx Hy
fK x K y fK x y H x y ǾxHy f Kx f Ky    
    º ‰ º ‰ º
  
= f is a group homomorphism.
f is clearly onto.
' Any element of G/H is of the form Hx, x ‰G
()fKx Hx =
ker { / ( ) } / , } /for f Kx x G f Kx H Kx x G Hx H H K        ^     = ‰  ‰  
= By First Isomorphism Theorem, G/K/H/K x G/H munotes.in

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52 ALGEBRA II
Theorem: Let 12 1 1 2 2,, ,GGb e g r o u p s a n d H GH G      Then
12 1 2HHG G a n d   €€12 1 2
12 1 2GG G G
HH H H
€x€€

Proof : Consider the map 12
12
12:GGfG GHH€l€
12 1 1 2 2(. ) ( , ) defined by f g g H g H g      

Then, for 12 1 2 1 2(, ) , ( )gg g g G G    aa‰€
12 1 2 1 1 2 2 1 1 1 2 2 2(( , ) ( , )) ( , ) ( , )fgg g g f g g g g H g g H g g       aa a a a a
1
11 1 1 2 2 2 2 11 2 2 11 2 2(, ) ( , ) ( )Hg H g H g H g Hg H g Hg H g      aa a a 
12 1 2
12
11 2 2
12(, )( , )
hom .
(, )fgg f g gfis a group omorphism
GGFor H g H gHH    
 
 aa 
=
‰€

11 2 2 1 2(, ) ( , )Hg H g f g g
fi s o n t o
=   
 
12 1 2 12 1 12 2 1 2
11 2 2 1 2 1 2(, ) k e r ( , )(, ) ( , )( )
,( , )gg f f g g H H H g H g H H
gH g H g g             
        +  +‰”  ” 
”‰ ‰ ” ‰€
= By First Isomorphism Theorem,
12 1 2
12 1 2GG G G
HH H H
€x€€
Correspondence Theorem: Let G, G be groups and f : G l G b e a
homomorphism of group G onto G.
Then
i] H < G ()fHG º 
ii] H < G 11() (){ :( ) }fHG fH x G f x Hº  ‰‰            munotes.in

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53Chapter 2: Isomorphism of Groups
iii] () HG f H G º
iv] 1() HG f H G  º
v] 1ker ( ( )). HG a n d H f H f f H„ º   
vi] The map H l f(H) is a 1-1 corespondence between the family of subgroups
of G containing ker f and the family of subgroups of G. Furthermore normal
subgroups of G correspond to normal subgroups of G.
Proof :
i] ,( ) ( ) ( ) eH f e e f H a n d f H      a ‰=  ‰ v 
,( ) , ( ) , ( ) , . For a b f H a f a b f b for some a b H          aa a a‰  ‰

11
11 1 1
1,( ) ( )
() ( ) ( ) ( ) ( ( ) ) ( )
() ()ab H and f ab f H
fa b fa fb fa fb ab
ab fHa n d fH G     
    
     
  
=‰ ‰
aa  
aa=‰ 
ii] 1()Let H G and H f H 
1
1,( ) ( ) ( )
() .Then e G and e f e f e H e f H
fH         

aa‰ º ‰ º ‰
=v 
Let 1,( ) ( ) , ( ) ,ab f H t h e nf a f b H a n dH G       ‰‰ 
111 1
11 1() ( () ) . . () ( ) , ( ) () ( )
() ()faf b H i e f a f b H f a b f a f b H
ab f H f H G          
      
 ‰‰  ‰
=‰ = 
iii] Let , HG
11 1
11,( ) .
() , () , ,
() () ( () ) () () ( 1 )
() ( ) ( )Let a G h f H
Then a f a h f h where a G h H
ah a f a f h f a f a f h f
fa h a fh a h a H 
        
   
      
aa‰‰
aa ‰ ‰
aa 
‰ ‰ '
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54 ALGEBRA II
iv] Let 1,( ) ( ) HG L e t a G hf H i e f hH          ‰‰ ‰ 
11 1
11 1() ( ) ( ) ( ) ( ) ( ) ( ( ) )
() ()faha f a f h f a f a f h f a H
aha f H f H G        
       
 =  ‰
=‰ = 
v] Let H < G and H „ kerf.
Then, 1(( ) ) .ffH H„ 
1
1
11
11
11
1
1(( ) ) ( ) ( )
() () () ()
() ( ) ( )
ker ,
,( ( ) )
(( ) ) .Let x f f H f x f Hfxf h f o r s o m e h H f x f h e
fxfh e a n d fx h e
xh f H xh h h H
xh hHa n d f f H H
ff HH  
      
    
      
        
  




‰= ‰
a = ‰= 
aa = 
=‰‡ =‰
= ‰ ‡
=
vi] Let H be a subgroup of G.
Then, 1()fH is a subgroup of G containing ker f
1(( ) ) .
() .ff H H
The map H f H is onto=
=l 
     
12 1 2
11
11 2 2
12 1 2() ( )
(( ) ) (( ) )
,k e r .For H H G f H f H
Then f f H H f f H H
HH f H H
The map is one oneº 

„= 
=    
     
      
  
Corollary: Let G be a group and N G.
Given any subgroup H of G/N, there is a unique subgroup H of G containing N
such that /, , / /HHN F u t h e r H G i f f HN GN      a 
Proof : Consider the homomorphism f : G lG/N defined by f(x) = Nx. f is an
onto homomorphism and ker f = N = By correspondence Theorem, there is a
unique subgroup H of G containing N such that f(H) = H/N = Hc
Further H // GH N G N” munotes.in

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55Chapter 2: Isomorphism of Groups
Proposition : 2nnAS f o r n p  then !.2nnR$  
Proof : Consider :{ 1 , 1 } ( )nSd e f i n e d b y s i g n         ‰l  ‰ T  T
12 12 1 2 1 2
12() () ( ) ( )( ) ( )
,nThen sign Sign Sign
for S         
 ‰T T  T T  T T  ‰T ‰T
TT‰

= ‰is a group homomorphism.
((12)) 1, ( ) 1,
ker { / ( ) 1 }
.,
/{ 1 , 1 }
() / ()2nn
nn
nn
n nis onto for I
SAASa n d b y F i r s t I s o m o r p h i s m T h e o r e m o f G r o u p s
SA
OS oA     
  
     

   ‰‰   ‰ 
‰ T‰ ‰ T  
=
x
=
() !()22n
nOS nOA  =
Note : We have seen that by Lagrange’s Theorem if G is a finite group and
H < G, then O(H)\O(G).
However, converse of Lagrange’s Theorem is not true is general.
We give an example below.
Example : A4 has no subgroup of order 6. (O(A4) = 12).
Proof : Suppose A4 has a subgroup H of order 6.
Then 4[: ] () / ( ) 1 2 / 62n AH O A o H       
4
4.
3HAAhas eight cycles
 =

Let T be a 3-cycle in A4.
If T/H, then H T vH.
If 22H Ht h e n H TŠ T T munotes.in

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56 ALGEBRA II
21
22 1... ,
,( ) .
,3 .
()Hi e Hw h i c h i s n o t t r u eIfH t h e n a H W h i c h i s n o t t r u e
Ht h u s a l l c y c l e s a r e i n H
OH      
         
     
  
=TT ‰ T‰
T‰  T ‰
=T‰
=p
A contradiction.
4A= has no subgroup of order 6.
2.7 Classification of Groups of Order b 7 Upto Isomorphism.
We note that group of order 1 is a the trivial group which is unique upto
isomorphism. Group of order 2, 3, 5, 7 are cyclic (as prime order) and are unique
upto isomorphism.
Let G be a group of order 4. Then, any element of G has order which is divisor of
4 i.e. 1, 2 or 4.
If G has an element of order 4, then G is cyclic.
Suppose G has no element of order 4.
Then any non-trivial element in G has order 2. Let G = {e, a, b, c} which e, a, b, c
and distinct elements 222abce
We find ab ev
,, ,ab e ab aa b a which is not true
ab a ab ae b e which is not true
ab b ab eb a e which is not true
ab c similarly ba e ba a ba b
ba c ab          
              
              
       
º  ºžžžº  ºžžžžº  ºžžž= v v v
= Ÿ' ¬­­­­­­­­­­­­ ž ­ ž ­ ž ­­ ž ®
Similarly, we can show ac = b = ca
bc = a = cb.
=Composition table of G is
e a b c
e e a b c
a a e c a
b b c e a
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57Chapter 2: Isomorphism of Groups
Thus, any non-cyclic group G of order 4 is of the above type (and is called the
Klein’s four group 4V)
Let G be a group of order 6. Then, order of any element of G divides 6.
? Order of any element in G is 1, 2, 3 or 6. G is a group of even order and so has
an element of order 2.
Suppose every non-trivial element of G has order 2. Then, 2,xex G 






1,, ( )For x y G xy xy?







11yx x y?

i.e. yx xy


G? is abelian
If ,, ,ab Ga eb ezz






then ^`,,, He a b a b

is a finite subset of G having closure
property.
e a b ab
e e a b ab
a a e ab b
b b ab e a
ab ab b a e
HG?


, By Lagrange’s Theorem () / ( )oH oG i.e. 4/6.
A contradiction.
G?has an element of order 3 or 6.
Case 1: If G has an element of order 6 then G is cyclic.
Case 2 : G has no element of order 6. Let aG

have order 3 and bG


have order 2. Then 22,, ( ( ) ( ) 3 ,( ) 2 ) be a ao a o a o bz











. If ab ba , then
6 oa b
which contradicts that G has no element of order 6.
   12 2,, ba ab ba e a a ba a b e ba a b a?z z z z z z ''























, ba b a e ba abzzz'








. munotes.in

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58 ALGEBRA II
We note 22,, ,, ,eaa ba bab are distinct
22 2 2, ab e b a ab a a b e ab a b e


















zz z z z z'''
2ab a b a e





zz'
^`22,, ,, , Ge a a b a b a b


?
2ba a b


?
Thus, 32 2,, Ga b abe b a a b



3 GS?|


? There are two non isomophic groups of order 6, one is cyclic, and the other is
isomorphic to S 3
Note : An abelian group of order 6 is cyclic.
2.8 Fundamental Theorem of Finite Abelian Groups
Let G be a finite abelian group of order n. Then
12 k nn n G|†††]] ]




     

where
1iinnfor 11ikdd




and the above decomposition is unique.
This is also expressed as.
Let G be a finite abelian group of order n > 1 and let 12
12 ...k
k npp pD DD

where
1,...k pp

are distinct primes, then
12 ....KGG G G|†††



where () 1i
iioG p i kD d d






and
1...
ii ip i p i iGr DD|† †]]




where 12 ... 1ii i rDD Dtt tt







and
1...ii r iDD D





.
We note that if ,...ikpp are primes dividing o(G), then in /ipn for . iikbb 
Recall that .. ( , ) 1mnm nZZ Z i f a n d o n l y i f g c d m n€x          
Example : Let us list all abelian groups (upto isomorphism) of order 180.
22180 2 3 5. ¸¸ munotes.in

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59Chapter 2: Isomorphism of Groups
Abelian groups.
180 4 9 5
90 2 9 5 2 2
60 6 5 3 2 3 2€€
€€ € €
€€ € € €]] ] ]
]] ] ] ] ]
]] ] ] ] ] ]    
            
             
Let us consider classification of abelian groups of order 234,,ppp   u p t o
isomorphism.
Partitions Abelian Groups
1] 2p 2 2p]
2 = 1 + 1 ]pp€]
2] 3p 3 3p]
3 = 2 + 1 3pp€]]
3] 4p 4 4p]
4 = 3 + 1 3pp€]]
4 = 2 + 2 22pp€]]
4 = 2 + 1 + 1 2ppp€€]] ] 
4 = 1+1+1+1 pppp€€€]]]]  
Example: Expressing abelian groups in the form 12...pp p nGG G  €€ €
And 12nn n kZZZ€€ €    where 1icnn for group of order 360.
360 = 3223 5 qq
30 6 2
2233
51] 2 3 5
235
23
211ppp A b e l i a n g r o u p s
€€
€€€
€]] ]
]]]]
]        
      
    
   munotes.in

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60 ALGEBRA II
2
60 6 2
222 3 322] 2 3
35
2
21 1 1pp p
€€
€€€€]] ]
]] ] ] ]           
     
  
  5€] 
22
180 2
42953] 2 3 5
23 5
21 1pp p 
€
€€€]]
]]]]         
   
 
32
3604] 2 3 5
23 5
11 1pp p 
]         
   

Thus, there are 4 non-isomorphic abelian groups of order 360.
An important consequence of the Fundamental Theorem of finite abelian groups.
Theorem: If G is a finite abelian groups of order n, and m is a positive divisor of
n, then G has a subgroup of order m. (This is converse of Lagrange’s Theorem)
Note : The above partition may not be unique.
2.9 Summary
1) Let G, G be groups. A map f : G Gl is called a homomorphism of group G
to group G if f(ab) = f(a) f(b) for each a, b ‰G
(i.e. f preserves group operation)
2) Let G, G be groups. A map f : G Gl is called an is isomorphism of group G
to group G i f i ) f ( a b ) = f ( a ) f ( b ) f o r f o r e a c h a , b ‰G (ie f is a group
homomorphism) and ii) f is bijective.
3) Any finite cyclic group of order n is isomorphic to Z ,nthe group of integer
residue classes modulo under addition.
4) Cayley’s Theorem : Every group is isomorphic to a group of permutations
5) An isomorphism of a group G onto itself is called an automorphism of G
6) Let G be a group and a .G‰ T h e m a p :aiG G l d e f i n e d b y
1()aix a x a f o r xG‰     is called the inner automorphism of G induced by a. munotes.in

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61Chapter 2: Isomorphism of Groups
7) For 11,.., ) .... ,nn gg G G   ‰€€ then 11(, . . , ) ) ( () , . . , ( ) ) .nn Og g l c mo g og   
8) Let G be a group. A subgroup H of G is called a normal subgroup of G if aHa
1 ˆ H for each aG‰
1(, )or aha H for each a G each h H        ‰‰ ‰
9) First Isomorphism Theorem : Let f : G Gl be a homomomorphism of
groups. If f is onto, G/kerf Gx (or in general G/kerf x Im f).
10) Second Isomorphism Theorem: Let H and K be subgroups of a group G and
K  G. Then, HHKHKK ˆ
11) Third Isomorphism Theorem of groups:
Let G be a group and H, K be normal subgroups of G If K H, then
(G/K) / (H/K) x G/H
12) There exist unique groups of order 1, 2, 3, 5 as they are prime and 2 groups of
order 4 and 6 upto isomorphism.
13) Fundamental theorem of finite abelian groups: Let G be a finite abelian
group of order n. Then
12 k nn n G|†††]] ]




     

w h e r e 1iinnfor
11ikdd




and the above decomposition is unique.

2.10 Unit and Exercises
1) Show that Z(G) is a normal subgroup of a group G
2) Show that a subgroup of index 2 is a normal subgroup of any group G
(Let H < G, [G : H] = 2, Let a v H, then H, Ha are distinct right cosets of H of G.
H, aH are distinct left cosets of H in G. G = H U Ha = H UaH.
(Ha = aH = G H)

= For x ‰G, x H‰º Hx = xH = H
x v H º Hx = xH = G (H). munotes.in

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62 ALGEBRA II
3) What is the order of 5 + <6> in the quotient group 18/6? .]
4) Let G=222{1 , , , } 1 ij k w h e r e i jk    oooo    
,, ij ji k jk kj i ki ik j         
Construct the composition table of G.Show that every subgroup of G is normal.
5) Show that H = {I, (12)} is a not a normal subgroup of S3 but {I, (123), (132)}
is a normal subgroup of S3.
6) If H and K are normal subgroups of a group then H KG‚
7) If H is a subgroup of a group G and K  G, then H K < G.
11 2 2
11 1 1
11 2 2 112 2 1 2
11
12 2 2(, , , , ,
() ,
)ii eH K F o r a bH K a h k b h kh H k K
hk h k hkk h hk h k
hh h k h H K        
   .
     
‰‰   ‰ ‰
¸  ‰
‰
8) If H, K < G, where G is a group and H  G then HKK‚
Miscellaneous problem on Group Theorem
1] Compute order of each element in the following groups
i) 3D- Dihedral group of order 6
ii) 4D- Dihedral group of order 8.
iii) u(30) iv) 4S v) 8
2] Let X = (1 2 3 4 .. 11 12) 12.S‰ for which integers I, 12,iii i s xbba 12
cycle ?.
3] If X = (1 2) (3 4) (5 6) (7 8) (9 10). Is there a n cycle
(1 0 )kns u c h t h a t xTp  T     for some positive integer k?
4] Show that 1{/ }01nHn  ¬­ž‰­ž­ž­Ÿ®] is a cyclic subgroup of GL2(\)
5] Let G be an abelian group prove or disprove
{: }nHx G xe f o r s o m e n‰  ‰ `       is a subgroup of G. munotes.in

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63Chapter 2: Isomorphism of Groups
6] Let G be a finite group of order 2,np G cannot have a subgroup H of order
n-1.
7] Let 12 1 2 2 1,.GGb e g r o u p s P r o v e G G G G      €x€
8] Show that the following are subgroups of 12GG€ where 12,GG are groups
then,
12 2 2 1 2 1 1
11 21] {( , ) : } 2] {( , ) : }
3]{( , ) : }eg g G ge g G
gg g Gw h e r e G G G                   
         ‰‰
‰ 
9] i] Prove that Z 2€] is not isomorphic to ].
ii] Prove that 2is not isomorptic c to € ]  
10] Find the subgroup of nS generated by {(1 2), (1 2) (3 4)}
11] Prove that 4S = <(1 2 3 4), (1 2 4 3)>
12] Find a group which contains a, b such that o(a) = , o(b) = 5 and o(ab) = 2.
13] Let G be a group and , HG If o(H) = 2, Prove that H C Z(G).
14] Show that 4S has a unique subgroup of order 12.
15] From the given pairs of subgroups, find isomorphic pairs, justify your answer
1 ] 42 2andN€]]  2 ] 39 2 7 and €]] ] 
3 ] and€]] ] 4 ] 35 1 5 and €]] ] 
5 ] (8) (10)andNN 6 ] (8) (12)andNN
7 ] ] and 2 ] (under additional)
16] Find a non-cyclic group of order 4 in ]4 10€]
17] Find all generators the following cyclic groups of
1] 30 20 2] / 30 3] (13)   ‚N]]] ]
18] Find a cyclic group of order 4 and a non-cyclic group of order 4 in 4S. munotes.in

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64 ALGEBRA II
19] Let 2{: , } { 2 : , }abGa b G a n d H a b a b Gba      ¬­ž‰   ‰ ­ž­ž­Ÿ® s h o w t h a t
GHx (operation being addition in both groups)
20) Prove that every subgroup of nD of odd order is cyclic.
21] Find orders of all elements in nAand nD
22] Let G be a finite group and NGif [G : H] and o(n) are relatively prime,
then for any (),oNxGx e i m p l i e s x N‰ ‰   
23] Let G be a group and N be a cyclic subgroup of G such that NG%.
If H < N, show that H < G
24] Let =
{ {: , , ,} {: }00 1ab abGa b d a d o a n d N bd              ¬ ¬­­žž‰ v  ‰­­žž­­žž­­Ÿ® Ÿ®\\
show that NG and G/N is abelian.
25] Prove or disprove: If G is a group and K < H < G.
KHa n dH Gt h e nK G   
24] Show that i) () { }nZDe i f n i s o d d 
ii) 2() { , } . ( ( ) ) .m
mZDe a w h e r e a e o a m     
25] Let 31
{0 1 : , , }
001ab
Hc a b c  ¬­ ž ­ ž ­ ž ­ ‰ž ­ ž ­ž ­­ žŸ®]
Is H abelian? Is 33() ? HS L] Justify your answer.
26] Prove that there is no homomorphism from 82 44 onto €€]] ]]   
27] Determine all group homomorphism from.
1 ] 20 8onto]] 2] 422to €]]] 
3] 42 .St o ] munotes.in

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65Chapter 2: Isomorphism of Groups
28] Suppose :( 3 0 ) ( 3 0 )N l N   is a group homomorphism and ker
{1, 1 1} . If (7) 7, find all elements of (30)N that map to 7.
29] Suppose : zG € l]     (G is a group) is a group homomorphism such that
(3, 2) a and (2,1) , b D e t e r m i n e (4, 4) (Assume operation in G is
addition).
30] Show that a group of order 65 is cyclic.
31] Let H, K be distinct subgroups of a group G of index 2 Prove that HKG ‚
and [: ]GH K ‚     In /GH K ‚ Cyclic? Justify your answer.
32] Let G be an abelian group of order 8. Prove or disprove. G has a cyclic
subgroup of order 4.
33] Classify upto isomorphism abelian groups of under 108.
34] If G is an abelian group of order 120 and G has exactly 8 elements of order 2.
Determine the isomorphism class of G.
35] Find the number of abelian groups (upto isomorphism) of order [1] 15 [2] 42
[3] 16 [4] 48.
36] Let G be an abelian group of order 16. and ,,ab G‰ such that o(a) = o(6) = 4
and 22abv then determine the isomorphism class of G.
™™™™™munotes.in

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66 ALGEBRA II UNIT 2
3
SYLOW THEOREMS

Unit Structure
3.0 Objectives
3.1 Centralizers , Normalizers and Stabilizers
3.2 Groups Actions
3.3 Orbits & Stabilizer
3.4 Sylow Theorems
3.5 Classification of Group of Order p] Where p is a Prime
3.6 Classification of groups of order 15 upto isomorphism
3.7 Summary
3.8 Unit And Exercises

3.0 Objectives
After going through this unit you shall come to know about
x The concept of group action and stabilizers
x The notion of the order and its relation with the order of the groups
x Classification of groups upto order 15 using various results of Sylov
theorems



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67Chapter 3: Sylow Theorems
3.1 Centralizers and Normalizers Stabilizer
We recall that centre of a group G is defined as
/


‰





ZG x G x g g x v \^

‰
gG
ZGG, Moreover, if
/GZ G is cyclic, G is abelian.
Let G be a group and A be a non-empty subset of G.
Definition : The centralizer of a ()GCA i n t h e g r o u p G i s d e f i n e d a s
\^1() /GCA gG g a g aaA

‰









‰
.
We note 1


”


gag a ga ag .
()GCA is the set of all elements in G which commute with every element of A.
Proposition : ()

GCA G
Proof : ()
‰







'GeC A e a a e a v

‰
aA
Let ,( )
‰
GxyCA then ,





xa ax ya ay v

‰
aA .
= For 11 1 1;, ( )  
‰






aA ya y a y a y a
111 11 1  



ay a a y a


11 1 1 1 11  













xya x y a x a y a a x a y a x y a x y
1()=
‰
GxyC A .
We note that if \^ \^
,



‰


n
G Aa aCa v

‰
`n .
When \^\^
,



GAaC a is denoted by

GCa .
Definition : The normalizer of GA
 in G, is defined as
\^1() /


‰





GNA g G g A g A .


munotes.in

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68 ALGEBRA II
Proposition : ()

GNA G
Proof : 1



eAe A
()G eN A=

‰

Let ,( )GxyA
‰
/ then 11xa x = A, ya y = A
.
=
11 1 1 1 1 1 1 1xy A y x x y y A y y x x y y A yy x x A xA      










1()
()G
GxyN A
NA G=
‰




Clearly, () ()aGCA N A


Again, if \^Aa

, we denote ()GNA by ()GNa and we shall drop a when there
is no ambiguity.
i) () ()GCa Z a


ii) \^()GCa a
iff () aZ G‰.
Examples :
1) If 3 GS

and (123)a. Then \^( ) , (123), (132)Ca I




I f \^,( 1 2 3 ) ,( 1 3 2 ) AI



.Then () , ()GGCA A NA G




.
2) If G is a group and H < G then H, ()GHNH‡ .
= For 11,xHx H x H x H
‰







.
3) IF G is a group and H < G then
()aHCH
ˆ
iff H is abelian
()G HCH v
‡

”
,( )G hH hCH

‰


‰

v”
1, hH a h a h v

‰




. aH
ah ha v

ˆ

”



, aH v

‰

i s a b e l i a n hH H

‰
”
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69Chapter 3: Sylow Theorems
3.2 Group Actions
Let G be a group and S be a non-empty set. Then map B:GS S
q

l
denoted by
(,)gS g S
l
¸is an action of G on S if it satisfies the following conditions.
1) e.S = S for each sS
‰

2)

12 1 2.. ..gg s g gs

for each sS
‰
and each 12,gg G
‰

S i s c a l l e d a G - s e t .
Example :
1) Group acting on itself by left multiplication g.a = a a for each , gG aG
‰


‰

i ) .ea e a a



for each aG
‰
.
i i )




12 12 12 1 2 1 2.. .. .. .ggaggaggaggagga






f o r e a c h
aG
‰
, for each 12,gg G
‰
.
2) Group acting on itself by conjugation 1.gag a g

for gG
‰
and aG‰

1 ) 1eae a e a






2 )


11 1
12 12 2 1 2 2 1.. .gg a gg a g g g a g g 






11
12 2 1gg ag g





1
12 12 12 . gg a gg gg a




3) A group G acts on ދ(G) (The set of all subsets of G) by conjugation.
1.gSg S g



for () SP G
‰
. As in (2), it is a group action.
4) Let n be a positive integer. Then n GS

by
.iiTT



for \^1, ....in
‰


. For \^1, ....in
‰



\^/inGS i i TT


‰




.
5) Let G be a group acting on a non empty set S.
T h e n , f o r , gG
‰
the map :gS S
l
defined by
ga ga


¸
is a permutation
of S and the map :n GS


l
d e f i n e d b y
gg T

i s a m o n o m o r p h i sm .
(Cayley’s Theorem).
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70 ALGEBRA II
3.3 Orbits and Stabilizers
If G is a group acting on a set S and sS‰
(is fixed) then, the stabilizer of s in G is
the set \^/SGgGgss


‰





.
SGG

.seS S
eG


=

‰

Let ,sxyG

‰
then ,xss yss






.



11 1
11.. ( . ) . .
.. .ysy y s y y Se ss
xy s x y s x s s 

















1
ssxyG G G=

‰
=



Orbit of sS
‰
is defined as \^./ sgs S

‰
and is denoted by orbit (S)
Proposition : Let G be a group and S be a subset of G. Then the number of
conjugates of a subset S is :( )a GN S

and the number aS
‰
is :( )GGC a

.
We first prove a lemma (The proof may be read quickly)
Lemma : Let G be a group acting on a set S. Then, the relation on S defined by
for ,ab S

‰
ab

iff . ag b

for some gG‰is an equivalence relation. For each
aS‰, the number of elements in the equivalence containing a is <>:aGG

, the
index of the stabilizer of a.
Proof : We first prove is an equivalence relation. For aS‰, e - a = a.
aa=


 ( is reflexive)
,,ab Sa b g G

‰




º


‰
 such that . ag b


11 1.. .gag g b g g b e b b º









baº


( is reflexive)
12 ,, b ,ab Sa b c g g G

‰






º


‰
 such that 12,. ag b bg c¸




12 1 2 1 2.( . ) . , aggcggcggG º






‰
acº


( is transitive)
<>\^/ ag a g G

¸
‰
(orbit of a). munotes.in

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71Chapter 3: Sylow Theorems
Now if <> bg aa


¸
‰
, then agG
is the left cost of aG in G and .a bg a g G
l

.
= We have a map from <>a
l left costs of aG in G defined by .a ga g G

.
The map is well defined for ..ga ba


11 1
1.. ( . ) ( ) . .
aa ahg a h h a hh a e a a
hgG h G g G 
º









º

‰
º


Similarly 1
aa ehG gG h g G


º

‰

1.. . hg a a h a g aº


º




and the map is one - one.
The map is clearly onto, for ,. .aa gG g aG g a g G
‰


‰

º


.
=The use of elements in the equivalence classes of a = no. of left costs of aG in
G.
= Elements in orbit of <>:a aa G



.
Proof of Proposition: \^1/( )sGGg G g S g S N S


‰








.
No. of conjugates of <>:( )G SG N S

when
\^ () () ()GG Sa N S N a C aº




.
=No. of conjugates of <>:a aa G




Class equation : Let G be a finite group and let 1....rIIbe representatives of the
distinct conjugacy classes of G which are not contained in ()ZG. Then
<>
1() : ( )r
Gi
iGZ G G C G






œ .
Proof : For gG
‰
, the conjugacy class of g is \^g iff () gZ G‰.
1xgx g v




xg x g g x v

‰

”



()xGg Z G

‰
” ‰

Let \^1 () . . .mZGz z

. Let 1,...rKK be conjugacy classes of G not contained in
centre, and ....nrIIbe representatives of conjugacy classes. Then, the conjugacy
classes are \^ \ ^\^11... , ...mr ez z K

, .
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72 ALGEBRA II
This partitions G
()11 11( ) ( )
gZGomr r
ii
ii iGk Z G G C g






=










œœ œ
Note :
:( )i GC gO G

Theorem : A group G of order npwhere p is a prime and 1n
p
has non-trivial
centre.
Proof : B y c l a s s e q u a t i o n .
()1:() ()
gZ Gir
i
iGG C g Z G

Š







œ
As () , ( ) ,iigZ G C g G
Š


v

Now , :() / ()ii GCg G Cg


 so each term in
()1:()
gZ Gir
i
iGCg

Š




œ is of the form
kp.
Now
()1:() ()
gZ Gir
i
iGG C g Z G

Š







œ
1p\, p \ : ( )n
r
i
iGp
GG C g





=




œ p\( )
() 1G
ZG=

;

=

v

G= has non-trivial centre.
Note : A group of order np, p prime, 1n
p
is called a p group. A subgroup whose
order is power of prime is called a p - subgroup.
Corollary: A group of order 2p is abelian.
Proof : By class equation, () 1ZG
v
2()ZGp o r p Ÿ
If 2()ZGp then ()ZG=G and G is abelian.
()ZGp then ()GZ G is cyclic
G= is abelian. munotes.in

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73Chapter 3: Sylow Theorems
3.4 Sylow Theorems
Definition : Let G be a finite group and let p be a prime divisor of G. If pk divides
G and pk+1 does not divide order Gthen any subgroup of G of order pk is called
Sylow p - sub group of G.
Sylow’s First Theorem :
Theorem : Let G be a finite group and p be a prime. If /kpG
t h e n G h a s a
subgroup of orderkp.
Proof : We prove the result by induction on G. If G=1, the theorem is trivially
true. We now assume that the theorem is true for all groups of order less than G.
If G has a proper subgroup H such that kp d i v i d e s H, then by induction
hypothesis H has a subgroup of order kp, which is a subgroup of G.
So, we now assume that kp does not divide the order of any proper subgroup of
G we next consider the class equation,
()() :()
aZ GGZ G G C A

‰





œ .
/, ( )kkpG pC G v





() aZ G

Š


p/ :( )GGv =





() aZ G

Š

= From class equation, it follows that /( )pZ G . From the structure theorem of
finite abelian groups Z(G) has an element x of order p, ()xZG
‰
xG =
Consider the factor group 1/, p /kGx Gx

.
= By induction hypothesis, /Gx has a subgroup of order 1kp. This subgroup
is of the form /Hx where HG and xH‰
.
1,k H px px




.
kHp=

and thus G has a subgroup of order kp. munotes.in

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74 ALGEBRA II
Corollary 1:
Cauchy’s Theorem : Let G be a finite group and p be a prime that divides the
order of G. Then, G has an element of order p.
Proof : By Sylow’s Theorem, G has a subgroup H of order p.
, Hp H


= is cyclic and Hx.
() .Ox p=



Corollary 2 : Let G be a finite group, ,1 ( , ) 1kGp m k p m



p



then G has a p-
sylow subgroup.
Definition : Conjugate Subgroups :Let H, K be subgroups of a group G. We say
that H, K are conjugates in G if there exists gG
‰
such that H = g k g.
Sylow’s Second Theorem :
Theorem : If H is a subgroup of a finite group G and H is a power of a prime P,
then H is contained in a sylow - p subgroup of G.
Proof : L e t K b e a S y l o w p - s u b g r o u p o f G . L e t \^ 12,, . . .n C

,

,
,

, be the
set of all conjugates of K in G.
Since conjugation is an automorphism, each member of C is also a Sylow p-
subgroup of G. Let S c denote the group of permutations of C.
For gG
‰
, define :gCC


l
by
1
gi iKg K g

.
Then,

11
gi g j i jKK g g g K g


º

,


11 1 1
ij i j gg K gg gg K gg º



º
,

, gG= is one one.
If 1,jj iKC h K h
‰

,

for some hG
‰



11 1 11 1
11 1 1 1()ii
iigg h h gg g g h h g g
gg h g h g g g   
   

, 

,



, 
, munotes.in

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75Chapter 3: Sylow Theorems
Where 11 1()iiKg h K g h C 



‰
.
() ,jg i i KC =
,


,
‰
g=is onto.
Thus, :gCC


l
is bijective and gcS 
‰
.
We define a map :c TG S

l
by ()gTg

.
For 11 11, ( ) ( )( ) ( ) ( )ig h i i i in T g h K K g h K g h g h K hg 
b

b











11() ( ) ( ) ()ig h i i gh K h g K TgTh T K


 








ghTg h TgTh =





.
T= is a group homomomorphism.
We next consider T(H).
H is a power of p. ()TH=
is also a power of p.
Now
() \( )TH iOrb K T H



()TH iOrb K= = power of p (or 1)
Now
() 1rTH iOb k

for some ,1ii n

b

b


gi iKK v ”



1
ii gH g K g K v

‰
”

 gH

‰


i gN Kv”

‰


iigHH N K H K

‰
”



”





&ii Hi sap g r o u pa x N k x H x k





=

‰



‰

º

‰
'
Now :() CG N K



.
:: ( ) ( ) : , :GK GNK NK K GK









i s n o t d i v i s i b l e b y p ( K i s s y l o w - p
subgroup.
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76 ALGEBRA II
Now, C is sum of no. of elements in the orbits.
:GK

is not divisible by p.
:()GNKº


is not divisible by p. Cº
is not divisible by p.
()TH i Co r b K

œ
= There is at least one orbit having only one element.
i H=


, for some I by
Sylow’s Third Theorem :
Theorem 3 : t G be a finite group and p be a prime dividing G. Then, the number
of sylow p subgroups of G is equal to 1 module p and divides G.
Furthermore, any two Sylow-p subgroups of G are conjugates.
Proof : L e t K b e a n y S y l o w - p - s u b g r o u p s o f G a n d \^ 2,, . . . .n CK K K K





be
the set of all conjugates of K in G.
We show 1n
w
mod p.
For gG
‰
, define :gCC


l
by
1.gi iKg K g




11 1 1 1 1
gi g j i j i j
ijKK g K g g K g g g K g g g g K g g
KK    

 º


º



º


g=
is one-one.
For 1,ii iKC K h K h
‰



for some hG
‰
.

11 1 1 1 1 1()ii igg h h gg g g h g h g     =
,

, 
,
1
jgg

, where
111
jjKg h K g h C

‰

()gjK 
 g= is onto.
:gCC =


l
is bijective. gcS =
‰
, where cS is the group of permutations of set C.
Let :c TG S


l
be defined by ()gTg

.
munotes.in

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77Chapter 3: Sylow Theorems
Then, for ,.gh G

‰

()gh Tg h


For


1 111,gh i i i in K g h K g h g h K hg 
b

b












11 1
ig i g h i g h i gh K h g h K h K K 





  


 D
() ( ) ( )gh g h
Tg h TgTh=

 
=



and T is a group homomorphism.
Consider() () 0 /()TK i TK r b K TK
.
But K is a power of p. ()KTK

()TK=
is a also a power of p.
() 1TK iorb K=


when 1i, and () pTK iorb K

otherwise
()
11( 1 )1 m o dn
Tk i
iCo r b K p n m p


=











œ
We next show every sylow p subgroup is a member of C. Suppose H is a sylow p-
subgroup of G which is not a member of C.
We consider T(H), then sum of the orbits size blender action of T(H) is sum of
terms each divisible by p () 11TH iorb k


v
' for any i.
mod nC p=

w



A contradiction.
HC=

‰

Now <>:( ) nG N K




nG=


Corollary 1 : A unique sylow p - subgroup is normal. munotes.in

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78 ALGEBRA II
Proof : If H is the only Sylow p - subgroup of a group G, then for each
1,gGg H g
‰



is also a sylow - p- subgroup of G.
1gHg H v=





gGG

‰
=
) 
Corollary 2 : A sylow - p - subgroup of a group which is normal is unique.
Proof : Let H be a sylow p - subgroup of a group G.
Then, any sylow-p-subgroup of G is conjugate to H, and is of the form 1gHg for
some gG‰
.
G)1gHg H=




.
H= is a unique sylow p - subgroup of G.
3.5 Classification of Group of Order p] Where P is A Prime
Theorem: Let p G

] where p is an odd prime. Then 2pp Go r G D
x



x
]
Proof : By Cauchy’s Theorem, G has subgroups of order 2 and p.
G=
has an element of order 2 and an element of order p.
,,
:2Ga b b p
Gb b G









=






= 
1
1,1kaba b
ab a b for some k k p
=‰  
=  
        
21 1 1 1 2 2
2() ( ) ( )
2xk k k
kbb a b aa b a a a b a aa b a
ab b      
      
= 
212(1 )( 1 ( 1 ) ( 1 ) 1kbe p k p k k k p     = º º  º  
11 1 1ko r k P ko r k p    º º 
Case 1: If k = 1,
1
222
paba b ab ba ab p and G is cyclic of order p
GZ         
=  = 
=x munotes.in

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79Chapter 3: Sylow Theorems
Case 2: if k = p - 1 11 paba b
1
1ppaba b
ba ab


Gi saD i h e d r a lg r o u p= 
Groups of order pq. where p,q are distinct prime .
Theorem : If G is a group of order p.q where p.q are primer, p < q and p does not
divide q-1, then G is cyclic and isomorphic to pqZ.
Proof : Let H be a sylow p subgroup of G and K be a sylow q subgroup of G.
The no of sylow p subgroups of G is of the form 1 + kp and divides pq.
But p x q-1
= k = 0 and H is the unique sylow.
p – subgroup of a . Ha=
Similarly, no of q – sylow subgroups of G are 1 + kq and divides pq.
k =
= K is the unique sylow subgroups of G. =. KG
H and K are cyclic subgroups of G
Let H = , k=
we show xy = yx.
11 11 1 1() , ( ) .xyxyyxyHxyxyk   ‰ ‰  
11{}xyx y h k e H K H HNK K
xy yx=‰ ‚ ‚  = ‚
=     
 xyp q G
But xy G=  
 ‡   

,pqxy G
Gi sc y c l i c G Z= 
== x
 
3.6 Classification 0f Groups 0f Order 15 Upto Isomorphism
We have already seen that groups of order 2, 3, 5, 7, 11, 13 are cyclic (prime order)
and are unique upto isomorphism.
A group of order 1 is trivial and is unique upto isomorphism. munotes.in

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81Chapter 3: Sylow Theorems
3) Class equation : Let G be a finite group and let 1....rIIbe representatives of
the distinct conjugacy classes of G which are not contained in ()ZG. Then
<>
1() : ( )r
Gi
iGZ G G C G






œ .
4) A group G of order npwhere p is a prime and 1n
p
has non-trivial centre.
5) Let G be a finite group and let p be a prime divisor of G. If pk divides G and
pk+1 does not divide order Gthen any subgroup of G of order pk is called Sylow
p - sub group of G.
6) Sylow’s First Theorem: Let G be a finite group and p be a prime. If /kpG

then G has a subgroup of orderkp.
7) Cauchy’s Theorem : Let G be a finite group and p be a prime that divides the
order of G. Then, G has an element of order p.
8) Let G be a group. A subgroup H of G is called a normal subgroup of G if aHa
1 ˆ H for each aG‰
1(, )or aha H for each a G each h H        ‰‰ ‰
9) Sylow’s Second Theorem : If H is a subgroup of a finite group G and H is a
power of a prime P, then H is contained in a sylow - p subgroup of G.
10) Sylow’s Third Theorem: If G be a finite group and p be a prime dividing G
. Then, the number of sylow p subgroups of G is equal to 1 module p and
divides G.
Furthermore, any two Sylow-p subgroups of G are conjugates.
11) A unique sylow p - subgroup is normal.
12) A sylow - p - subgroup of a group which is normal is unique.
13) Let p G

] where p is an odd prime. Then 2pp Go r G D
x



x
]
14) If G is a group of order p.q where p.q are primer, p < q and p does not divide
q-1, then G is cyclic and isomorphic to pqZ.
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82 ALGEBRA II
3.8 Unit And Exercises
1) For the given group G and the H subgroup of G. Show that ()GCH H

and
()aNH G

.
i ) \^22
4,1 , , , GD H b a a b







where 42 3ae b b a a b







i i ) \^ 3,1 , GS H








.
2) Find all groups (up to isomorphism) of order 99.
Answer : Let G be a group of order 99. Let H be a sylow 3 subgroup of G and
K be a sylow 11 subgroup of G Then, no of sylow 11 subgroups of G is
congruent to 1 mod 11 and divides 99 = K is unique sylow subgroup of G.
.
,{ }KG
Similarly H G H K e
   =
‚

It follows that elements from H and K commute and therefore GH Kq
33 3 9 9Gi sa b e l i a n
GZ Z o r GZ=
x€ x 
  
3) Determine all groups of order 66 upto isomorphism.
Answer: Let G be a group of order 66. Let H be a sylow 3 subgroup of G and
K be a sylow 11 subgroup 8 G.
Then, 1 is the only positive divisor of 66 which is congruent to 1 mod 11,
Therefore , . KG
() 3 3 ( 3 3 3 1 1 3 1 1 1 ) HK G and O HK and
HK is cyclic     
  =   q q 
=
Let HK = < x >, HK has index 2 in G . HK G=
Let 2. yG a n d y ‰
1yxy x  ‰ 
Let 113 2 ,iiyxy x for some i for to Now x x      munotes.in

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83Chapter 3: Sylow Theorems


211,( )iiii i iyx x y x y x y y xy x x     
2,ixe 
233 \ 1
11/ 1 11/ 1i
io ri =
= 
,1 , 1 1 1 , 1 1 3 3 1 .ii i = o r i           =    o  
1, 10, 23 32.io r     =
= T h e r e a r e a t m o s t 4 g r o u p s o f o r d e r 6 6 . w e o b s e r v e t h a t
66 33 11 3 3 4,,ZDD Z a n d D Z €€     a r e o f o r d e r 6 6 , a n d n o t w o o f t h e m a r e
isomorphic.
4) Show that the only group of order 255 is 255.Z
Proof:- Let G be a group of order 255.
2 5 5 = 3 . 5 . 1 7 .
Let H be a 17 sylow sub group of G Then, number of 17 sylow subgroups of G is
congruent to 1 mod 17 and divides 255.
The 17 sylow subgroup is unique.(Divisors of 255 are 1,3,5,15,51,85)
17 ()
()HG HZ
NH G 
 =x
=

17 () /() () ( 1 7 ) 1 6NH CH d i v i d e s A u t H A u t U     =   ]
Since () /() /() / 1 6NH CH G CH   we have () /() 1NH CH 
() GC H=
= every element of G commutes with every element of H.
()
17 ( ) , ( ) 255HZ G
ZG ZG
=‡
=
/( ) 1 5 , 5 , 3
/( )GZ G
GZ G i s c y c l i c  
 =
=or 1 munotes.in

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84 ALGEBRA II
Gi sa b e l i a n
Gi sc y c l i c=
= 
 
5) Show that a group of order pqr, where p,q,r are distant primer is cydic.
6) If (G) = 36 and G is non-abelian, probe that either G has more than one 2 sylow
subgroup or more than one 3 sylow subgroup.
7) Show that a group of order 56 has a proper non-triral normal subgroup.
8) Let G be a group of order 60. If the sylow 3 subgroup is normal, show that the
5 sylow subgroup is also normal.

™™™™
munotes.in

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85Chapter 4: Introduction to Ring UNIT 3
4
INTRODUCTION TO RING

Unit Structure
4.0 Objective
4.1 Introduction
4.2 Ring
4.3 Characteristic Of Ring
4.4 Subring
4.5 Summary
4.6 Unit and Exercise
4.0 Objective
The objective of the this unit is to introduce the concepts of
x Ring integral domain and fields with example.
x The characteristic of ring.
x How to check a given subset of ring is subring?
4.1 Introduction
Basically there are two operations ‘ +’ and ‘ .’ which we can apply on scalars &
vectors. We know how to add and multiply two scalars. In vector space, we saw
two operation vector addition and scalar multiplication. In group theory we have
seen a set with one operation whether addition or multiplication satisfying certain
property. So for we had not seen any set with both operation addition as well as
multiplication. Ring is one such algebraic structure which has both these operation.
Loosely speaking ring means closed structure that is why (may be) an algebraic
structure which is closed with addition and vector multiplication is called as ring.
munotes.in

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86 ALGEBRA II
4.2 Ring
A ring R is set together with two binary operations + (addition) and ‘ ¸‘
(Multiplication) satisfying
i) (R, +) is an abelian group.
ii) `.’ is associative : i.e.

ab c a bc¸¸

¸¸ for all a, b, c, ‰R.
iii) The distributive laws hold in R.
i.e. for all
,,abc R a b c ac bc


‰





<<< and
ab c a b a c



<< <
Before we see example of ring let understand versions terms involved in ring with
the help of following remarks.
Remarks:
1) If R is ring, then (R, +) is abelian group. Multiplication (''¸) n e e d n o t b e
commutative. When ''¸ is commutative we say R is commutative Ring.
2) As (R,+) is abelian group,  o n e l e m e n t s a y ` 0 ’ , s u c h t h a t
aooaa v . aR‰ T h i s e l e m e n t ` 0 ’ i s c a l l e d a s z e r o e l e m e n t o r
additive identity of R. A ring may, may not have multiplicative identity. But
when it has, i.e. if ring R has an element 1 such that 11aa a¸

¸  for all a
‰ R then such element `1’ is called as unity in R & R is said to be ring with
unity. Note that 01 .v Throughout the course we assume R is ring with unity
otherwise
3) For each a in ,Rb R‰ such that 0 ab such element is called additive
inverse of ‘ a’ and is unique.
4) A non zero element ‘a’ of R is said to be unit if there exists b in R such that
aºb = 1 . b is said to multiplicative inverse of a. Multiplicative inverse of
element may or may not exists.
5) An element aR‰ is said to be zero divisor if there exists bo
v
such that
ab o¸

. Note that zero divisor cannot be unit.
(as if a is unit and zero divisor then ,1bc s u c h t h a t ab ba    ¸   ¸ and ac 
1 bac bc c c             º¸¸¸º¸ º a contradiction.
6) A commutative ring (with unity) is said to Integral domain (ID) if it has no non
zero zero divisor. (i.e. in R the only zero divisor is zero element 0). munotes.in

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87Chapter 4: Introduction to Ring
One more way is there to define ID is that, whenever ab

 in ID a  º
or b = 0. N o t t h a t i f R i s n o t ID a b o v e r e l a t i o n m a y n o t b e s a t i s f i e d a s i n
62. 3 0  ] but 23 0 .vv    
In case of ID cancellation law hold.
as if .. ( ) ( ) ab a c ab ac a b c a or b c since ID            º º   º   
if a o b cvº

  ab ac b c   =º  if aov
7) In the definition of units we have seen that every element of ring R need not be
unit.
A commutative ring in which every non zero element is unit is called as field.
A commutative ring R (with unity) is said to be field if for all a v, in R, there
exists b in R such that a.b = 1.
They are some standard examples of ring which are easy to show. We will list
then without proving.
1)
Set Commutative
Ring Zero
element Unity Zero
Divisor ID Units Field

,,¸] Yes 0 1 No Yes 1o No.

,,
 ¸_ Yes 0 1 No Yes all non Zero
element Yes

,,¸\ Yes 0 1 No Yes All non Zero
element Yes

,,†
< Yes 0 1 No Yes all non zero
element Yes
()nM\ No. Null
Matrix Identify
Matrix Yes No ()
..n AM
st A 
 ‰
v\
No. <><><>,
&xxx
_
\
] Yes Zero
Polynomial 1 No Yes Non Zero
Constant
Poly. No

2) IF R is a ring than the set {( , ): , } RR a ba bRq ‰    with + and ¸ define as
(,) (, ) ( , ) & (,)(, ) ( , )ab ad a cb d ab cd a cb d          ¸ is not integral
domain as if a,b ‰ R, ,( , 0 ) . ( 0 , ) ( 0 , 0 ) ab t h e n a b         vv  Hence not ID.
(a,0) is zero element of this ring and if 1 is unity in R. then (1,1) is unity in R
x R. If R is commutative, R x R is also commutative. munotes.in

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88 ALGEBRA II
3) Let R = ] [i] = {: , , 1 }ab i a b i‰   ]     is also ring with 0 as zero element
and 1 as unity. This is commutative ring with no non zero divisor and hence
integral domain. Note that only 1o and iohas multiplicative inverse (as ( o
1) (o1) = 1, i (-i) = 1). So it is not field.
This ring is known as ring of Gaussian integers.
These are some important example of rings. Many examples can be found in
reference books. We end this section by given two more important class of ring.
4) Consider {0,1, .. 1}n n ]    the set of residue classes modulo n. define + and
¸as abab and .ab a b   Then n] is ring.
Some important results aboutn].
i) Every non zero element of n] is either zero divisor or unit.
Proof: - Let n a ‰]such that 0av suppose a is not zero divisor.
We will show that a is unit.
Claim: (a,n) = 1 . ((a, n) means G.C.D. of a and n)
Suppose (a, n) = d, d > 1 . Then d/a and d/n.
Let ,0nnbba n d bd  ‰ v]
and
0.na aab a ndd d        ¬ ¬ ¬ ­ž ­­žž ­  ž ­­žž ­ž ­­žž ­ž Ÿ® Ÿ® Ÿ®
º a is zero divisor which is not the case.
Hence (a, n) = 1 .
Therefore there exists x, y ‰ ] such that
1 ax yn 
11 ax yn ax= º 
yn o

' x=is multiplicative inverse of a.
=a is unit.
Hence if a is not zero divisor then a is unit.
(ii) If n is prime then n] is field.
' n is prime, therefore
,1an

, for all a, 1 < a < n .
Hence for any a in n] multiplicative inverse of a exists. munotes.in

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89Chapter 4: Introduction to Ring
5) Polynomial Ring : The polynomial plays very important role in Algebra.
Finding root of polynomial is one of the central problems of Algebra. Set of all
polynomial over \with respect to operation polynomial addition and
multiplication from a ring. Let we recognize it’s important and discuss it
separately.
Definition: Let R be a commutative ring. The set of formal symbols
1
11 [] { . . : : ,nn
nn o iRxa x a x a x a a R n
    

‰
      is non negative integer} is
called the ring of polynomials over R in indeterminate x.
Remark:
i) Let 1
11 () . .nn
nn ofxa x ax a x a       
   and
1
1 () . .mm
mm o gx b x b x bx b        
Then
1
11 1 1 () () ( ) ) . . ( )sssss s o ofxg x ab x a bx a b x ab         
       
where s = max {m, n}
and 1
11 () .() . .mn mn
mn mn ofxg x c x c x c x c 
        
were i
ii k k
koca b
œ
 
Definition: Let 1
11 () . .nn
nn ofxa x a x a x a   
    
If 0nv then n is called degree of f(x), and na is called leading coefficient of f(x).
If 1na, then polynomial is called monic polynomial. The degree of polynomial
is denoted by deg
fx.The polynomial
0 fx

has no degree and is called zero
polynomial. The polynomial
0fxa

is called constant polynomial.
If R is commutative ring with 0 as zero element and 1 as unity then <>Rx is also
commutative ring with zero element as zero polynomial 0 and unity as constant
polynomial
1 fx

. If R is integral domain then <>Rx is also. But <>Rx is not
field even though R is field as the only units in <>Rx are constant polynomial.
In this next unit we have one separate chapter for polynomial ring where we discuss
one important concept of ring irreducible polynomial.
munotes.in

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90 ALGEBRA II
Let we discuss some theorems giving relationship between integral domain and
fields.
Theorem 1: Every field is integral domain.
Proof : Let R be a field and let ,ab R
‰
such that
01 ab




l

Let assume . a  vWe will show b = 0 .
Since . a  v and R is field. xR = ‰ such that 1 ax xa¸



¸ .
Multiply both side of (1) by x we get
0xab x¸¸

¸
()xa b   =
bb     º º 
And as R field, therefore R is commutative ring.
Therefore R is integral domain.
Note : Every field is integral domain. But every integral domain need not be field.
For example ] which integral domain but not field.
A ring R is said to be finite ring if R is finite Set.
Theorem 2: Every finite integral domain is field.
Proof : Let R be finite integral domain.
Therefore R is commutative ring.
Let a ‰R be such that a v
then 23,,, . . . .aa a R ‰   But as R is finite all this indices may not be distinct.
That is there exist ijv such that ijaa
let i > j then ..ijjjaa a a   
1ija  º
Also note that i – j > 0
Therefore 1 ijaR ‰ such that 1.1ij ijaa a 


Therefore 1 ija is multiplicative inverse of a. a v
Therefore all non zero element of R has multiplicative inverse.
Therefore R is a field . munotes.in

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91Chapter 4: Introduction to Ring
4.3 Charctristic of Ring
Let R be a finite ring. Then (R, +) is a finite abelian group. If O(R) = m then for all
a ‰R, ma = 0 . Hence for finite ring ( R, +) there exist m
‰
` such that 0 ma
¸



for all a ‰ R.
Note that such number is not unique. If m is one such number then, all multiple of
m satisfies same condition. We are interested smallest positive integer n such that
0 na¸

, such number is called characteristic of ring.
Definition: Let R be a ring. A least positive integer n such that 0 na¸

for all a
‰ R is called characteristic of R & is de noted by char R. If no such number exists
then char 0R

.
For example char
] char
_ char 0

\, char .nn]
Characteristic of infinite ring is obviously zero. But characteristic of finite ring is
also not easy to find. The theorem given below helps to find characteristic of ring.
Theorem 3: The characteristic of ring R with unit 1 is n if and only if n is least
positive integer such that 10n¸

.
Proof : Let assume that char R = n
= n is least positive integer such that n.a = o for all a ‰R.
= 10n¸


Conversely let n is least positive integer such that 10n¸

.
To prove that char R = n.
Let a‰\ be arbitrary.
Consider
11 0 0 na n a n a a¸

¸
¸

¸¸

¸

and n is least such positive integer.
= char R = n .
Theorem 4: The characteristic of integral domain is (hence field) is either zero or
prime.
Proof: Let R be an integral domain.
If char R = 0 Then nothing to prove.
Hence assume char R = n, n v 0
To prove n is prime number,
Let assume n = x.y where 1 < x, y < n
' char R = n
= for each ,0 aR n a
‰

¸


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92 ALGEBRA II
i.e. n.a2 = 0
i.e. (xy)a2 = 0
= (xa) (ya) = 0
= x.a = o or y.a = o ( since R is integral domain)
which is not possible as x, y < n and char R = n.
Hence no such x, y exists
= n is prime
Corollary 1: Characteristic of a finite field is prime.
Proof: Let F be a field.
(F, +) is a finite group.
Hence order of F exist & Let o(F) = m
10m=
¸

.
Let n be least positive integer such that 10n¸

as F is field and hence integral
domain. Therefore by above theorem n is prime.
4.4 Subring
Subrings are non empty subset of a ring which itself is ring with respect to the
operation of ring.
So if we want to prove a non empty subset S o f r i n g R i s s u b r i n g , o n e h a s t o
prove all the property of ring, which is a lengthy procedure. The following theorem
gives a easy method to determine whether a given subset of Ring R is subring or
not.
Subring Test:
Theorem 5: Let s be a non empty subset of ring R. Then S is subring of R if for any
a, b ‰ S,
a-b ‰ S and ab S¸‰
.
Proof: Let S be subring of R.
Then S be itself a ring.
Therefore for any a, b ‰ S, a-b ‰ S and a.b ‰S munotes.in

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93Chapter 4: Introduction to Ring
Conversely let assume that S be non empty subset of R, such that a – b and a. b ‰
S whenever
a, b ‰ S.
To prove S is subring.
' for a, b ‰ S a-b ‰ S
=(S, +) is subgroup of (R, +)
= (S, +) is abelian group.
For ,,ab S a b S
‰

¸
‰

S is closed under multiplication .
Also as multiplication ( ''¸) distribute over addition in R and therefore in S also.
Therefore S is a ring.
Hence S is subring of R
Eg:- 1) Let R = 2()M]
i) 1 :,aaSa bbb£² ¯¦¦¦¦¡°‰¤»¡°¦¦¢±¦¦¥¼]    
Let ,,aa xxAB S a n d l e tA Bbb yy    ¯ ¯¡°¡° ‰ ¡°¡°¢±¢±
then A – B = 1axazSybyb ¯¡° ‰¡°¢±
and AB = 1ax ay ax aySbx by bx by ¯¡° ‰¡°¢±
1S= is subring of R
ii) 2 :,aa bSb aab a  £² ¯  ¦¦¦¦¡°‰¤»¡°¦¦¢±¦¦¥¼]
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94 ALGEBRA II
Let 2aa b xx yAand B Sab a z y x    ¯  ¯ ¡° ¡° ‰¡° ¡°¢± ¢±
2ax abxyABSabx y ax ¯  ¡° ‰¡° ¢± 
2.() () () ().() () () ()ax a b x y a x y a b xABSab x a xy abxy a x ¯  ¡°‰¡°   ¢±        
2S= is subring
2) Let R€€]]]
{( , , ) / }Sa b c R a b c‰     
Take a = (1, 2, 3), b = (4, 2, 6) then a, b ‰ S
a.b = (1, 2, 3) ¸(4, 2, 6) = (4, 4, 18)
but 4 + 4 v 18 =(4, 4, 18) ŠS
Hence S is not subring.
4.5 Summary 1) A ring R is set together with two binary operations + (addition) and ‘ ¸‘
(Multiplication) satisfying
i) (R, +) is an abelian group.
ii) `.’ is associative : i.e.

ab c a bc¸¸

¸¸ for all a, b, c, ‰R.
iii) The distributive laws hold in R.
i.e. for all
,,abc R a b c ac bc


‰





<<< and
ab c a b a c



<< <
2) Every field is integral domain.
3) Every finite integral domain is field.
4) Smallest positive integer n s u c h t h a t 0 na¸

, such number is called
characteristic of ring.
5) The characteristic of ring R with unit 1 is n if and only if n is least positive
integer such that 10n¸

. munotes.in

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95Chapter 4: Introduction to Ring
4.6 Unit and Exercises
1) Show that
q
]]under component wise addition and multiplication is a
ring. Is it an integral domain? Justify.
2) Let
,R
¸ be a ring with multiplicative identity 1.R Show that
1,,R
€
:
is a ring where, 1Rabab
€






and ababa b





:
3) Show that the ring
( i ) \^ :,, da b d a b   ¯ ‰¡°¢±]] d is an integer is integral domain.
( i i ) {: , } , da b d a b         ¯ ‰¡°¢±__ d is an integer is field.
4) Let R = {0,2,4,6,8} under addition and multiplication modulo 10. Prove that
R is field.
5) Let F be a field of order 2n. Prove that char F = 2 .
6) Let D be an integral domain of characteristic P. For any ,xyD
‰
show that
( i ) ()pppxyx y   (ii) ()nnnpppxyxy   for all n ‰N
( i i ) F i n d e l e m e n t x a n d y i n a r i n g o f c h a r a c t e r i s t i c 4 s u c h t h a t
444() .xyx yv 
7) Let F be a field and let K be a subset of F with at least two elements. Prove
that K is subfield. (i.e. prove that for any
1,, 0 i n , a n dab F b K a b a b K
‰


v






¸
‰
)
(This is known as subfield Test)
8) An element a ‰ R is said to be nilpotent element if there exist n
‰
` such
that 0na

. Prove that a ‰ R is nilpotent then 1 – a is unit in R.
(Hint : 11nnao a=  
9) An element a ‰ R is said to be idempotent if 2. aa Prove that
(i) The set of idempotent of a commutative ring is closed under
multiplication.
(ii) The only idempotent in an integral domain is 0 and 1.. munotes.in

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96 ALGEBRA II
10) Determine all zero- divisors, units & idempotent elements in (i) 18]
(ii) 36q]] (iii)
q
]_.
11) Let d be a positive integer. Prove that [] { , }da b d a b      ‰ __ is a
field.
12) In the following examples, show that S is a subring of the given Ring R.
( i ) 2 :( )abSa c b d Rcd£² ¯¦¦¦¦¡°   ¤»¡°¦¦¢±¦¦¥¼\         0
(ii) 22:, ( )abSa b Q R Mba£² ¯¦¦¦¦¡°‰ ¤»¡°¦¦¢±¦¦¥¼_       
13) Determine which of the following are subrings of
,,

¸_. Justify your
answer.
( i ) :, , (,) 1 ,aSa b a b b i s o d db£²¦¦¦¦‰ ¤»¦¦¦¦¥¼]     
( i i ) :, , (,) 1 , 0 ,aSa b a b b b i s e v e nb£²¦¦¦¦‰ v¤»¦¦¦¦¥¼]    
(iii) :, , 0 , (,) 1 ,aSa b b a b a i s o d db£²¦¦¦¦‰ v¤»¦¦¦¦¥¼]    
( i v ) \^2: Sx x‰ _   
( v ) \^:, Sx x x o‰  _   
14) Let ,,zwHz wwz   £² ¯¦¦¦¦¡°‰¤»¡°¦¦¢±¦¦¥¼\ S h o w H is a non commutative subring of
2M \ i n w h i c h e v e r y n o n z e r o e l e m e n t h a s a n i n v e r s e w i t h r e s p e c t t o
multiplication. ™™™™ munotes.in

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97Chapter 5: Ideal and Quotient Ring UNIT 3
5
IDEAL AND QUOTIENT RING

Unit Structure
5.0 Objective
5.1 Ideal
5.2. Quotient Ring
5.3 Types of Ideal
5.4 Unit Exercises
5.0 OBJECTIVE:
x The objective of this chapter is to introduce the concept of
x Ideal and its importance.
x Theorem used to determine whether given set is ideal or not.
x Quotient ring
x Types of ideal i.e. Prime and maximal ideal.
x Their relation with Quotient ring.
5.1 Ideals
Our aim is to define something similar to quotient space (in case of vector space)
and quotient group (in case of group). Note that if W is any subspace of vector
space we can very well defined quotient space which itself a vector space. To define
quotient group of group G w e n e e d s p e c i a l t y p e o f s u b g r o u p c a l l e d a s n o r m a l
subgroup. In group theory we have seen that if H is normal subgroup of G then G/H
is defined.
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98 ALGEBRA II
Quotient ring is also similar to quotient space (or group). It is denoted as R/I where
R is ring and I must be similar to normal subgroup (not just subring).
This leads to concepts of ideal.
Definition (Ideals): Let R be a ring a subring I of R is said to be ideal of R if for
, , a n d a xxR aI x a
‰


‰


‰
*
T h a t i s i d e a l s a r e t h o s e s u b r i n g o f R which absorbs the element of R.
I f I is proper subset of R then I is said to be proper ideal of R.
A ring having no proper ideals is called as simple ring.
Ideal Test: A non empty subset I of ring R is an ideal of R if
(i) a-b ‰ I whenever a, b ‰ I
(ii) (ii) ,xaax I¸
¸
‰
whenever a ‰ I and x ‰ R.
Example :
1) For any , nn
‰


]]is ideal of ] as n]are only subgroup of ].
2) Let <>Rx


\the set of all polynomial with real coefficient.
Let A be subset of all polynomials with constant term zero. Then A is ideal of R.
As let f(x), g(x) ‰ A
Therefore (0) 0 (0) (fg  since constant term of f & g is zero)
() () () ( 0 ) 0 therefore if h z f x g x then h        
Also for any <>()kx x
‰
\
(0) (0) (0) 0 0kf k ¸

¸

, there fore () ()kxf x I
‰

Therefore I is ideal of \.
Note that I 1 {( ) : ( ) . . . . }n
nfxR f xa x a x ‰             
1
1 {( ) : ( ) ( . . . ) }n
nfxR f x x a x a‰             
{( ) : ( ) .( ) }fxR f x x g x
x‰ 
        
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99Chapter 5: Ideal and Quotient Ring
3) Let R be ring of the real valued functions of real variable. The subset of S of all
differentiable functions is a subring of R but not an ideal of R.
xR
‰
, x i s n o t d i f f e r e n t i a b l e , x ‰ S , s i n c e x i s d i f f e r e n t i a b l e b u t xxS¸
Š
as xx¸ is not differentiable hence S is not an ideal.
4) Let R be a ring and a ‰ R. Then, \^: aR ar r R




‰
is ideal of R, known as
principal ideal of R generated by a.
Principal ideal plays very important role in ring theory. An integral domain in
which every ideal is principal ideal is known as principal Ideal Domanin which
we will see in coming chapters.
Theorem 1: L e t I, J be any ideal of ring R.
Then I + J = {x + y : x, ‰ I, y ‰ I}, IJ =
1:,k
nn n n
nxyx I y J
£²¦¦¦¦‰‰ ¤»¦¦¦¦¥¼œ
    
(i.e., IJ is sum of finite product of element of I and J), and IJ

 is ideal of R.
Proof:
(i) Let a, b ‰ I + J
11 2 2 ax y a n d bx y=      
then 12 1 2() ( ab x x y y I J    ‰   
Also for any 11 1 1 ,. ( ) rR a r x yr x r y rI J  
‰


  ¸   ‰ 
11 ,) as I J are ideals therefore x r I and y r J         ‰‰
Similarly r.a ‰ I + J
Hence I + J is ideal.
(ii) Let a, b ‰ I, J
11nm
ii j j
ijax y a n d b p q
    

œœ where ij i jxpI a n dyqJ ‰‰       
then a – b =
11nm
ii j j
ijxyp q
œœ
  1
11()nm
ij j
ljxyp q I J
      
  ‰œœ
Let r ‰ R
ar = 11
11()nn
ii
Lixyr x ry I J
¬­ž­‰ ž­ž­žŸ®œœ
       (, )Since I is ideal therefore xr I   ‰
Similarly ra ‰ IJ.
Therefore IJ is ideal munotes.in

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100 ALGEBRA II
Theorem 2: A Field is simple ring. That is the only ideal in a Field is {0} and F
itself.
Let A be any non zero ideal of field F.
Let a ‰ A, a v 0. F is field therefore  b ‰ F
Such that a.b = 1
But A is ideal also.
?1.ab A  ‰ (Since I is ideal, r.a ‰ I for r ‰ F)
? for any x ‰ R, F x.1 = x ‰ A
?FAˆ
. FAº
5.2. Quotient Ring
Let R be a ring and I be its ideal define R/I = {x + I : x ‰ R}
Define `+’ and `.’ as
(x+I)+(y+I) = (x + y) + I a n d (x + I)(y +I) = x y + I
We claim that R / I is a ring with respect to above operation, called quotient ring.
(i) First we prove that ‘+’ and ‘ ˜’ defined as above is well defined.
Let assume that ''xIxI a n d yIy I    
'' therefore x x I and y y I            º‰  ‰ (By definition of cosets)
''xxyyI  º ‰''() ( )xyxyI  º ‰
'1 ' ''( ) ( ) ( ) ( )xyIxyIx IyIx IyI          º  º       
? + is well defined.
also as '',xxyyI   ‰
'',, therefore let x x ayybfor some a b I      ‰
? ''xxa a n d yyb    munotes.in

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101Chapter 5: Ideal and Quotient Ring
?''() ()xyxayb    '' ' 'xyxbyaa b  
? '' ''xyIxyxbyaa bI       ''xyI  (Since I is ideal therefore
'',xbyaa b I    ‰ hence''.xbyaa bII     
?''() () ( ) (xIyI x Iy   ,   
? ‘¸’ is well defined.
As (, )R is abelian group
(/ ,)RI= is also abelian group.
and





xIyI zI x yIzI x y zI       
Similarly



xIy I z I x y z I       
? ‘˜’ is associative.











() ( ) ( )
()xI yI zI xy I zIxyzI x zyzIx z IyzI
xI zI yI zI   
     
 ¸  
¸       
  
Therefore ‘ ˜’ is distribute over ‘+’.
Hence (R / I, +, ˜) is ring.

Remark :
1) If I is subring only, R/I is not defined.
2) If 0 is zero element of R then 0 + I = I is zero element of R/I.
3) If R is commutative ring then quotient ring R/I is also commutative.
4) If R is ring with unity 1, then R/I is also ring with unity 1 + I.


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102 ALGEBRA II
Examples :
1) \^/4 4 , 1 4 ,2 4 ,3 4










]] ] ] ] ]
2) \^ 2/ 6 6 , 2 6 , 4 6




]] ]] ]] ]
3) Let R = ][i], the ring of Gaussian integer and I = <2-i>
t h e n R/I = {a+bi+<2-1> : a + bi ‰R}
= 2 – i ‰ <2 – i> = I
= 2 – i + I = I in R/I 20ii n R =     ,
24 1 /ii n R i n RI     ,  = = 
50 / 1 35 1 in R I Let i R       = =  ‰
u s i n g 2 = i , a n d 5 = 0 w e g e t
13+ 51i + I = 3 + 5 q 2 + (5 q10+1) (2) + I
= 3 + 2 + I = 5 + I ( S i n c e 5 = 0 i n R / I ) = I
Thus using 2 = i, 5 = 0 in R/I any element a + bi + I of R/I is reduced to one
of the element I, 1 + I, 2 + I, 3 + I or 4 + I .
Also note that all this elements are distinct as additive order of 1 + I is 5
Hence R/I = {I, 1 + I, 2 + I, 3 + I, 4 + I}
4) Let 2() :,,,abRM a n d S a b c dcd     £² ¯¦¦¦¦¡° 
‰ ¤»¡°¦¦¢±¦¦¥¼]] then S is ideal of R.
Let we find R/S:
T a k e o n e e x a m p l e . C o n s i d e r o n e l e m e n t
34
17AS ¯¡°‰¡°¢± 
34
17ASS ¯¡° ¡°¢±  12 4
11 6S ¯¡°¡°¢± 
10 2 4 10 2 4,11 0 6 11 0 6SS s i n c e S          ¬  ¯   ¯  ¯   ¯  ­ ž ¡° ¡ ° ¡° ¡ ° ­    ‰ ž ­ ž ¡° ¡ ° ¡° ¡ ° ­ žŸ® ¢± ¢ ± ¢± ¢ ± munotes.in

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103Chapter 5: Ideal and Quotient Ring
This means that it lmSnp ¯¡°¡°¢± is element of R/S then the only choice of l, m ,n , p
are 0 and 1 Hence there total 421 6 element in R/S, therefore |R/S|=16.
5) Let <>x\ d e n o t e t h e r i n g o f p o l y n o m i a l s w i t h r e a l c o e f f i c i e n t a n d l e t
21 Ix   

22[] / 1 { () 1 : () [] } RR x x f x x f x R x      ‰     
Let us see what quotient ring R is;
By Division algorithm, for any
<>fxx‰
\ <> ,( ) ( )pxa n d rxi n x  \
Such that2() ()( 1 ) () () 0fx p x x r xw h e r e r x      ¸   
2deg ( ) deg ( 1) . . ( ) 0 ( ) , , .rx x i erx o r rx a x ba b R          
     ‰



2() 1 ()fxI p x x r xI=         

2
22 2() ( 1 ) ()
() , 1 , () ( 1 ) () ( 1 )px x I rx I
rx I S i n c e x I t h e r e f o r e px x I px x II        
                       
  ‰  ‰   

2[] / 1 { ., } Rx x a x b a b   , =      ‰\\
5.3 Types of Ideal
Prime Ideal: Let R be a commutative ring. An ideal P of R is said to be prime ideal
if whenever a, b ‰ P º either a ‰ P or b ‰ P.
Maximal Ideal: Let R be a commutative ring An ideal M of R is said to be maximal
ideal if (i) M v R (ii) MTRˆˆ  then either M = I or I = R.
Before we see example of prime and maximal ideal we just prove two theorems.
These theorem use to characterise ideal with the help of quotient ring.
Theorem 3: (P is prime /RPº
is integral domain)
Let R be a commutative ring. An ideal P of R is prime ideal if and only if R/P is
integral domain. munotes.in

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104 ALGEBRA II
Proof: Let R be commutative ring and P is prime ideal of R.
To prove R/P is Integral domain.
Let / aaP a n d bb p R P ‰ 
Such that ./ab P i n R P     (P is zero element in R/P)
.. ( ) ( ) / ..ie a P b P Pi nR Pie a b P P            ab Pº‰ 
But P is prime ideal
Therefore ab P a P‰º

‰
or bP‰aPP o r bPP  º


Hence if .ab P a P o r b Pº           
Conversely let assume that R/P is integral domain To prove P is prime ideal.
Let ,ab R
‰
such that ab P‰º ab + P = P º (a +P) (b + P) = P
But R/P is integral domain.
() . ()aPbP P aPP o r bPP ap o r bp        º     º ‰ ‰
.ab P a Po rb P=‰ º ‰ ‰       
= P is Prime ideal.
Theorem 4: M is Maximal ideal ”R/M is Field.
Let R be a commutative ring with unity. An ideal M of R is maximal ideal of R if
and only if R/M is field.
Proof: Let R be a commutative ring with unity and M is Maximal ideal of R.
To prove /RM is field. Let / aMR M
‰
such that aM M
v aMºŠ
Consider the ideal {:: } aR a r r R‰   of R.
= M is ideal, aR is ideal.
Then N = M + aR is also ideal such that MR  1 ˆˆ
But M is maximal ideal and M z N.
Therefore NR, hence 11RN  ‰º ‰
,xMr R   = ‰ ‰ such that 1xar  munotes.in

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105Chapter 5: Ideal and Quotient Ring
() 1 () 1xa r M M xM a rM      0=  =  
() () 1 ( )aMrM M x MxM M=    ‰=  '       
Hence a + M has multiplicative inverse.
= Every non zero element of R/M has multiplicative inverse.
= R/M is field.
Conversely, let assume that R/M is field.
To prove M is Maximal ideal.
Let assume that N is any ideal of R such that MR  1 ˆˆ
then N/M is ideal in R/M
but R/M is field & hence the only ideals in R/M is zero ideal which M or R/M itself.
/
//If N M M N M
if N M R M N Rº
º      
    
If MR  1 ˆˆ either M = N or N = R
= M is maximal ideal of R/M
Remark :
1) Since every field is integral domain therefore if M is maximal ideal º R/M is
field º R/M is integral domain º M is prime ideal
Hence every maximal ideal is prime ideal. But every prime ideal need not be
maximal ideal which we will see in the examples let see some examples of prime
and maximal ideals.
Examples:
1) Let R = ][x] and I = = set of all polynomial with constant term zero.
Claim : I is prime ideal.
Let

fxh x x¸
‰


() []. . () () .()
(0) (0) 0 0 0there exist g x x s t f x g x x g x
fh g=‰ 
= ¸

]    
 
(0) 0 (0) 0fo r hº      (']is integral domain)
() ()fxx o r h xx º‰ ‰    
[]xis prime ideal in x =  ]    
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106 ALGEBRA II
2) Let R = ][x] and M = = set of all polynomial with even constant term.
Note that M = = {x.f(x) + 2g(x) : f(x), g(x) ‰ ] [x]}
Claim : M is maximal ideal in R
Clear MRv
Let N be any other ideal of R such that MR  1 ˆˆ
we will prove that M = N or N = R
Let assume that MNv
Therefore we prove that N = R
,( ) ( )MNf x N s u c h t h a t f x Mv=  ‰ Š'         
= constant term of f(x) i.e. f(o) is not even
1
11 () . .nn
nn oLet f x a x a x a x a
              where 0ais odd.
1 .. 2 1( )n
no oax a x b a i s o d d    '             = g(x) + 1
where 1 () . . 2n
no gx a x ax b M N       ‰ ˆ
() , ()gx N f x N=‰ ‰    
1( )( )
() , 1 ()fxg xNfor any k x R k x N    
         =  ‰
=‰ ¸ ‰
RN B u t NR
NR 
=ˆ ˆ
=
º M is maximal ideal of R
(i) Note that ,2IxM x

‡


= is not maximal ideal.
Hence prime ideal need not be maximal ideal.
(ii) By same argument we can show for any prime is maximal ideal
of ][x]
(iii) Since there are infinitely many prime, hence there are infinitely many
maximal ideal in ][x]
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107Chapter 5: Ideal and Quotient Ring
3) Let p is prime, then p] is prime ideal of ].
As . Let x y p‰]xy p k f o r s o m e k     º¸ ‰ ]
/. / /px y px o r py  ==12,.xpk or y p k x p or y p    º  º‰ ‰ ]]
. pi s p r i m e i d e a l  =]
Also note that ]/p] is ]p which ring of residue modulo p. And as p is prime it
is field. Hence p] is maximal ideal also. Thus in ], prime and maximal ideal are
some.
4) Let {( , 0): } Ra n d I a a     €  ‰]] ] consider R/I,
Let (, ) (, )ab R t h e nab I ‰ (( , 0) (0, )) ( , ) ( , )ab I a I a b I                   
(0, ) ( ( , ) , ( , ) )bI S i n c e a I a I I              ‰   '
/{ ( 0 , ): }RI b I b        =  ‰ ]
Let ((x, y) + I) . ((p, q) + I) = I ( I is zero element in I)
(, ) (, ) . .xp yq I I xp yq I y q         º  º‰ º 
0( )
(2, ) ( , )yo r q i s I n t e g r a l d o m a i n
Io r p I      
       º 
º‰ ‰']

(, ) ( , )xII o r p II            º   
Hence R/I is Integral domain
= I is prime ideal
' (1, 1) is unity in €]]
and R/I = {(0, b) + I : b ‰ ]} does not contain unity hence it is not field.
= I is not maximal ideal
5) Let us find maximal ideal of
( i ) 8] (ii) 10] ( i i i ) 12]
Let us see lattice of ideals for 8]
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108 ALGEBRA II
I) 8] 8]
2
4
0
Clearly 2is maximal ideal of 8].
II) 10]
2 5
0
Clearly 2 and 5 is maximal ideal of 10].
III)
C l e a r l y 2 and 3 is maximal ideal of 12].
Thus in general for any prime divisor p of n,

is maximal ideal of .n]
5.4 Unit and Exercises
1) Let S = {a + bi : a, b ‰], b is even}. Show that S is subring of <>i] but
not on ideal.
2) Check whether the following set I is ideal of ring R?
){ ( , ) : } ){ ( ,) : }
){ ( 2 , 2 ) : , }iI a a a R i iI a a a R
iii I a b a b R            
       ‰  q   ‰  q
‰  q]] ] ]] ]
]] ]
(In all three problems R is ring under com ponent wise addition and multiplication.)
munotes.in

Page 109

109Chapter 5: Ideal and Quotient Ring
(iv) 10 0 {( ) [ ] : ( ) . . 3 | }n
n If x x f x a x a x a a          ‰    ] []Rx ]
(v) 10 {( ) [ ] : ( ) . .n
nIfx K x fx ax ax a        ‰     ]
^10 .. 0naa a     and <>Rx

]
(vi) :: , ,oa abIa R Ra b doo od£² £ ²¬ ¬¦¦ ¦ ¦¦¦ ¦ ¦­­žž‰  ‰­­¤» ¤ »žž­­žž­­¦¦ ¦ ¦Ÿ® Ÿ®¦¦ ¦ ¦¥¼ ¥ ¼\       
(vii) {4 : , }, [ ]Ia b i a b Ri ‰  ]]    
3) Determine the number of element in quotient ring
(i) 3 ] / 9 ] ( i i ) <>/3ii


]
4) Let 22Ii

. Determine R = ][i]/<2+2i>.
What is characteristic of R? Is R is Integral domain? Is I = <2 + 2i> is prime
ideal.
5) Show that the following ideals are prime ideals in the given ring
( i ) 2
2 1[ ]xxi n x ] 
(ii) I = {(3x, y) : x, y ‰ ]} in ] x ] under component wise addition and
multiplication.
6) Show that <1 – i> is maximal ideal in ][i]. (Show that ][i]/<1-i> is field.
Note that i =1 and 2 = 0 in ][i]/<1-i>)
7) Let R be the ring of continuous functions from to\\. Show that A = {f
‰R / f(0) = 0} is a maximal ideal of R.
8) Prove that every prime ideal is maximal ideal. ™™™™munotes.in

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110 ALGEBRA II UNIT 3
6
RING HOMOMORPHISM AND
ISOMORPHISM
Unit Structure
6.0 Objective
6.1 Ring homomorphism & isomorphism
6.2 Fundamental theorem of isomorphism
6.3 Ring of fraction, Quotient field
6.4 Summary
6.5 Unit and Exercises
6.0 Objective
After going through this unit you shall come to know about
x The concept of ring homomorphism and isomorphism.
x Fundamental Theorem of isomorphism and its application.
x Method to construct ring of fraction and Quotient field.
One way to study property of a ring is to examine its interaction with other ring by
finding some relation between them. This relation is something which must
preserve the operation of the respective ring. Such relation is called as ring
homomorphism.
6.1 Ring Homomorphism
Let R, S be any two ring. A map f:R is said to be ring homomorphism from R
to S if for any ,

‰
ab R


ff fab a b








and


ff fab a b





i.e. f is
operation preserving mapping.
Properties of Ring Homomorphisms :
Let :


l
RS be a ring homomorphism. Let A be a subring of R and B is an ideal
of S.
munotes.in

Page 111

111Chapter 6: Ring Homomorphism and Isomorphism
1. For any
‰
rR a n d a n y p o s i t i v e i n t e g e r

,




nn rn r a n d






n nrr .
2.

\^: 




‰
Aaa A is a subring of S.
3. If A is ideal and  onto S, then
A is an ideal.
4.
\^1/( )


‰



‰
Br Rr B is an ideal of R.
5. If R is commutative, then
R is commutative.
Proof :
1)

... 







nr r r r (n times)

() 








rr n r
2)

... 





nrr r r (n times)

() ( () )nrr r r 








Let ,( )xyA

‰


,ab A=



‰
such that

,xayb










xya b a b A a b A =











‰





















‰
'

A=
 is subring
3) :RS


l
is onto. = for any , yS xR
‰



‰
such that
xy


To prove:
A is ideal of S. From (2),
A is subring.
Hence it is enough to show that for any

,,xAySxyA
‰


‰

¸
‰
 .

xAa A

‰

=


‰
' such that
,, axyS 



‰

 is onto, bR

‰
such
that
by

.
Hence



.xyab a b A






‰

(Since in ideal A ideal,
therefore ab ‰ A)

A=

is ideal of S.
munotes.in

Page 112

112 ALGEBRA II
4) \^1() / ( )Br Rr S



‰




‰

Let

1,( ) ,xyB x y B

‰

=


‰
as B is ideal

xyB =
 

‰


1()xyBxyB=




‰

=



‰

Hence 1()B is subring.
Let rR
‰
be arbitrary.
rS =

‰

As B is ideal of S



therefore xrB x rB =

‰


‰


1, th e r e f o r e ( )xrB x r B=
‰

‰


Similarly we can show that 1.( )rx B
‰

Hence 1()B is ideal of R.
V) Let R is commutative ring.
= for any ,, ..xyR x y y x
‰










..
.xy yxxyy x=



= 

 
R º
 is commutative ring.
Kernel of Ring homomorphism: Let :RS


l
be a ring homomorphism then
Kernel of , denoted as ker  and is defined as
\^/0 Ker r R r



‰





.
Note that ker  is ideal of R.



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Page 113

113Chapter 6: Ring Homomorphism and Isomorphism
Example:
1) Let :,abRa bba£² ¯¦¦¦¦¡°





‰
¤»¡°¦¦¢±¦¦¥¼]. Let :R


l
] b e d e f i n e d b y
ababba¬ ¯­ž¡°­ 




ž­ž¡°­žŸ®¢± then  is ring homomorphism.
As let ab x yABba yx ¯  ¯¡° ¡°






¡° ¡°¢± ¢±
then
ax byABa x b yby ax¬ ¯ ­ ž¡° ­














ž ­ ž¡° ­ žŸ®¢±


() ()ab xy
AB












.ab xy a xb ya y b xABba yx b xa ya xb y ¯  ¯   ¯ 

¡° ¡° ¡ °




¡° ¡° ¡ ° ¢± ¢± ¢ ±

.( ) ( )ABa x byaybx =









……………… (1)
Also

ab x yABba yx¬ ¬ ¯ ¯­­žž¡°¡°­­ 






žž­­žž¡°¡°­­žžŸ® Ÿ®¢±¢±

() ab xy a xa yb xb y

  






() ()ax by ay bx





……… (2)
From (1) and (2)


.ABA B 



Hence  is homomorphism.
Let us find Ker ,
Ker 0ab abRba ba£² ¬  ¯  ¯¦¦­ ¦¦ ž ¡° ¡° ­ 


‰






ž ¤» ­ž ¡° ¡° ­ ¦¦ žŸ® ¢± ¢±¦¦¥¼
/0 /ab abRa b Ra bba ba£² £ ² ¯  ¯¦¦ ¦ ¦¦¦ ¦ ¦¡° ¡°

‰


 


‰


¤» ¤ »¡° ¡°¦¦ ¦ ¦¢± ¢±¦¦ ¦ ¦¥¼ ¥ ¼
:aaaaa£² ¯¦¦¦¦¡°



‰
¤»¡°¦¦¢±¦¦¥¼\ 11 11/11 11aa R£² £ ² ¯  ¯¦¦ ¦ ¦¦¦ ¦ ¦¡° ¡°



‰

¤» ¤ »¡° ¡°¦¦ ¦ ¦¢± ¢±¦¦ ¦ ¦¥¼ ¥ ¼ munotes.in

Page 114

114 ALGEBRA II
2) Define
:m o dn kk n 


l







]] . Then  i s a n h o m o m o r p h i s m
known as natural homomorphism.
3) Let R be commutative ring of prime characteristic P.
Define :RR


l
as
pxx 

is ring homomorphism
as
12 112 2 1() () . . .
ppppp pp p p p
CC Cxyx y x x y x y x y y
 
















 =
Each of
111p
C ip

b

b


is multiple of p and char R = p 10ppi iCx y= for all
i, 11ip
b

b










..pp
p ppxyx y x yxyx y x y x y=
























Therefore  is ring homomorphism, known as Frobenius homomorphism.
4) Let R be a ring and A be its any ideal.
Define a map :/fRR A

l
a s ()frr A



, then f is ring homomorphism
known as a natural homomorphism.
Ring Isomorphism: A ring homomorphism :RS


l
which is one-one and onto
is known as ring isomorphism.
If :RS


l
is ring isomorphism we say that ring R and S is isomorphic to each
other and this is denoted by RS
x
.
Isomorphic rings are exactly similar in terms of property of element and their
behavior. In Other words isomorphic rings are nothing but different way of looking
of same ring.
Theorem 1. First Isomorphism Theorem of Ring:
Statement: Let :RS


l
be a onto ring homomorphism then /k e r ( )RR



x

.
Proof : Let Ker  = W.
To prove
/RW R
x

C l e a r l y WK e r


 is ideal of R.
Define a map
:/fRW R
l
 as () ( ) ,frW r rR



‰
. munotes.in

Page 115

115Chapter 6: Ring Homomorphism and Isomorphism
Claim :
1) is well defined one - one map.
Let assume that
s, , rW W r sR







‰
rs W K e r”



‰



0 rs ”



( B y
Definition of Ker )


0 rs ”





(By definition of homomorphism)



() rs f r w f r W ”



 ”



= f is well defined one-one map.
2) f is onto.
Let () yR
‰


Therefore there exist xR‰ such that
xr 

.
Therefore there exist /xWR W


‰
such that

fxW xy



.
Therefore f is onto.
3) f is homomorphism.
Consider
()fxWyWfxyW  

 

xy

 (By definition of f)


xy



(is homomorphism)
() ()fxW f yW 






()fxWyWfxyW 



xy



xy



() . ()fxW f yW 




'f is homomorphism
Hence f is isomorphism.

/RWR =
x

munotes.in

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116 ALGEBRA II
Theorem 2. Second Isomorphism Theorem for Rings :
Statement: L e t A b e a s u b r i n g a n d l e t B b e a n i d e a l o f R. Then
\^:,ABa b a A b B




‰


‰
i s a s u b r i n g o f ,RA B
 i s a n i d e a l o f A a n d
() / /ABBA AB

x

Proof : Let we give sketch of the proof.
To prove this we use first isomorphism theorem of rings.
Let xAB
‰


,,
()xab aA b BxBabBa bB aB b B=




‰


‰

=





















‰
'
Hence define a map
:/AABB 


l
 such that ()aa B



.
You prove this map  is onto homomorphism.
Then apply first isomorphism theorem, according to which

/k e r /AABB

x 

.
Now prove that ker AB



.
Theorem 3. Third Isomorphism Theorem of Ring:
Statement : Let I and J be ideals of R with IJ
ˆ
. Then /JI is an ideal of /RI
and (/ ) / (/ ) /RI JI RJ
x
.
Proof : This proof is also similar to proof of second isomorphism theorem.
Here define :/ /RI RJ


l
as ()rI rJ



.
Show that  is onto homomorphism.
Then by first isomorphism theorem //RI K e r RJ


x
.
Show that / Ker J I


.
The first isomorphism theorem has lot of application. One of the application is to
prove that a field contains p] or _. We prove this by following steps.
munotes.in

Page 117

117Chapter 6: Ring Homomorphism and Isomorphism
Theorem 4 : Let R b e a r i n g w i t h u n i t y e . T h e m a p p i n g : R 

l] g i v e n b y

. nn e

is a ring homomorphism.
Proof : Let ,, , 0mn mn

‰



]

() . . . ( ) m n m n e e e e m n times



  




.. ..
mt i m e s nt i m e see e e e



  

  
 


me ne m n  
Let assume ,0mn



Then Let , mp nq





 where ,0pq

.
() () ( ) () ( )
() ( ) ( )
() ( ) ( ) ( )mn mn e pq e pq e
peq e
pe qe m e n e m n=










































































Similarly if 0m
p
0n

we can prove () ( ) ( )mn m n



 .
Also () () () ( . ) () ( )mn mn e mn e e me ne








.mn 
 
= i s r i n g h o m o m o r p h i s m .
Corollary 1: Let R is the ring with unity and the characteristic of R is 0n

, then
R c o n t a i n s a s u b r i n g i s o m o r p h i c t o n]. If the characteristic of R i s 0 , t h e n R
contains a subring isomorphic to ].
Proof: Let \^: Sk e k




‰
]
Claim : S is subring of R.
Let ,ab S

‰

,nm=


‰
] such that an e

and bm e


() ab n em e nm es= 


‰
.( ) ( )( ) ( ) .ab n e m e n e m e n m e e n m e S









‰

Hence S is subring of R.

Define : S 


l
] such that
nn e

.
Then  is onto and above theorem  is homomorphism.
Hence by first isomorphism theorem /.Ker S

x
]
If Char R = n munotes.in

Page 118

118 ALGEBRA II
Then Ker 
\^/0 mm

‰




] \^/0 mm e
‰



]
\^
:/ mn m c h a r R n

‰








]' n

/
nnS
S=

x

=

x
]
]
Hence R contains subring isomorphic ton].
If char R = 0 ºThere exist no integer n such ne=0 .
\^/0 Ker m m



‰




] \^ \ ^/0 0 ( S i n c e 0 ) mm e c h a r R

‰










]
/k e r S
S=



x

=

x
]
]
Hence R contains a subring isomorphic to ].
Corollary 2: A field contains p]or_.
Proof: Characteristics of a field is either zero or prime. Let R be a field.
Case I: Char R = p, p is prime.
Then by previous corollary R contains a subring isomorphic to p], p] is field,
as p is prime
Hence R containsp].
Case II: Char R = 0
Then subring S of R is isomorphic to ].
Let \^1.: , , 0 Ta b a b S b





‰


v

The T is subfield of R and isomorphic to _.
Hence a field contains p] or _.



munotes.in

Page 119

119Chapter 6: Ring Homomorphism and Isomorphism
6.3 Ring Of Fractions
The aim of this section is to prove that a commutative ring R is always a subring of
a larger ring _ in which every non zero zero divisor of R is unit in _. In case of
integral domain such ring is field and called as field of fraction or Quotient field.
Construction of quotient field _ from ring R is exactly as construction of _ from
].
Theorem 6 : Let D be an integral domain. Then there exists a field F (called the
field of fraction of D) that contains a subring isomorphic to D.
Proof: Let S be the set of all formal symbols of the form a/b, where ,ab D

‰
and
0bv. Define an equivalence relation w on S by //ab cd w
ifad bc

. Let F
be the set of equivalence classes of S u n d e r t h e r e l a t i o n w a n d d e n o t e t h e
equivalence class that contains /xy by<>/xy. We define addition and
multiplication on F b y <> <>
// /ab cd a d b c b d  ¯







¢± a n d <><><>/./ /ab cd a cb d




.
We will prove F is field with respect to the operation and

¸ define above.
To prove this first of all the operation and

¸ is well defined.
Let assume that 1
1aa
bb

and1
1cc
d d

.
To prove 11
11aca c
bd bd






1
11
1
1
11
1aaab a bbb
cccd c ddd



º









º


'

' To prove 11
11aca c
bd bd





i.e. to prove 11 11
11ad bc a d c d
bd bd




i.e. to prove that
11 11 11() ad bc b d a d c d bd




.

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Page 120

120 ALGEBRA II
i.e. to prove that 11 11 11 11adb d bcb d a d bd c d bd






In LHS put 11ab a b

and 11cd c d

we get RHS.
Hence + is well defined.
Similar To prove 11
11..ac a c
bd bd

w h i c h i s o b v i o u s a s 11
11 and aa c c
bd bd


11
11ac a c
bd bdº



It is trivial to prove that (F, +) is abelian group with zero element as 01ªº¬¼ and
additive inverse of abªº¬¼ as abªº¬¼.
Multiplication is obviously distributive over addition and associative.
The unity element of F is 11ªº¬¼and multiplicative inverse of non zero element
abªº¬¼ is baªº¬¼ for , 0ab
v
.
Hence F is field. Finally let we define a map :DF‡o as 1xx ªº‡ ¬¼then ‡
is ring isomorphism from D to D‡as,
 >@>@>@/1 /1 /1 /1 /1xyx y x y x y ‡    ªº¬¼  xy ‡ ‡
>@>@>@  /1 /1 /1xyx y x y x y‡ ‡ ‡
^` : 0Ker x D x‡  ‡
>@^`^`^`:/ 10 : 0 0 xDx xD x  
?By first isomorphism theorem,
 /
()DK e r D
DD‡|‡
?| ‡
Hence F contains a subfield isomorphic to D.

munotes.in

Page 121

121Chapter 6: Ring Homomorphism and Isomorphism
Examples :
1) Show that the ring 2ªº¬¼] a n d H a r e i s o m o r p h i c w h e r e
2:,abHa bba­½ªº ®¾«»¬¼¯¿] under addition and multiplication of 22u matrixes.
Solution : :2Hªº‡o¬¼] as 22ababbaªº‡ «»¬¼
Claim :
1) ‡ is well defined, one-one map
L e t 22ab x y
ba yxªº ªº «» «»¬¼ ¬¼
& axby ?œ
22 ab xyœ 
22 2 2 ab xy ab xy ?‡  ‡  œ  
?‡ is well defined one-one map.
2) ‡ is onto
For any 2,2 2xyHx yyx ¯ ¯ ¡°
‰





‰
¡°¢± ¡°¢±] such that

22xyxyyx ¯¡° 


¡°¢±. Hence  is onto.
3) ‡ is homomorphism.
22 ( ) 2 ab xy ax byªº‡   ‡   ¬¼
 22 () 2
2
22ab x y ax by
ba yx by a
ab xyªº ªº ªº   «» «» «» ¬¼ ¬¼ ¬¼
‡  ‡ 
 22 2 ( ) 2 ab xy a x b y a yb xªº‡   ‡¬¼ munotes.in

Page 122

122 ALGEBRA II
222
2ax by ay bx
ay bx ax byªº «»¬¼
22 2 2 2222ab x y a xb ya yb xab xyba yx a y b x a x b y ªº ªº ª º‡  «» «» « » ¬¼ ¬¼ ¬ ¼
= 22 ab xy‡ 
Hence ‡ is homomorphism.
?‡ is isomorphism.
2 H ªº?|¬¼] .
2) Let :,abRa bba­½ªº ®¾«»¬¼¯¿]. Let :R‡o] defined ababba§·ªº‡ ¨¸«»¬¼©¹. Show
that ‡ is a ring homomorphism. Determine Ker ‡. Is Ker ‡ a prime ideal? Is it a
maximal ideal? Justify.
Solution: Let abAbaªº «»¬¼ and xyBRyxªº «»¬¼






ab x y a xb yABa x b yba yx b ya x
ab xy A B¬  ¯  ¯   ¯  ­ ž ¡° ¡° ¡ ° ­ 












 ž ­ ž ¡° ¡° ¡ ° ­ žŸ® ¢± ¢± ¢ ±





























ab x y a xb ya y b xABba yx a yb xa xb y¬  ¬ ¯  ¯   ¯  ­­ žž¡° ¡° ¡ ° ­­ 







žž ­­ žž¡° ¡° ¡ ° ­­ žž  Ÿ® Ÿ ®¢± ¢± ¢ ±








. ax byaybx a xybyxa b xyAB 





 



 






=
is homomorphism
Ker\^:( )0 AR A 


‰






\^ :0 : 0abAR AR abba£² ¬ ¯ ¦¦­ ¦¦ ž¡°­ 

‰







‰






ž ¤» ­ž¡°­ ¦¦ žŸ®¢± ¦¦¥¼ .aaaaa£² ¯¦¦¦¦¡°



‰
¤»¡°¦¦¢±¦¦¥¼]

munotes.in

Page 123

123Chapter 6: Ring Homomorphism and Isomorphism
= First Isomorphism theorem
/RK e r

x
] as ] is Integral domain, /RK e r
is integral domain.
=Ker
 is prime ideal.
Also as ] is not field. =/RK e r
 is not field.
=Ker
 is not maximal ideal.
3)Prove that
2M \ contains a subring that is isomorphic to ^.
Solution :
2 :,,,abMa b c dcd£² ¯¦¦¦¦¡°







‰
¤»¡°¦¦¢±¦¦¥¼\\ .
Define a map
2 : M 
l
^\ as
abai bba ¯¡° 




¡°¢±.
Claim :  is homomorphism.

() ai bci d aci bd





  () ac bd
bd ac ¯ ¡°
¡°¢±

()abcdai b ci dba d c ¯  ¯¡° ¡°






 ¡° ¡°¢± ¢±






ab cd a c b d a d b cai b ci dba d c a d b ca c b d
ai b ci d ¯  ¯   ¯   ¡° ¡° ¡ ° 





¡° ¡° ¡ ° ¢± ¢± ¢ ±












































Hence  is homomorphism.
Ker
\^ :0 ai b ai b 




‰
†





\^ \ ^00:: 0 , 0 000abai b ai b a bba£²  ¯   ¯ ¦¦¦¦ ¡° ¡ ° 



‰
†







‰
†







¤»¡° ¡ ° ¦¦¢± ¢ ± ¦¦¥¼
= By First isomorphism theorem

2 .Ker MGx
^ \i.e.
2M
x

^\ .
Hence
2M \ contains a subring that is isomorphic to .
4) Is 2] is isomorphic to 3] munotes.in

Page 124

124 ALGEBRA II
Solution : No. Let :2 3


l
]] is isomorphism &
2 a 

.





42 222 2 2 2 a =















….. (I)
Also





2 242 . 22 . 2 2 a 













…………. (II)
Therefore From (I) & (II).
22 aa


220 2 0 aa a aº




º



0aº


or 2a if 0a

then


2. 2 0nn 




.
Hence  is zero map which is not isomorphism.
0 =
v
2a=

but then 3aŠ
].
Hence 2] is not isomorphic to 3].
5) Is 2 ¯
¡°¢±] isomorphic to 5 ¯
¡°¢±] .
Solution : N o , 2 ¯
¡°¢±] i s n o t i s o m o r p h i c t o 5 ¯
¡°¢±] . As if
:2 5 ¯ ¯



l

¡°¡°¢±¢±]] is isomorphism
1 a 

.
() (. 1 ) () . ( 1 ) . ()xxx a x        =G G G G  G
() . ()xax   ºG  G
º Hence a is multiplicative identity in 5. ¯¡°¢±]
1a=
,( )for any a a a=‰ G  ]      
Let assume

25 G B  C




2 222 : 22 5 x        =  G  G   C
2225 2 5x =   C  BC     
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125Chapter 6: Ring Homomorphism and Isomorphism
Comparing we get B
C = o
ºB = o or C = o
if B = o 225 º  C 22/5 ºC 
But no such C in ] exists
if C = o then 22 B 
again so such Bexists.
Hence no such G exists
2] =G ]>  is not isomorphic to 5. ¯¡°¢±]

6.4 Summary
1) First Isomorphism Theorem of Ring: Let :RS


l
be a onto ring
homomorphism then /k e r ( )RR



x

.
2) Second Isomorphism Theorem for Rings: Let A be a subring and let B be an
ideal of R. Then \^:,ABa b a A b B




‰


‰
is a subring of ,RA B
 is
an ideal of A and () / /ABBA AB

x

3) Third Isomorphism Theorem of Ring: Statement : Let I and J be ideals of R
with IJ
ˆ
. Then /JI is an ideal of /RI and (/ ) / (/ ) /RI JI RJ
x
.
4) Let D be an integral domain. Then there exists a field F (called the field of
fraction of D) that contains a subring isomorphic to D.




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126 ALGEBRA II
6.5 Unit and Exercises
Exercises :
1) Let :,abRa bba£² ¯¦¦¦¦¡°
‰¤»¡°¦¦¢±¦¦¥¼\   under addition and multiplication of 2 x 2
matries. Prove that R is isomorphic to †.
2) Show that the map 2 :[ ] ()xM Gl\\  defined as
1(. . )o n
on
oaaaa x a xoa¬­ž­ G   ž­ž­žŸ®   is ring homomorphism. Find ker G.
3) Let :,abRa b dod£² ¯¦¦¦¦¡°‰¤»¡°¦¦¢±¦¦¥¼]    . Show that :RGl q]]      defined by
(, )abadod¬ ¯­ž¡°­Gž­ž¡°­žŸ®¢±  is a ring homomorphism. Find Ker G.
4) Let Gbe a ring homomorphism from a commutative ring R onto a commutative
ring S and let A be un ideal of S.
(i) If A is prime in S, show that 1() { : ( ) }AxR x AG ‰ G ‰   is pprime in R.
(ii) If A is maximal in S, show that 1()AG is maximal in R.
5) Let <>i
] = {a + bi : a, b ‰]}. Show that the field of quotients of <>i
] is ring
– isomorphic to <>\^:/ ,irs i r S








‰
__ (Hint: Let F is field containing ] and i,
then for any 1 ,, , aa Fa    ‰v ‰] Hence F contain _ and i. Also prove that for
any ,0 [ ] , [ ]ab iab i cd i i icd i     v ‰ ‰]_ .
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127Chapter 7: Euclidean Domain, Principal Ideal
Domain Unique Factorization UNIT 4
 7
EUCLIDEAN DOMAIN, PRINCIPAL IDEAL
DOMAIN UNIQUE FACTORIZATION DOMAIN

Unit Structure
7.0 Objective
7.1 Introduction
7.2 Prime and irreducible element
7.3 Euclidean domain (ED)
7.4 Principal ideal domain (PID)
7.5 Unique factorization domain (UFD)
7.6 Summary
7.7 Unit and Exercises
7.0 Objective The Objective this chapter is to make you understand
x Prime and irreducible elements and difference between them.
E u c l i d e a n d o m a i n (ED)
x Principal ideal domain (PID)
x Unique factorization domain (UFD)
x Difference between E D, PID, UFD


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128 ALGEBRA II
7.1 Introduction
Primes plays central role in theory of integers. Lots of famous theorems are there
for prime. For example, there are infinite numbers of primes; there are infinite
numbers of prime of the type 4n-1 (or 4n+1). Euclid lemma which states that p is
the prime, plab then either pla or pla, etc.
There some conjectures which are simple to state but yet not proved. One of them
is twin prime theorem. Prime like 3 and 5, 11 and 13, 17 and 19 are called twin
prime. Twin prime theorem states that “there are infinitely many twin primes”. This
conjecture yet to be proved.
We want to introduce same notion of prime to general ring. That is we want find
element in general ring which has property similar to prime of integers.
Note that in case of integer we define prime as positive integers which are divisible
by 1(unity) and itself but in case of general ring this definition may not be
appropriate. As if u is unit in ring R and a
a|b then ua|b. Hence in a ring if there more than one unit then definitely there more
than two divisors.
Hence we need different approach to define prime. One such approach given by
Euclid lemma.
7.2 Prime and Irreducible Element
Definition: Let R be an integral domain. A non zero non unit element p of R is said
to be prime element in R if whenever pa b pao rpb  ¸º .
Definition: A non zero non unit element P o f i n t e g r a l d o m a i n R i s s a i d t o b e
irreducible element if whenever p=a˜b then either a is unit or b is unit.
Two elements a and b of integral domain R is said to be associates of each other if
they differ by unit (i.e., a=ub for some unit u). The other way to say same thing is
a/b and b/a.
Theorem 1: In an integral domain prime element are irreducible.
Proof: Let R be an integral domain and let pR‰ be prime element in R. To Prove
P is irreducible.
Let p = a ˜b
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129Chapter 7: Euclidean Domain, Principal Ideal
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Hence To prove either a or b is unit.
Since . pp pa b º pao rpbº  (Since p is prime element)
If pa then a = kp for some kR‰
pa b= 1 pk p b k b    º º  bº is unit in R
Similarly if pb then we can prove a is unit in R. Hence whenever p = ab˜ º
either a or b is unit.
Therefore p is irreducible.
Remark:
1. In case of ], the prime and irreducible are same.
2. The above theorem say that prime and irreducible are same in an integral
domain. But converse need not be true.
That is there are integral domain in which irreducible need not be prime.
Let see one such example. Consider the ring d ¯¡°¢±] where d is square free integer.
(That is d is not divisible by square of any number).
We define a function :Nd ¯l¡°¢±]] define as
22. Na b d a b d  this
function is called as norm function.
This function has some trivial property which is easy to prove (of course involve
some calculation)
1.


. NxyNx Ny for all ,xyd ¯‰¡°¢±] .
2. If u is unit in d ¯¡°¢±]if and only if N(u) = 1
(This is clear because the only unit in d ¯¡°¢±] are 1+ and 1+ , + i in case d = i )
3. If ,xyd ¯‰¡°¢±] such that x|y then

.NxNy
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130 ALGEBRA II
4. If pd ¯‰¡°¢±] is prime then N(p) is prime number (This can be proved from
III property and Euclid Lemma)
Now consider particular example 3 ¯¡°¢±]
Consider 13 3 ¯‰ ¡°¢±]  
Claim : 13 is irreducible element
Let 13xy for some ,3xy ¯‰¡°¢±]

13NN x y=  



11 3 NxNy=   


Nx . Ny = 4 =






1 4 or 4 1, or 2 2 Nx = , Ny= N x = , Ny=N x = , Ny= ง
But in this case either x or y is unit in 3 ¯¡°¢±]
Hence 13become irreducible.
The other possible case is

2, 2 Nx Ny 
Let 33 xab ¯ ‰ ¡°¢±] then
2232 Nx a b   but this is not possible
for any value of ,.ab‰] Hence this case is not possible.
Thus we have shown that 1+ -3 is irreducible.
Also

1+ -3 1- -3 =4=2.2
Thus
1+ -3 2.2
but 1+ -3 2 as if 1+ -3 2 then


2 = 1+ -3 a+b -3

2 = a - 3 b + a + b - 3º a-3b = 2 º and abo
But the above equations have no integer solution 1 + -3 † 2
Hence 1+ -3 2 2 ¸but
1+ -3 † 2
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131Chapter 7: Euclidean Domain, Principal Ideal
Domain Unique Factorization
3. If p is a prime element in an integral domain R pj is prime ideal.
as let ab p¸‰ ab k p p ab  º¸ º ¸ pao rpbº [Since p is prime element]
or ap bpº‰ ‰  
ab p a p    =¸‰ º‰ or bp‰
p= is prime ideal.
Conversely, let p is prime ideal.
Let pa b a b p   ¸º¸‰ apº‰  or bp‰ (Since p is prime ideal)
pao rpbº  pºis prime element.
There are different classes of ring. Let we study them. In first year we study division
algorithm in ] and in []x\. Actually they are Euclidean algorithm. The integral
domain having Euclidean algorithm is called Euclidean domain (ED) . Similarly in
F.Y.B.S.c we study the fundamental theorem of Arithmetic (i.e. Unique
factorization theorem) applicable in ] and [] .x\
The integral domain having this property is called unique factorization domain
(UFD). The class of integral domain in which every ideal is principal ideal is called
principal ideal domain (PID). We will study these class of ring one by one.
7.3 Euclidean Domain (ED)
An integral domain R is said to be Euclidean domain (ED). If there exist a function \^:0dR lƒ] such that
(i)

dab da ¸ for all ,ab R‰.
(ii) For any ,,ab b o ‰v\ there exist ,prR‰ such that a = bp+r with r=0 or


< . dr db 
Note: Such function d is called Euclidean function.


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132 ALGEBRA II
Example:
1. ] is an Euclidean domain.
I n ], the Mod Function | | is an Euclidean Function and division algorithm
is Euclidean algorithm.
2. <>x\, the ring of polynomial over \ is Euclidean domain.
I n [] ,x\the degree function, degree of polynomial is Euclidean Function and
division algorithm of polynomials are Euclidean algorithm.
3. Any Field is by default Euclidean domain. In a field multiplication is Euclidean
Function. If ,,ab Fb o‰v  then
1ab b a this is nothing but Euclidean
algorithm.
4. Consider the ring of Gaussian integers<>\^:, ia b i a b ‰]]     .
Then ][i] is Euclidean domain.
Define the function <>\ ^:Ni U o l]`   as
22.Nai b a b =   
Then for any <>,xyi‰]



.NxyNxNy 
To See Euclidean algorithm.
Let xai b  and <>yci d i  ‰ ] s u c h t h a t . ci d v C o n s i d e r t h e
Quotient .xy
Let sxity  where ,.st‰_
Let m, n be the integers closest to s and t respectively such that 1s2mb and
12nt b

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133Chapter 7: Euclidean Domain, Principal Ideal
Domain Unique Factorization
Then s+itx
y

mms i nnt i    



mi n sm i tn    ¯ ¢±
Then


xmi n sm i tn y    ¯ ¢±
We claim thath the division algorithm of the definition of a Euclidean domain is
satisfied si satisfied with q = m + ni and


rm i n s mi t n    ¯ ¢±
Clearly, q belongs to Z[i], and since r = x - qy , so does r. Finally




() Nr N s m it q ni Ny     ¸



22smt n N y   


11
44Ny Ny¬­ž ­ž­žŸ®  
H e n c e f o r a n y <><> ,, 0 q , rxyiyi   ‰v  ‰]] S u c h t h a t qrxy w i t h


rNNy.
Hence <>i] has Euclidean algorithm. <>i=] is Euclidean domain.

These are some example of Euclidean domain. Let us understand the importance
of being Euclidean domain.
Theorem 2: In an Euclidean domain every ideal is principal ideal (i.e generated by
single element).
Proof: Let R be an Euclidean domain and let I be any non zero ideal of R.
Let d be an Euclidean function of R choose aI‰such that d(a) is minimum.
Claim : Ia.
Let bI‰ be arbitrary. Then by Euclidean algorithm there exist ,prR‰. Such that
b = ap + r w i t h r = 0 o r

. dr da  I f r v t h e n rba p I ‰ 

since is ideal I but then d(r) < d(a) will contradiction as a is the element of I
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134 ALGEBRA II
rb a q  == baº‰
but bI‰ be arbitrary.
Hence Iaˆ
IaIis principal ideal
=
=
The Integral domain in which every ideal is principal is known as Principal ideal
domain (PID). Thus above theorem says that every Euclidean domain is principal
ideal domain.
Now we see this class of ring in details.
7.4 Principal Ideal Domain
An Integral domain R is said to be principal ideal domain (PID) if every ideal of R
is Principal ideal.
Example:
1. Every field is by default Principal ideal domain.
S i n c e o n l y i d e a l i n f i e l d F is zero ideal and F itself. (Since fields are simple
ring). The zero ideal is generated by zero elements and F is generated by unity.
2. ] is PID.Since any ideal of ]are of the form ,.mm ‰]]
3. <>x\is PID.One can argue like, as <>x\ i s E u c l i d e a n d o m a i n , e v e r y
Euclidean domain is PID, hence <>x\ is PID. But we can prove directly that <>x\ i s P I D . P r o o f i s e x a c t l y s i m i l a r t o t h e w a y w e p r o v e “ I n E u c l i d e a n
domain every ideal is principal”. We request the students to understand the
similarity between two proofs.
Proof: Let I be any nonzero ideal of <>x\ Since Let
fxI‰ such that f(x) is
monic and deg f(x) is minimum.


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135Chapter 7: Euclidean Domain, Principal Ideal
Domain Unique Factorization
Claim:-
If x
Let
gxI‰ be arbitrary.
Let <>x\is Euclidean domain, then by division algorithm, there exist
pxand

<> rx x‰\ s u c h t h a t



gxp xfxr x    w i t h
rx o or deg


rx d egfx ˜  .
If
rx O vthen



rxgxp xfxI  ‰    (Since I is ideal)
But this is contradiction to minimality of deg
fxas

,fxIfx ‰ is monic &
deg
fx is minimum.
Hence
rx 



gxp xfx = 


gx f x I f x   º‰ º ˆ

If x=
Hence I is principal ideal.
= Every ideal of <>x\ is principal Hence <>x\ is PID.
4. <>x]is not PID.
Consider the ideal

\^:0 If x f    LVHYHQ i.e. I is ideal of <>x] with even
constant term.
\^ 12 .. : ,n
no o Ia x a x a a a  ‰ ]       LVHYHQ

\^1
1 .. 2 : 2 , 2n
no o oxax a b a b x         
Let assume that
,2xfx for some
<> .fxx‰]



<>xf xxfx g x g x x
   =‰
= ‰ ] IRUVRPH
= 1 = deg x = deg




fxgx d e g f x d e g gx     
ºdeg

1 fx d e g g x    but as degree is non negative number,
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136 ALGEBRA II
Hence either

1, deg f x deg g x       
or

,1 deg f x deg g x     
 .
If deg f(x) = 1, then f(x) = ax + b , for some ,ab‰] b u t t h e n

22fxa x b ‰º ‰      is not possible. Hence
. degfx1v

deg f x   =
fx ºis constant polynomial.
Let

2fxhx   for some

hx x ‰] t h e n

2f1h1 


f1 2 1       ,º  Š ' b u t
fx2 º    ( A s f(x) i s c o n s t a n t p o l y n o m i a l b u t
then
x2gx  which is nonsense.
Hence
,2xfx is not possible.
= I is not principal ideal.
A ring being Principal ideal domain has lots of advantage. We see them one by one.
We already seen “In an integral domain primes are irreducible”. But we also seen
the example of integral domain where irreducible are not prime. The one advantage
of being PID is that irreducible are also prime.
Theorem 3: I n P r i n c i p a l i d e a l d o m a i n i r r e d u c i b l e a r e p r i m e ,
Proof: Let R be principal ideal domain, and let r be irreducible element of R.
To prove r is prime.
Let ,, .rb c cb R   ‰
Clearly r is non zero, non unit element.
To prove rb or rc consider the ideal \^ /,Ixr byxyR  ‰   Since R is PID Id = for some dR‰
Since rI d‰
ra d = for some a in R. (as r is irreducible )
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137Chapter 7: Euclidean Domain, Principal Ideal
Domain Unique Factorization
If d is unit then I = R and as 1,RR‰=  B C ‰ such that 1 rbB C



, cc rb c r c r r b c r b c      = B  C = B ºC
rc r b c r c=B Cº 
Now if a is unit then rd! ! and as . bd br r b‰º ‰º 
=r is prime
1. In short we say that in PID, prime and irreducible are same.
2. We had seen that in 3 ¯¡°¢±], irreducible are not prime from this and above
theorem we conclude that 3 ¯¡°¢±] is not PID.
3. One more advantage of being PID is existence of GCD . Before we see let recall
the definition.
Definition:
(i) Let R be a ring. We say that element a of R divides bR‰or b is divisible by
a if there exist cR‰ such that b = ac and this we denote by ab.
(ii) Let R b e a r i n g ,.ab R ‰We say that dR‰ i s g r e a t e s t c o m m o n d i v i s i o r
(GCD) of a and b if (i) da and db (ii) If dR‰ such that 'da and 'db
then 'dd . It is denoted by
,.ab
In short the greatest common divisor of a, b at is the largest among all common
divisor.
In school, even if in F.Y.B.Sc. we seen how to find GCD of two positive integer.
The Euclidean algorithm is powerful technique to find GCD. We also seen how to
GCD of two polynomial with the help of division algorithm.
Do you think GCD of any two number exists in all integral domain. To surprise
you the answer is no.


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138 ALGEBRA II
Let see this example.
Let a4 &
b=2 1+ - 3 in -3 ¯¡°¢±]
as



a = 2 2 = 1+ -3 1- -3
b = 2(1 3) then 2a n d 2aband
1+ -3 a and
1+ -3 b
=Both 2 a n d
1+ -3 a r e c o m m o n d i v i s o r o f a & b. But 2† 3 )   and
†2 1+ -3 as both are irreducible.
Like this you can see so many example of non existence of GCD, Note that -3 ¯¡°¢±]
is not PID.
So we hope you can guess the second advantage being yes. PID guarantees the
existing of GCD.
Theorem 4: Let R Principal ideal domain. Then for any ,ab R

‰
, greatest common
divisor of a and b exists.
Proof: R is PID and ,.ab R‰ Consider the ideal \^:,Iax byxyR  ‰ As R
is PID,
= there exist d ‰R such that I=< d >.
Claim: d is gcd of ,.ab
(1) ,ba I d   ‰'
11 1 1&, , & aa dbb d a bR d ad b=  ‰º       
dd I‰'
,mn R= ‰  such that d=am + bn .
Now let assume that 1dR‰ such that 1da and 1.db
1 R ºB C‰  such that 11. ad a n d bdB C   
then
1 11 1 dda m b n m d n d m n d d  B C B  C º 
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139Chapter 7: Euclidean Domain, Principal Ideal
Domain Unique Factorization
Hence d is GCD of a and b.
1. Note that the GCD d of a and b is generator of ideal generated by a and b. This
may the reason why GCD is denoted by (,)ab .
2. Every Euclidean domain is PID. Hence in Euclidean domain also GCD of any
two elements exists. In fact Euclidean algorithm is best way to calculate the
GCD.
3. We have shown that Euclidean domain º PID but PID does not mean
Euclidean domain. For example it shown that the ring
,11 9|,2Ra b a b  £²¦¦  ¦¦ R ‰R ¤»¦¦¦¦¥¼] is PID but not Euclidean domain,
but detail is beyond the scope of syllabus. So we skip that.
Now let we move to next advantage of being PID. That is PID every ascending
chain of ideal is finite.
Theorem 5: Let R b e P I D a n d 12 3 n II I  ,‡‡‡ ‡ "" b e s t r i c t l y i n c r e a s i n g
chain of ideals. Then this chain must be at finite length.
Proof: Let nnIU I

Claim: I is ideal of R.
12 3
,n
nnII I
Let a b I U I  ,
‡‡‡ ‡
‰""

,pq = such that 1 ,,ppq aIb Ib I    ‰‰‰
if let pq p q    v then pqII‡

,qq q ab I a b I I  º‰ º  ‰ 'LVLGHDO nnabIU Iº‰    
Let r ‰ R then clearly ar and ra ‰ pI
.nnar and ra I U I   =‰ 
Hence I is ideal.
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140 ALGEBRA II
But as R is PID, therefore there exist d ‰ R such that I = d
Since nndd I U I‰  
=there exist k such that kdI‰
but then I = k dIˆ nknUI Iºˆ
Hence 12 3 n II I  ,‡‡‡ ‡ ""
= chain of ideal is finite.
This is important theorem which helps us (in future) to prove that PIDs h a v e
property called unique factorization as a product of irreducible. The class of ring
having this property is called as unique factorization domain (UFD). Let we study
this in details.
7.5 Unique Factorization Domain (UFD)
An integral domain R is said to be unique factorization domain (UFD) if
(i) Every non zero non unit element R can be expressed as a product of irreducible
of R.
(ii) The factorization into irreducible is unique up to associates and the order in
which the factors appear.
Now let we prove U. PID FDº
Theorem 5: Every principal ideal domain is Unique factorization domain.
Proof: Let R b e a p r i n c i p a l i d e a l d o m a i n a n d l e t oa b e a n y n o n z e r o n o n u n i t
element of R.
We will prove that oa can be expressed as product of irreducible
Claim: oa has at least one irreducible factor.
If oa is itself irreducible, then we are done. So assume 11 oaa b where neither of
11ab are unit and 1ais non zero.
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141Chapter 7: Euclidean Domain, Principal Ideal
Domain Unique Factorization
If 1a is irreducible then 1a become irreducible factor of oa.
So assume 12 2aa b where neither of 2aand 2b are unit. It 2a is irreducible
then we done otherwise we continue some step so that sequence of element nnab 
of R such that 1 .nn naa b
Thus we have 012 .... aaa  
vvv‡‡‡
Thus we have strictly increasing sequence of ideal in R. But R PID. There for this
Chain must be finite.
That is there exist raR‰ such that 12 ...r aa a ‡‡
In particular rais irreducible factor of .oa
Thus every non zero non unit element of R has irreducible factor.
Let 01 1ap q


where 1p is irreducible if 1qis also irreducible we get 0aas product
of irreducible.
Let assume 1q i s n o t i r r e d u c i b l e . c l e a r l y 1q is not unit otherwise 0abecome
irreducible which is not the case.
Hence let 12 2qp q

where 2pwhere 2qis irreducible & 2q is not unit. If 2q is
also irreducible then 01 11 2 2ap qp p q



, product of irreducible and we done. If
not we continue same process. Thus again getting ascending chain of ideal
01 ... aq c
‡

‡ . Being PID this chain must be finite. Hence mqR

‰
such that
01 ... ,mm ac q CC q q






being irreducible as product of irreducible.
Thus every non zero non unit element of R can be written as product uniqueness.
Let 12...k ap p p



12...r qq q


where ipand jqare irreducible in R for all i and
j.
12 1 2
112
112... ...
...
...kr
k
rpp p q q p
pp p p
pq q q=













=




'
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142 ALGEBRA II
And as irreducible are prime in PID.
1jpq=

for some j. Without loss of generality let 11pq

11 1qp a=


but both 1pand 1q are irreducible 1aº
is a unit.
1p=is associate of 1q.
12 1 2 2 1 2 2 1 2 ... ... ... ... ... ...kr k r k rpp p q q q p paqq p paqq=







º







º








Similarly canceling 23,. . .pp

and if Kr

we get
12 1 1. . . . . .kk r aa a q q







1º as product of irreducible.
This is contradiction.
Hence rk

and 'ips are associate of 'jqs . Hence proved.
Example :
1) A field <>,,Fx
]\ as being PID, there are UFD.
One can prove directly. In case of field, there is no non zero non unit element.
Hence by default field are UFD.
I n c a s e o f <>
and x


]\ w e c a n p r o v e u n i q u e f a c t o r i z a t i o n i n p r i m e
(irreducible polynomial) as in F.Y.B.Sc.
2) A part from this one can show that it D is UFD then <>Dx
also.
Thus as ] is UFD, there fore <>x
] also.
T h i s t h e o r e m i s p r o v e d i n n e x t c h a p t e r .
Remark : Since PID UFD
º
but converse need not be true that is UFD
ºPID
.
For example we seen that <>x] is not PID but it is UFD as ] is UFD.
Being UFD will also has lots of benefit. One such benefit is given in following
theorem.
Theorem : In unique factorization domain irreducibles are primes.
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143Chapter 7: Euclidean Domain, Principal Ideal
Domain Unique Factorization
Proof :
Let R be a UFD and aR
‰
be irreducible.
To show that a is prime.
As a is irreducible. a=is non zero non unit.
Let ab c
for some ,bc R

‰
.
=bc =ka for some kR
‰
.
As R is UFD, therefore ,,bck can be written as product of irreducible.
Let 123 1 2 1 2 ... , ... , ...jn n bb b b bcc c c kk k k















A w h e r e ,ijbc
and pkare
irreducibles in R for all ,ij

and p then
bc ka

12 12 12... ... ... ...mn bb b cc c kk k aº














A .
As factorization is unique upto associates. Hence a must be associates of some ib
or jcif a is associate of ibthen iaband hence ab if a is associate of some jc
then jac
and hence ac.
or ab c ab ac=


º





a=
is prime.
Thus we see in case of PID and UFD primes and irreducibles are essentially same.
This is the one reason why interval domain like -3 ¯
¡°¢±] is no UFD (Note that
,-3 1 + -3 ¯

¡°¢±] is irreducible but not prime).
One more advantage of being UFD is that it also guarantees the existence of GCD
of any two element of UFD. This is given as an exercise.




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144 ALGEBRA II
7.6 Summary
1) Let R be an integral domain. A non zero non unit element p of R is said to be
prime element in R if whenever pa b pao rpb  ¸º .
2) In an integral domain prime element are irreducible.
3) An integral domain R is said to be Euclidean domain (ED). If there exist a
function \^ :0dR U l] such that
( i )

dab da ¸ for all ,ab R‰.
(ii) For any ,,ab b o ‰v\ there exist ,prR‰
such that a = bp+r with r=0 or

<. dr db 
4) In an Euclidean domain every ideal is principal ideal (i.e generated by single
element).
5) An Integral domain R is said to be principal ideal domain (PID) if every ideal
of R is Principal ideal.
6) Let R b e P I D a n d 12 3 n II I  ,‡‡‡ ‡ "" b e s t r i c t l y i n c r e a s i n g c h a i n o f
ideals. Then this chain must be at finite length.
7) An integral domain R is said to be unique factorization domain (UFD) if
(i) Every non zero non unit element R c a n b e e x p r e s s e d a s a p r o d u c t o f
irreducible of R.
(ii) The factorization into irreducible is unique up to associates and the order
in which the factors appear.
8) Every principal ideal domain is Unique factorization domain.
9) In unique factorization domain irreducibles are primes.


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145Chapter 7: Euclidean Domain, Principal Ideal
Domain Unique Factorization
7.7 Unit and Exercises
1. Define
(i) Euclidean domain
( i i ) P r i n c i p a l i d e a l d o m a i n . S h o w t h a t a E u c l i d e a n d o m a i n i s a
p r i n c i p a l i d e a l d o m a i n .
2. Show that the following rings are Euclidean domain:
(i) The ring []i], ring of Gaussian integer
(ii) The polynomial ring F[x], where F is a Field.
3. Show the polynomial ring []x\is OID.
4. Prove or disprove
I f F is PID then F[x] is also PID.
5. Show that every ascending chain of ideals
12 1 ... ...nn II I I    ‡‡‡ ‡ in a PID R is finite.
6. Prove that in a PID R, an element a‰R is prime
i f a n d o n l y i f a is irreducible.
7. Show that a PID is a UFD.
8. Show that any prime element in integral domain is irreducible.
Is converse true? Justify your answer.
9. Explain why -5 ¯¡°¢±] is not PID.
10. Show that any two elements a & b in a PID R have a G CD which can be
expressed in the form abM Nwhere ,,, R MN‰
11. Show that every irreducible element in a UFD is prime.
12. Is 5 ¯¡°¢±] is UFD? Justify your answer.
13. Show that ]is PID but <>x] is not.
( H i n t : s h o w t h a t ( 2 , x ) i s n o t p r i n c i p a l i d e a l ) munotes.in

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146 ALGEBRA II
14. Let Rbe an integral domain in which every non zero, non unit
element can be expressed as a product of irreducible and every
i r r e d u c i b l e e l e m e n t i s p r i m e . S h o w t h a t R is UFD.
15. (Hint : Note that factorization is unique upto associate.
Take two factorization of an element & use the fact that irreducible
is prime.)
16. Let \ be an UFD. Show that for any ,, ,ab R‰ gcd of a and b exist.
(Hint : let 12 1 212 1 2 ... , ..m mm kkkk ap P p bpp p     A AA where all andij m A
need not be non zero then take 212 ....nn n ikk dp p p  where \^max ,ii inmA  
then show that d is gcd of a and b.

™™™™
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147Chapter 8: Irreducibility in Polynomial Ring UNIT 4
8
IRREDUCIBILITY IN POLYNOMIAL RING

Unit Structure
8.0 Objective
8.1 Introduction
8.2 Definition and example of irreducible and reducible polynomial.
8.3 Gauss lemma.
8.4 Eisenstein’s Criterion.
8.5 Summary
8.6 Unit and Exercises

8.0 Objective
This chapter makes you to understand
x Irreducible, reducible polynomials
x Classification of irreducible polynomials in \[x] and []x†
x Various criteria to check irreducibility.
x Gauss lemma and Eisenstein’s criteria.
8.1 Introduction
In the chapter of ring we had deals with polynomial ring in detail. In high school
students spend much time factoring polynomials and finding their roots. This is
what we going to learn in this chapter but in abstract manner. Let us understand
FIRST which polynomial we can factorize and which cannot. We start with
definition. munotes.in

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148 ALGEBRA II
8.2 Definition and Example of Irreducible and Reducible
Polynomial
Definition: L e t D be an integral domain. A polynomial p(x) in D[x] is said to
be irreducible polynomial if whenever p(x) = f(x) g(x) then either f(x) or g(x) is unit
(i.e. constant polynomial) in D[x].
Definition: A n o n z e r o – n o n u n i t p o l y n o m i a l o f D[x] which is not irreducible
is said to reducible .
Example:
1. The polynomial 223xis irreducible in []x_ but reducible in \ [x].
2. The polynomial 21x is irreducible over \ but reducible over ^.
3. The polynomial 21x is irreducible over 3] but reducible over 5]
Note that \^3 0,1, 2 ]
2fx x    then



01 , 12 25 m o d 3 2 .ff f           
but if we take element from 5]




01 1224 1 5 0 m o d 5fff         

2
5
220
Hence 2 1
1f
xx i n
x 
  
=

=] LVURRW
LVUHGXFLEOH
4. The only irreducible polynomial in \[x] are linear polynomial or the
polynomial224xbx c b c o   VXFKWKDW .
Proof: Clearly the linear polynomial ax + b, ,ab‰\are irreducible polynomial
because they cannot be factorize.
Similarly for polynomial 22,xbx c if b 4c o   then it has complex root
hence cannot be factorize. Now we will prove polynomial of any other degree must
be reducible. Let f(x) is any polynomial of degree n, n>2.
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149Chapter 8: Irreducibility in Polynomial Ring
Case I: deg f(x) is odd.
As complex roots are always come with conjugate pair and number of roots is equal
to degree of polynomial, therefore a polynomial of odd degree must have one real
root.
Let B be real root of f(x).



fxx g x  =  B where degree of
gxn 1 f=is reducible.
Case II: deg f(x) is even.
In worst case let assume all roots of f(x) are complex.
Let a + ib be one root of f(x) then aíib must be other roots.





.fxx a i b 2 a i b g x =         








2 2xai bx ai b g x x a b g x      


2 2andxab g x =    are two factors of
<>fxi n x\

fx = is reducible.
5. The only irreducible polynomial in ^ are linear polynomials.
T h e o t h e r w a y o f s a y i n g t h i s i s e v e r y p o l y n o m i a l o v e r
O n e o f i m p o r t a n t p r o b l e m i n m a t h e m a t i c s i s t o f i n d r o o t o f p o l y n o m i a l s . T h e s e
roots have different meaning and application in different context. Lots of
techniques are developed in order to find the root of polynomial.
S u p p o s e w e h a d a p p l i e d o n e t e c h n i q u e t o f i n d r o o t . B u t w e h a v e n ’ t g o t , t h e n
we apply second technique to find root but we havn’t got, then third, and so on.
And then we came to know that this polynomial has no solution.
I f i n a d v a n c e w e g e t t o k n o w t h a t t h e p o l y n o m i a l h a s n o r o o t o r i n o t h e r w o r d
it is irreducible then lots of our efforts can be saved. So we are going to see
some test to check whether given polynomial are reducible or not.
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150 ALGEBRA II
Reducibility test for Degrees 2 and 3.
Theorem 1: Let F be a field. If
<>fxF x‰ and
degfx2 or 3 then f(x) is
reducible over F if and only if f(x) has a zero in F.
Proof : Suppose that


fxg x h x
Where

<> ,.gxh x F x ‰ 
as


deg deg degfxg xh x   =
as
deg orfx2 3   
= one of the polynomial g(x) or h(x) must be of degree one.
Let assume deg g(x) = 1.

,0 gx a x b a     = v
then clearly 11.xba F ‰   is root of g(x) and hence of f(x).
=Conversely let assume that f has root Į in F then f(x) = (x- Į) g(x)
where deg g(x) = 1 or 2 = f(x) is reducible.
The above theorem is particularly used when the underlying Field is .p] because
in this case, we can check for reducibility of f(x) by simply checking that f(a) = 0
or not for a = 0, 1, …., p-1.
Note that the polynomials of degree larger than 3 may be reducible over a field,
even though they do not have zeros in field. For example in
_[x], t h e p o l y n o m i a l 4221 xx is equal to
221, x b u t h a s n o z e r o s i n
_.
To see next tests for inedibility we need following definition.
Content of Polynomial
Definition: The content of a non zero polynomial 1
1 .. ,nn
nn oax a x a
    
where ia‰] is the greatest common divisor of 11,, . . , .nn oaa a a  
Definition: A polynomial 1 .. ,n
no iax a x a a   ‰ ]    i s s a i d t o b e primitive
polynomial if its content is one. munotes.in

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151Chapter 8: Irreducibility in Polynomial Ring
8.3 Gauss Lemma
Theorem 1: T h e p r o d u c t o f t w o p r i m i t i v e p o l y n o m i a l s i s P r i m i t i v e .
Proof: Let f(x) and g(x) be primitive polynomials and suppose that f(x)g(x) is not
primitive. Let p i s p r i m e d i v i s o r o f c o n t e n t o f f(x) g(x). L e t

,fxg x a n d


fxgx be the polynomials obtained from f(x), g(x) and f(x) g(x) by reducing
the coefficient modulo p
then

<> ,pfxg x x ‰]  and

<>,pfxg x x   ]LQ
( ' Each coefficient of f(x) g(x) is divisible by p)

( ' p is prime, p=] is integral domain and hence <>px] is integral domain.)
ºEither p divides coefficient of f(x) or p divides coefficient of g(x).
Which is contradiction as f(x) and g(x) is primitive.


.fxgxi s p r i m i t i v e =
Theorem 2: L e t
<>fxx‰]. If f(x) is reducible over _,
then it is reducible over ].
Proof: Suppose that f(x) = g(x) h(x), where g(x) and h(x) ‰>@x_.
We may assume that f(x) is primitive because we can divide both f(x) and g(x) h(x)
by content of f(x). Let a and b be the least common multiple of the denominators
of the coefficients of g(x) and h(x) respectively.
Then ab f(x) = ag(x) bh(x) w h e r e

<>,. ag x bh x x ‰] 
Let 12and cc be the content of ag(x) and bh(x) respectively.


11 ag x c g x=  and

21 . bh x c h x  where
1gx a n d h1(x) are primitive
polynomial in <>x]



12 1 1 .. ab f x c c g x h x=   
Note that f(x) is of content 1, Hence content abf(x) is ab.

<>


.pfxgx i n x
fx o r g x    
         º
º ]
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152 ALGEBRA II
By Gauss lemma

11.gxhx is also primitive.




12 1 2 12
12 1 1 12.
.cc g xg xi s cc
abf x c c g x h x ab c c       
         =
= º FRQWHQWDW






<>
<>11 11.,
.fxg x h x w h e r e g x h x x
fx x= ‰
=]
]     
 LVUHGXFLEOHRYHU
Theorem 3: Let P be a prime and suppose that
<>fxx‰]  with deg
.fx 1
Let
fx be a polynomial in <>px] obtained from f(x) by reducing all coefficients
of f(x) modulo p. If f(x) is irreducible over p] and deg

degfxf x  then f(x)
is irreducible over _.
Proof: Since by previous theorem we know that if f(x) is reducible over _ then it
is reducible over ].
Hence let assume f(x) = g(x). h(x) where g(x), h(x) <>x‰].
Let


,a n dfxg x h x  be the polynomials obtained from f(x), g(x) and h(x) by
reducing all the coefficients modulo p. And
fx is irreducible over .p]
Since deg f(x) = deg
fx


deg deg ( ) deggx gx f x =    and


deg deg deghx hx f x     
But


.fxg x h x 
This means that

gxa n d hx are proper factor
fx, but this is contradiction
to our assumption that
fx is irreducible over .p]
Hence our assumption that f(x) reducible over _is wrong.
()fx =is irreducible over _.


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153Chapter 8: Irreducibility in Polynomial Ring
Example:
Let
3221 3 2 7 fx x x x     
Then over
32
2 1 fx x x      ]
S i n c e

01 1ff  
t h e n
2 fx ] LVLUUHGXFLEOHRYHU
and as deg f= deg f

fx = is irreducible over _.
8.4 Eisenstein’s Criterion
Theorem 4: L e t
<>1
1 ..nn
nn ofxa x ax a x
    ‰ ]      if there is a prime p
such that 2
1 †, , . . . , b u t †nn o oPaP a P a P a   then f(x) is irreducible over _.
Proof: Let assume that f(x) is reducible over _. Then by previous theorem f(x) is
reducible over ] also.
Let assume that f(x) = g(x) h(x) where

1d e g , d e g .gxh x n    
Let

11 .. and .. .rk
ro ko gx b x b x b hx c x c x c             
Since f(x) = g(x). h(x)
and .nr k oo oab c ab c=    
Since .oo oooPap b cp b o r p cºº  (Since P is prime)
but 2
0†pa, p does not divide both &.oobc Hence Let assume 0†opbb u t p c
Also .& .nr kr kpap b cp b p c  ºº
†† † †nr kr kpa pb c pba n d pc ºº
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154 ALGEBRA II
Therefore there exist least integer t such that †tpb but the
11 ...tt o t o t
tab cb c b c
as p a    

Now by choice of t, p must divide each term of right hand side except the first one.


11 .... . . .t
to t t o pa b c b c i e p bc     Which is not possible as ,.topb pc
Hence f(x) is irreducible over _.
Example: Irreducibility of thp Cyclotomic polynomial for any prime p,
The thpCyclotomic polynomial.

12 111p
pp
pxxx x xx   !! is irreducible over _.
As consider the polynomial





1
11.. 1 1111. .11pp p
pp
pxp x p xxfxx x p x pxx
   
         
Note that p divides all coefficient of f(x) except for leading coefficient 1 and also
2p does not divide constant term which p.
Hence by Eisenstein’s criteria f(x) is irreducible over Q.
But if


. Ox gxhx  were a non trivial factorization of
pOxover _ then

(1 ) (1 ) (1 )p fx O x g x h x     will we a non trivial factorization of f(x).
Since this is impossible, hence we conclude that
pOx is irreducible over _.
Let us understand the importance of irreducible polynomials.
1. The irreducible polynomial plays exactly same role as prime plays in case of
integer. For example the fundamental theorem of arithmetic (or unique
factorization theorem) irreducible polynomial are treated as prime.
2. One more reason which makes irreducible polynomial important, which is
given in terms of next theorem. munotes.in

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155Chapter 8: Irreducibility in Polynomial Ring
Theorem 5: Let F be a field and let
<> px Fx ‰ Then
px is a maximal ideal
in F[x] if and only if p(x) is irreducible over F.
Proof: Let assume that
px is maximal ideal in F[x].
Let assume that p(x) is reducible then p(x) = f(x) . g(x)




px f x px c f x   º‰ º
but
px is maximal ideal.
Hence it either

px f x  or
<>fxF x if

px f x  then

px and
fx are associate of each that therefore


px f xgx  implies
g(x) is unit.
And if
<>fxF x this implies f(x) is unit.

px= is irreducible.
Conversely, let p(x) is irreducible.
Let assume I is ideal of F[x] such that
<>px I Fx ˆˆ
As F[x] is PID,
Therefore there exists
<>gxF x‰ such that

Ig x





.[ ] px gx px f x gx f x Fx     º‰ º ‰ IRUVRPH
Now as p(x) irreducible.
() () ()px f x gx=  implies either f(x) is unit or g(x) is unit.
If f(x) is unit p(x) and g(x) are associates of each other.


px gx I=
If g(x) is unit then

Ig x F x 
Hence
px is maximal. munotes.in

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156 ALGEBRA II
Corollary 1: L e t F b e a f i e l d a n d l e t


<> ,px ax bx Fx    ‰. If p(x) i s
irreducible over F and





,( ) .pxax bx t h e n pxaxo r p xbx 
Proof: Let since p(x) i s i r r e d u c i b l e , <>()Fx
px=is a field and therefore
integral domain.
Let

axa n d bx  b e t h e i m a g e s o f a(x) a n d b(x) u n d e r t h e n a t u r a l
homomorphism from F[x] to <>.()Fx
px Since


pxaxbx .


.0 ax bx  = in <>
()Fx
px, but then
0 ax  or

<>Fx0S i n c e i s i n t e g r a l d o m a i n

bx¬­ ž ­ ž ­­ žŸ®.
Therefore it follows that



.pxaxo r pxbx 
Now we conclude this chapter by proving <>x] is unique factorization domain.
Theorem 5: Every non zero non unit polynomial in <>x] c a n b e w r i t t e n a s


12 1.. ...sm bb b p x p x  where '
ibs are irreducible polynomials of degree zero and
'
ips are irreducible polynomials of positive degree.
More over if



12 1 12 1.. ... .. ..sm k bb b p x p x cc c g x q x      where ''
ijbsa n dc s
are irreducible polynomial of degree zero and

and 'ijpxs q x s are irreducible
polynomial of positive degree then s = k, m=t and after renumbering


. andij i jpxq x b c    
Proof: Let
fx be a nonzero non unit polynomial from <>.x] If deg
, fx  
then
fx is constant and we are through, by Fundamental Theorem of Arithmetic.
If deg
, fx let b d e n o t e t h e c o n t e n t o f
fx, and let 12...sbb b b e t h e
factorization of b a s a p r o d u c t o f p r i m e s . T h e n

12 1... ,sfxb b b f x  w h e r e

1fx belongs to <>,x]is primitive and deg

1 deg .fxf x munotes.in

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157Chapter 8: Irreducibility in Polynomial Ring
Thus to prove the existence portion of the theorem it suffices to show that a
primitive polynomial
fx o f p o s i t i v e d e g r e e c a n b e w r i t t e n a s a p r o d u c t o f
irreducible polynomials of positive degree. We proceed by induction on deg
.fx
If deg
1, fx  then
fx is already irreducible and we are done.
Now suppose that every primitive polynomial of degrees less than deg
fx can
be written as product or irreducible of positive degree. If
fxis irreducibles,
there is nothing to prove. Otherwise Let


.fxg x h x  where both
gx
and
hx are primitive and have degree less than that of
.fx Thus by induction
both
gx and
hx can be written as a product of irreducibles of positive degree.
Clearly then
fx is also such a product.
To prove the uniqueness portion of the theorem, suppose that





12 1 1 2 k 1... ... c c ... c ...sm tfxb bb x x q xq x      pp
where 'ibs a n d 'jcs are irreducible polynomial of zero and
'ipxs  a n d

'jqx s are irreducible polynomials of positive degree.
12 12 Let .. & ...sk bb bb cc c c        
Since

' 'andijpxs q xs  are primitive therefore by Gauss Lemmas



12 ...m pxp x p x   and

1...t qx qx  are primitive.
Hence both b and c must equal plus or minus the content of
fx and therefore
are equal in absolute value. It their follows from the fundamental Theorem of
Arithmetic that s = k and after renumbering ib = c and1 < i < k.io
Thus by cancelling the constant terms in the two factorizations for we get,





12 1 ... ....mt pxp x p x qx qx       .
By viewing
ipx and
jqx a s e l e m e n t o f _[x] & noting that



112pxpx p x ….
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158 ALGEBRA II
Therefore


11 ....t px qx qx , and by corollary of previous theorem

1 j px q x f o r s o m e j . b y r e n u m b e r i n g w e g e t ,

11px qx



11qx fx px º but as both
1qx and
1px are irreducible therefore

fx is unit in []x_say1f


11 1 , qx fpx=   b u t b o t h
1qx and
1px are primitive which implies
11 f r, so 11() ()qx px r.
Also after canceling we get




23 2 mtpxp x p x q x q x       "" . After repeating above argument if m
< t then we get 1 on left side and non constant polynomial in right which leads to
contradiction. Hence m = t and

iipx qx  .
8.5 Summary
1) Let D be an integral domain. A polynomial p(x) in D[x] is said to be irreducible
polynomial i f w h e n e v e r p(x) = f(x) g(x) t h e n e i t h e r f(x) or g(x) i s u n i t ( i . e .
constant polynomial) in D[x].
2) A non zero – non unit polynomial of D[x] which is not irreducible is said to
reducible .
3) The content of a non zero polynomial 1
1 .. ,nn
nn oax a x a
    where
ia‰] is the greatest common divisor of 11,, . . , .nn oaa a a  
4) Gauss Lemma: The product of two primitive polynomials is Primitive.
5) Eisenstein’s Criterion: Let
1
1nn
nnfxa x ax  
  <>..oax   ‰ ] if
there is a prime p such that 2
1 †, , . . . , b u t †nn o oPaP a P a P a   then f(x) is
irreducible over _.
6) Let F be a field and let
<> px Fx ‰ Then
px is a maximal ideal in F[x]
if and only if p(x) is irreducible over F.

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159Chapter 8: Irreducibility in Polynomial Ring
8.6 Unit and Exercises
1) Show that 2x+1 a n d 2x+x+4 are irreducible polynomial in =>[@.
Show that <>

=[
[ and <>


=[
[[ are fields having 121 elements.
Solution :- Let
21 fx x 




0f= 1, f 1 = 2, f 2 = 5, f 3 = 10
f4= 1 7 = 6 (mod 11),
f5 = 2 6=4 (mod 11),
f6= 3 7 = 4 (mod 11) ,
f7 = 5 0 = 6 (mod 11),

I  (mod 11),
f9= 5 (mod 11),

f10 = 2 (mod 11).
Hence f(x) is irreducible over <>11x]
fx =is maximal ideal in <>11x]
<>11
()xfx=˜§] is field.
Now for any
<>11 hx x‰], by division algorithm




2hx = gx x+ 1 + rx
where
rx oor deg
rx = 1

11 ,, rx a x ba b=  ‰ ]      


h2x = g x x + 1 + ax + b =




22 211 1 hx x gx x a x b x       2ax + b + x +1 <><>\^2 11
11 1: ,()xax b x a b xfx    =    ‰˜§]] munotes.in

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160 ALGEBRA II
= There are 11 choice for a and 11 choice for b. Hence total 121 polynomials are
there of the form ax + b .
Hence <>11
2121.
1x
x
]
Similarly we can prove that <>11
2121.
4x
xx
]
2) Show that
<>2
5 3x +4x+3 x ]ෛ f a c t o r s a s

3x+2 x+4 a n d


4x+1 2x+3  Is <>5x] a UFD?
Justify your answer. Is the above factorization is unique upto multiplication by a
unit?
Solution:
234 3 fx x x  


03 1 1 0 0ff        

22 232 4 3 21 28 3 1 48 3 43 ( m o d 5 )xx x x x xx x x         =      
Similarly

2241 23 8 1 43 3 43 ( m o d 5 )xx x x x x               
Hence




234 3 3 2 4 4 1 2 3xx x x x x         
=Since

3x+2 x+4 and

4x+1 2x+3 are two factors of 234 3xx .
4 1(mod 5)
1 4 (mod 5)w
'  
   

41 4 ( m o d 5 )
4( m o d 5 )xx
x=   
      
   
4x+1 =is associate of
< >5x+4 in x ] Similarly 2x+3 = - (3x+2) (mod 5).
H e n c e 2 x + 3 i s a s s o c i a t e o f ( 3 x + 2 ) i n <>5x]
= Hence the above factorization is unique upto multiplication by a unit.
5'] is field <>5x=] us UFD. munotes.in

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161Chapter 8: Irreducibility in Polynomial Ring
3. Determine which of the following polynomials are irreducible in the indicated
rings.
( i ) <>2
2 1xxi nx ] 
( i i ) <>4
52 xx i n]
( i i i ) <>4210 1xxi n x  ]
4. Show that if
fx is maximal ideal in <>x\, then f(x) is irreducible.
5. Let F be a field. Show that the <>
/ Fx f x is field if and only if f(x) is
irreducible over F.
6. Show that the only maximal ideals of <>x\ are of the form
(x-a), 2ao r x b x c‰ \  where b,c ‰\ with 24 ba cR
7. Show that maximal ideals of <>x^ are x-D where .B‰^
8. Show that if D is UFD then D[x] is also (Proof is on similar line of proving
<>x] is UFD)
™™™™
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