Groups | Mathematics Optional Notes for UPSC PDF Download

 Page 1


Edurev123 
Modern Algebra 
1. Groups 
1.1 If R is the set of real number and R
+
is the set of positive real numbers, show 
that R under addition (R,+) and R
+
under multiplication (R,+) are isomorphic. 
Similarly, if Q is the set rational numbers and Q
+
the set of positive rational 
numbers, are (Q
,
+
) and (Q
+
,+) isomorphic? Justify your answer. 
(2009 : 4+8=12 Marks) 
Solution: 
Let R be the set of real number and R
+
be the set of positive real number. 
We have to show 
(R
,
+)?(R
+
,·) 
Define ?? :R?R
+
as 
?? (?? )=?? ?? ; where ?? >0 
We will show ?? is one-one. 
Consider, 
ker??? ?={?? ??? |?? (?? )=1}
?={?? ?R|?? ?? =1}
?={?? ?R|?? =log
?? 1
}
?={?? ?R|?? =0}
?={0}
 
??? is 1-1. 
We will show ?? is homomorphism. 
Let ?? ,?? ??? 
Consider ?????????????????????????????????????? (?? +?? )=?? ?? +?? 
=?? ?? ·?? ?? =?? (?? )·?? (?? ) 
Page 2


Edurev123 
Modern Algebra 
1. Groups 
1.1 If R is the set of real number and R
+
is the set of positive real numbers, show 
that R under addition (R,+) and R
+
under multiplication (R,+) are isomorphic. 
Similarly, if Q is the set rational numbers and Q
+
the set of positive rational 
numbers, are (Q
,
+
) and (Q
+
,+) isomorphic? Justify your answer. 
(2009 : 4+8=12 Marks) 
Solution: 
Let R be the set of real number and R
+
be the set of positive real number. 
We have to show 
(R
,
+)?(R
+
,·) 
Define ?? :R?R
+
as 
?? (?? )=?? ?? ; where ?? >0 
We will show ?? is one-one. 
Consider, 
ker??? ?={?? ??? |?? (?? )=1}
?={?? ?R|?? ?? =1}
?={?? ?R|?? =log
?? 1
}
?={?? ?R|?? =0}
?={0}
 
??? is 1-1. 
We will show ?? is homomorphism. 
Let ?? ,?? ??? 
Consider ?????????????????????????????????????? (?? +?? )=?? ?? +?? 
=?? ?? ·?? ?? =?? (?? )·?? (?? ) 
??? is homomorphism. We will show ?? is onto, i.e., we have to find for any positive real 
number ' ?? ' some real number ?? such that 
?? (?? )?=?? ??.?? .,?????????????????????????????????????????? ?? ?=?? ???? ?????????????????????????????????????????????? ?? ?=?? 
On taking log both sides 
????????????????????????????????????????????????? =log
?? ??? 
???????????????????????????????????????????? (?? )=?? ????? 
Hence, ?? is onto. 
????????????????????????????????????????(R
,
+)?(R
+
,·) 
Let ?? be the set of rational numbers and ?? +
be the set of positive rational number. 
If ?? is homomorphism from ?? to ?? +
, then 
?? (?? ,?? )=?? (?? )?? (?? )??? ,?? ??? 
And if image of 1 is known then the image of every element will be known. 
????????????????????????????????????????? (?? )=?? ?? where??? =?? (1) 
If ?? =1,??????????????????????????? (?? )=1                       
??? is trivial homomorphism. 
If ?? ?1,??????????????????????????? (?? )=1 
then                              ?? (?? )=?? ?? ??? +
??? ??? 
which is a contradiction. 
Hence, only trivial homomorphism is possible. 
?????????????????????????????????????(?? ,+)?(?? +
,.) 
1.2 Determine the number of homomorphisms from the additive group ?? ????
 to the 
additive group ?? ????
. ( ?? ?? is the cyclic group of order ?? ). 
(2009 : 12 Marks) 
Solution: 
Let ?? :Z
15
?Z
10
 be a homomorphism. 
As Z
15
 is a cyclic group of order 15 . 
Z
15
=??? 
Under homomorphism, if element 1 will be mapped then remaining elements will get 
mapped themselves ( ??? is cyclic) 
Page 3


Edurev123 
Modern Algebra 
1. Groups 
1.1 If R is the set of real number and R
+
is the set of positive real numbers, show 
that R under addition (R,+) and R
+
under multiplication (R,+) are isomorphic. 
Similarly, if Q is the set rational numbers and Q
+
the set of positive rational 
numbers, are (Q
,
+
) and (Q
+
,+) isomorphic? Justify your answer. 
(2009 : 4+8=12 Marks) 
Solution: 
Let R be the set of real number and R
+
be the set of positive real number. 
We have to show 
(R
,
+)?(R
+
,·) 
Define ?? :R?R
+
as 
?? (?? )=?? ?? ; where ?? >0 
We will show ?? is one-one. 
Consider, 
ker??? ?={?? ??? |?? (?? )=1}
?={?? ?R|?? ?? =1}
?={?? ?R|?? =log
?? 1
}
?={?? ?R|?? =0}
?={0}
 
??? is 1-1. 
We will show ?? is homomorphism. 
Let ?? ,?? ??? 
Consider ?????????????????????????????????????? (?? +?? )=?? ?? +?? 
=?? ?? ·?? ?? =?? (?? )·?? (?? ) 
??? is homomorphism. We will show ?? is onto, i.e., we have to find for any positive real 
number ' ?? ' some real number ?? such that 
?? (?? )?=?? ??.?? .,?????????????????????????????????????????? ?? ?=?? ???? ?????????????????????????????????????????????? ?? ?=?? 
On taking log both sides 
????????????????????????????????????????????????? =log
?? ??? 
???????????????????????????????????????????? (?? )=?? ????? 
Hence, ?? is onto. 
????????????????????????????????????????(R
,
+)?(R
+
,·) 
Let ?? be the set of rational numbers and ?? +
be the set of positive rational number. 
If ?? is homomorphism from ?? to ?? +
, then 
?? (?? ,?? )=?? (?? )?? (?? )??? ,?? ??? 
And if image of 1 is known then the image of every element will be known. 
????????????????????????????????????????? (?? )=?? ?? where??? =?? (1) 
If ?? =1,??????????????????????????? (?? )=1                       
??? is trivial homomorphism. 
If ?? ?1,??????????????????????????? (?? )=1 
then                              ?? (?? )=?? ?? ??? +
??? ??? 
which is a contradiction. 
Hence, only trivial homomorphism is possible. 
?????????????????????????????????????(?? ,+)?(?? +
,.) 
1.2 Determine the number of homomorphisms from the additive group ?? ????
 to the 
additive group ?? ????
. ( ?? ?? is the cyclic group of order ?? ). 
(2009 : 12 Marks) 
Solution: 
Let ?? :Z
15
?Z
10
 be a homomorphism. 
As Z
15
 is a cyclic group of order 15 . 
Z
15
=??? 
Under homomorphism, if element 1 will be mapped then remaining elements will get 
mapped themselves ( ??? is cyclic) 
Suppose, 
?? (1)=?? 
As we know, if ?? is homomorphism from ?? to ?? '
 then ?? (?? (?? )/?? (?? ) where ?? ??? . 
As ?? (1)=??¨. 
??????????????????????????????????????????????????????????????????? (?? )/?? (?? )=15 
And order of element divides order of group 
??? (?? )/10 As ?? (?? )/15 
?? (?? )/15??? (?? )lg.c.d?(15,10)?? (?? )15 
???????????????????????????????????????????????????? (?? )=1 or ?? (?? )=5 
If ?? (?? )=1. Then it is trivial homomorphism. And if ?? (?? )=5. 
Note : ln?Z
?? , number of elements of order ?? =?? (?? ) ; provided ?????? . 
? In Z
10
, number of elements of order 5=?? (5)=4. 
? We have 4 possibilities for ?? . 
Total number of homomorphism =4+1=5. 
1.3 Show that the alternating group on four letters ?? ?? has no subgroup of order 6 . 
(2009 : 15 Marks) 
Solution: 
Consider the alternating group ?? 4
. 
?? (?? 4
)=
???(?? 4
)
2
=
14
2
=12 
We show although 6|12,?? 4
 has no subgroup of order 6 . Suppose ?? is a subgroup of 
?? 1
 and ?? (?? )=6. 
By previous problem the number of distinct 3-cycles in ?? 4
 is 
1
3
·
4!
(4-3)!
=
4·3·2·1
3·1
=8 
Again, as each 3-cycle will be even permutation all these 3-cycles are in ?? 4
. Obviously 
then, at least one 3-cycle, say s does not belong to ?? (?? (?? )=6) . 
Now, ?? ??? ??? 2
??? , because if ?? 2
??? . 
Page 4


Edurev123 
Modern Algebra 
1. Groups 
1.1 If R is the set of real number and R
+
is the set of positive real numbers, show 
that R under addition (R,+) and R
+
under multiplication (R,+) are isomorphic. 
Similarly, if Q is the set rational numbers and Q
+
the set of positive rational 
numbers, are (Q
,
+
) and (Q
+
,+) isomorphic? Justify your answer. 
(2009 : 4+8=12 Marks) 
Solution: 
Let R be the set of real number and R
+
be the set of positive real number. 
We have to show 
(R
,
+)?(R
+
,·) 
Define ?? :R?R
+
as 
?? (?? )=?? ?? ; where ?? >0 
We will show ?? is one-one. 
Consider, 
ker??? ?={?? ??? |?? (?? )=1}
?={?? ?R|?? ?? =1}
?={?? ?R|?? =log
?? 1
}
?={?? ?R|?? =0}
?={0}
 
??? is 1-1. 
We will show ?? is homomorphism. 
Let ?? ,?? ??? 
Consider ?????????????????????????????????????? (?? +?? )=?? ?? +?? 
=?? ?? ·?? ?? =?? (?? )·?? (?? ) 
??? is homomorphism. We will show ?? is onto, i.e., we have to find for any positive real 
number ' ?? ' some real number ?? such that 
?? (?? )?=?? ??.?? .,?????????????????????????????????????????? ?? ?=?? ???? ?????????????????????????????????????????????? ?? ?=?? 
On taking log both sides 
????????????????????????????????????????????????? =log
?? ??? 
???????????????????????????????????????????? (?? )=?? ????? 
Hence, ?? is onto. 
????????????????????????????????????????(R
,
+)?(R
+
,·) 
Let ?? be the set of rational numbers and ?? +
be the set of positive rational number. 
If ?? is homomorphism from ?? to ?? +
, then 
?? (?? ,?? )=?? (?? )?? (?? )??? ,?? ??? 
And if image of 1 is known then the image of every element will be known. 
????????????????????????????????????????? (?? )=?? ?? where??? =?? (1) 
If ?? =1,??????????????????????????? (?? )=1                       
??? is trivial homomorphism. 
If ?? ?1,??????????????????????????? (?? )=1 
then                              ?? (?? )=?? ?? ??? +
??? ??? 
which is a contradiction. 
Hence, only trivial homomorphism is possible. 
?????????????????????????????????????(?? ,+)?(?? +
,.) 
1.2 Determine the number of homomorphisms from the additive group ?? ????
 to the 
additive group ?? ????
. ( ?? ?? is the cyclic group of order ?? ). 
(2009 : 12 Marks) 
Solution: 
Let ?? :Z
15
?Z
10
 be a homomorphism. 
As Z
15
 is a cyclic group of order 15 . 
Z
15
=??? 
Under homomorphism, if element 1 will be mapped then remaining elements will get 
mapped themselves ( ??? is cyclic) 
Suppose, 
?? (1)=?? 
As we know, if ?? is homomorphism from ?? to ?? '
 then ?? (?? (?? )/?? (?? ) where ?? ??? . 
As ?? (1)=??¨. 
??????????????????????????????????????????????????????????????????? (?? )/?? (?? )=15 
And order of element divides order of group 
??? (?? )/10 As ?? (?? )/15 
?? (?? )/15??? (?? )lg.c.d?(15,10)?? (?? )15 
???????????????????????????????????????????????????? (?? )=1 or ?? (?? )=5 
If ?? (?? )=1. Then it is trivial homomorphism. And if ?? (?? )=5. 
Note : ln?Z
?? , number of elements of order ?? =?? (?? ) ; provided ?????? . 
? In Z
10
, number of elements of order 5=?? (5)=4. 
? We have 4 possibilities for ?? . 
Total number of homomorphism =4+1=5. 
1.3 Show that the alternating group on four letters ?? ?? has no subgroup of order 6 . 
(2009 : 15 Marks) 
Solution: 
Consider the alternating group ?? 4
. 
?? (?? 4
)=
???(?? 4
)
2
=
14
2
=12 
We show although 6|12,?? 4
 has no subgroup of order 6 . Suppose ?? is a subgroup of 
?? 1
 and ?? (?? )=6. 
By previous problem the number of distinct 3-cycles in ?? 4
 is 
1
3
·
4!
(4-3)!
=
4·3·2·1
3·1
=8 
Again, as each 3-cycle will be even permutation all these 3-cycles are in ?? 4
. Obviously 
then, at least one 3-cycle, say s does not belong to ?? (?? (?? )=6) . 
Now, ?? ??? ??? 2
??? , because if ?? 2
??? . 
Then,?????????????????????????????????????????????????????????????? 2
??? 
??????????????????????????????????????????????????????????????????????? ??? 
???? (???? )=
?? (?? )·?? (?? )
?? (?? n?? )
=
6.3
1
=18, 
not possible as ???? ??? 4
 and ?? (?? 4
)=12. 
1.4 Let ?? =?? -{-?? } be the set of all real numbers omitting -1 . Define the binary 
relation * on ?? by ?? *?? = ?? +?? +???? . Show (?? ,*) is a group and it is abelian. 
(2010 : 12 Marks) 
Solution: 
Given : Binary relation * as 
?? *?? =??? +?? +???? ,where??? ,?? ??? ??? 
Closure : Let ?? ,?? ??? 
?????????????????????????????????????????????????????? *?? =??? +?? +???? 
if ?? +?? +???? =-1, then 
?? +?? +???? +1 =0
? (1+?? )+?? (1+?? ) =0
? (1+?? )+(1+?? ) =0
? either 1+?? =0??? =-1
 or 1+?? =0??? =-1
 
? both ?? ,?? ??? 
? ??????????????????????????????????????????????????????? ?-1 and ?? ?-1
? ???????????????????????????????????? +?? +???? ?-1 for any ?? ,?? ??? ? ???????????????????????????????????? +?? +???? ?-1 for any ?? ,?? ??? ? ???????????????????????????????????????????????? *?? ??? 
So, closure is satisfied. 
Associative : Let ?? ,?? ,?? ??? 
???????????????????????????????????????(?? *?? )*?? ?=(?? +?? +???? )*?? ?=?? +?? +???? +?? +(?? +?? +???? )?? ?=?? +?? +???? +?? +???? +???? +?????? ?=?? +?? +?? +???? +???? +???? +?????? 
???????? ,????????????????????????????????? *(?? *?? )=?? *(?? +?? +???? ) 
??????????????????????????????????????????????????????????????=?? +?? +?? +???? +?? (?? +?? +???? ) 
Page 5


Edurev123 
Modern Algebra 
1. Groups 
1.1 If R is the set of real number and R
+
is the set of positive real numbers, show 
that R under addition (R,+) and R
+
under multiplication (R,+) are isomorphic. 
Similarly, if Q is the set rational numbers and Q
+
the set of positive rational 
numbers, are (Q
,
+
) and (Q
+
,+) isomorphic? Justify your answer. 
(2009 : 4+8=12 Marks) 
Solution: 
Let R be the set of real number and R
+
be the set of positive real number. 
We have to show 
(R
,
+)?(R
+
,·) 
Define ?? :R?R
+
as 
?? (?? )=?? ?? ; where ?? >0 
We will show ?? is one-one. 
Consider, 
ker??? ?={?? ??? |?? (?? )=1}
?={?? ?R|?? ?? =1}
?={?? ?R|?? =log
?? 1
}
?={?? ?R|?? =0}
?={0}
 
??? is 1-1. 
We will show ?? is homomorphism. 
Let ?? ,?? ??? 
Consider ?????????????????????????????????????? (?? +?? )=?? ?? +?? 
=?? ?? ·?? ?? =?? (?? )·?? (?? ) 
??? is homomorphism. We will show ?? is onto, i.e., we have to find for any positive real 
number ' ?? ' some real number ?? such that 
?? (?? )?=?? ??.?? .,?????????????????????????????????????????? ?? ?=?? ???? ?????????????????????????????????????????????? ?? ?=?? 
On taking log both sides 
????????????????????????????????????????????????? =log
?? ??? 
???????????????????????????????????????????? (?? )=?? ????? 
Hence, ?? is onto. 
????????????????????????????????????????(R
,
+)?(R
+
,·) 
Let ?? be the set of rational numbers and ?? +
be the set of positive rational number. 
If ?? is homomorphism from ?? to ?? +
, then 
?? (?? ,?? )=?? (?? )?? (?? )??? ,?? ??? 
And if image of 1 is known then the image of every element will be known. 
????????????????????????????????????????? (?? )=?? ?? where??? =?? (1) 
If ?? =1,??????????????????????????? (?? )=1                       
??? is trivial homomorphism. 
If ?? ?1,??????????????????????????? (?? )=1 
then                              ?? (?? )=?? ?? ??? +
??? ??? 
which is a contradiction. 
Hence, only trivial homomorphism is possible. 
?????????????????????????????????????(?? ,+)?(?? +
,.) 
1.2 Determine the number of homomorphisms from the additive group ?? ????
 to the 
additive group ?? ????
. ( ?? ?? is the cyclic group of order ?? ). 
(2009 : 12 Marks) 
Solution: 
Let ?? :Z
15
?Z
10
 be a homomorphism. 
As Z
15
 is a cyclic group of order 15 . 
Z
15
=??? 
Under homomorphism, if element 1 will be mapped then remaining elements will get 
mapped themselves ( ??? is cyclic) 
Suppose, 
?? (1)=?? 
As we know, if ?? is homomorphism from ?? to ?? '
 then ?? (?? (?? )/?? (?? ) where ?? ??? . 
As ?? (1)=??¨. 
??????????????????????????????????????????????????????????????????? (?? )/?? (?? )=15 
And order of element divides order of group 
??? (?? )/10 As ?? (?? )/15 
?? (?? )/15??? (?? )lg.c.d?(15,10)?? (?? )15 
???????????????????????????????????????????????????? (?? )=1 or ?? (?? )=5 
If ?? (?? )=1. Then it is trivial homomorphism. And if ?? (?? )=5. 
Note : ln?Z
?? , number of elements of order ?? =?? (?? ) ; provided ?????? . 
? In Z
10
, number of elements of order 5=?? (5)=4. 
? We have 4 possibilities for ?? . 
Total number of homomorphism =4+1=5. 
1.3 Show that the alternating group on four letters ?? ?? has no subgroup of order 6 . 
(2009 : 15 Marks) 
Solution: 
Consider the alternating group ?? 4
. 
?? (?? 4
)=
???(?? 4
)
2
=
14
2
=12 
We show although 6|12,?? 4
 has no subgroup of order 6 . Suppose ?? is a subgroup of 
?? 1
 and ?? (?? )=6. 
By previous problem the number of distinct 3-cycles in ?? 4
 is 
1
3
·
4!
(4-3)!
=
4·3·2·1
3·1
=8 
Again, as each 3-cycle will be even permutation all these 3-cycles are in ?? 4
. Obviously 
then, at least one 3-cycle, say s does not belong to ?? (?? (?? )=6) . 
Now, ?? ??? ??? 2
??? , because if ?? 2
??? . 
Then,?????????????????????????????????????????????????????????????? 2
??? 
??????????????????????????????????????????????????????????????????????? ??? 
???? (???? )=
?? (?? )·?? (?? )
?? (?? n?? )
=
6.3
1
=18, 
not possible as ???? ??? 4
 and ?? (?? 4
)=12. 
1.4 Let ?? =?? -{-?? } be the set of all real numbers omitting -1 . Define the binary 
relation * on ?? by ?? *?? = ?? +?? +???? . Show (?? ,*) is a group and it is abelian. 
(2010 : 12 Marks) 
Solution: 
Given : Binary relation * as 
?? *?? =??? +?? +???? ,where??? ,?? ??? ??? 
Closure : Let ?? ,?? ??? 
?????????????????????????????????????????????????????? *?? =??? +?? +???? 
if ?? +?? +???? =-1, then 
?? +?? +???? +1 =0
? (1+?? )+?? (1+?? ) =0
? (1+?? )+(1+?? ) =0
? either 1+?? =0??? =-1
 or 1+?? =0??? =-1
 
? both ?? ,?? ??? 
? ??????????????????????????????????????????????????????? ?-1 and ?? ?-1
? ???????????????????????????????????? +?? +???? ?-1 for any ?? ,?? ??? ? ???????????????????????????????????? +?? +???? ?-1 for any ?? ,?? ??? ? ???????????????????????????????????????????????? *?? ??? 
So, closure is satisfied. 
Associative : Let ?? ,?? ,?? ??? 
???????????????????????????????????????(?? *?? )*?? ?=(?? +?? +???? )*?? ?=?? +?? +???? +?? +(?? +?? +???? )?? ?=?? +?? +???? +?? +???? +???? +?????? ?=?? +?? +?? +???? +???? +???? +?????? 
???????? ,????????????????????????????????? *(?? *?? )=?? *(?? +?? +???? ) 
??????????????????????????????????????????????????????????????=?? +?? +?? +???? +?? (?? +?? +???? ) 
???????????????????????????????????????????????????????????????=?? +?? +?? +???? +???? +???? +?????? 
as??????????????????????????????????????(?? *?? )*?? =?? *(?? *?? ) 
? Associative property is satisfied. 
Identity: 
Let                                        ?? *?? =?? =?? +?? +???? 
??????????????????????????????????? +?? +???? =?? ?????????
??????????????????????????????????????? (1+?? )=0 ?????????
 as???????????????????????????????????????????????????? ?-1??? =0?????????
 
??? =0 is an identity and as ?? ??? . 
? identity exists. 
Inverse : 
?????? ???????????????????????????????????? *?? ?=0=?? +?? +????
?? (1+?? )?=-?? ??? =
-?? 1+?? ??????????????????????????????????????(?? ?-1)?????????????
 
Also, 
-?? 1+?? ??? 
? Inverse exists. 
As closure, associative property, identity, inverse conditions are satisfied. ?(?? ,*) is a 
group. 
Now,                                    ?? *?? =?? +?? +???? 
???????????????????????????????????????????????????? *?? =?? +?? +???? =?? +?? +???? 
As                                          ?? *?? =?? *?? 
?(?? ,*) is abelian. 
1.5 Show that a cyclic group of order 6 is isomorphic to the product of a cyclic 
group of order 2 and a cyclic group of order 3 . Can you generalize this? Justify. 
(2010 : 12 Marks) 
Solution: 
Let ??? 6
 is cyclic group of order 6