Number of homomorphisms between two arbitrary groups
$begingroup$
How many homomorphisms are there from A5 to S4 ?
This is how I tried to solve it.
If there is a homomorphism from A5 to S4 , then order of element of S4 should divide the order of its preimage. Now what are the possible order of elements in S4.1,2,3 and 4. Since A5 contains (12345), which is of order 5.. what could be image of (12345). Definitely Identity element which is of order 1. Similarly all 5 cycles must be mapped to identity. There are 24 elements of 5 cycles. 24 elements out of 60 are mapped to identity .. now only two types of homomorphisms are possible. either 30:1mapping or 60:1 mapping. Consider (12)(34) which belongs to A5. It's image can be element of order 2 or identity.there are 15 elements of order 2 . suppose these 15 elements are mapped to some element 'g' of order 2 of S4, you need another 15 elements to get mapped to 'g' to have 30 :1 mapping. Other type of elements left in A5 is of order 3. None of them can be mapped to g. hence 15 elements of order 2 should be mapped to identity .. so , (24+15=39) elements mapped to identity.As mentioned earlier it should be 30 :1 or 60:1 mapping. So it must be 60:1 mapping.Hence a trivial homomorphism. Answer is 1.
I wanted to know is there any other technique which can be used to find number of homomorphism in the above question ?
In general, how to find number of homomorphism between any two arbitrary groups ?
abstract-algebra group-theory finite-groups permutation-cycles
asked Dec 20 '18 at 18:23
Bharadwaj RcrBharadwaj Rcr
315
$endgroup$
add a comment |
$begingroup$
How many homomorphisms are there from A5 to S4 ?
This is how I tried to solve it.
If there is a homomorphism from A5 to S4 , then order of element of S4 should divide the order of its preimage. Now what are the possible order of elements in S4.1,2,3 and 4. Since A5 contains (12345), which is of order 5.. what could be image of (12345). Definitely Identity element which is of order 1. Similarly all 5 cycles must be mapped to identity. There are 24 elements of 5 cycles. 24 elements out of 60 are mapped to identity .. now only two types of homomorphisms are possible. either 30:1mapping or 60:1 mapping. Consider (12)(34) which belongs to A5. It's image can be element of order 2 or identity.there are 15 elements of order 2 . suppose these 15 elements are mapped to some element 'g' of order 2 of S4, you need another 15 elements to get mapped to 'g' to have 30 :1 mapping. Other type of elements left in A5 is of order 3. None of them can be mapped to g. hence 15 elements of order 2 should be mapped to identity .. so , (24+15=39) elements mapped to identity.As mentioned earlier it should be 30 :1 or 60:1 mapping. So it must be 60:1 mapping.Hence a trivial homomorphism. Answer is 1.
I wanted to know is there any other technique which can be used to find number of homomorphism in the above question ?
In general, how to find number of homomorphism between any two arbitrary groups ?
abstract-algebra group-theory finite-groups permutation-cycles
asked Dec 20 '18 at 18:23
Bharadwaj RcrBharadwaj Rcr
315
$endgroup$
4
$begingroup$
$A_5$ is simple.
$endgroup$
– Lord Shark the Unknown
Dec 20 '18 at 18:26
3
$begingroup$
So $A_5$ is simple and since order of $A_5$ is greater than order of $S_4$ , It cannot be one-one mapping. So trivial
$endgroup$
– Bharadwaj Rcr
Dec 20 '18 at 18:40
1
$begingroup$
The question has been answered here already.
$endgroup$
– Dietrich Burde
Dec 20 '18 at 19:15
1
$begingroup$
List all the functions between the two sets of group elements and count which ones are group homomorphisms.
$endgroup$
– Somos
Dec 20 '18 at 19:26
1
$begingroup$
The exact number of homomorphisms between two arbitrary groups is rarely of much interest, although for some reason it seems to be a popular homework problem. It is of course interesting to know hwether two groups are isomorphic, or whether one is isomorphic to a subgroup or quotient of the other, but the exact number of homomorphisms has no particular significance.
$endgroup$
– Derek Holt
Dec 21 '18 at 8:31
add a comment |
$begingroup$
How many homomorphisms are there from A5 to S4 ?
This is how I tried to solve it.
If there is a homomorphism from A5 to S4 , then order of element of S4 should divide the order of its preimage. Now what are the possible order of elements in S4.1,2,3 and 4. Since A5 contains (12345), which is of order 5.. what could be image of (12345). Definitely Identity element which is of order 1. Similarly all 5 cycles must be mapped to identity. There are 24 elements of 5 cycles. 24 elements out of 60 are mapped to identity .. now only two types of homomorphisms are possible. either 30:1mapping or 60:1 mapping. Consider (12)(34) which belongs to A5. It's image can be element of order 2 or identity.there are 15 elements of order 2 . suppose these 15 elements are mapped to some element 'g' of order 2 of S4, you need another 15 elements to get mapped to 'g' to have 30 :1 mapping. Other type of elements left in A5 is of order 3. None of them can be mapped to g. hence 15 elements of order 2 should be mapped to identity .. so , (24+15=39) elements mapped to identity.As mentioned earlier it should be 30 :1 or 60:1 mapping. So it must be 60:1 mapping.Hence a trivial homomorphism. Answer is 1.
I wanted to know is there any other technique which can be used to find number of homomorphism in the above question ?
In general, how to find number of homomorphism between any two arbitrary groups ?
abstract-algebra group-theory finite-groups permutation-cycles
asked Dec 20 '18 at 18:23
Bharadwaj RcrBharadwaj Rcr
315
$endgroup$
How many homomorphisms are there from A5 to S4 ?
This is how I tried to solve it.
If there is a homomorphism from A5 to S4 , then order of element of S4 should divide the order of its preimage. Now what are the possible order of elements in S4.1,2,3 and 4. Since A5 contains (12345), which is of order 5.. what could be image of (12345). Definitely Identity element which is of order 1. Similarly all 5 cycles must be mapped to identity. There are 24 elements of 5 cycles. 24 elements out of 60 are mapped to identity .. now only two types of homomorphisms are possible. either 30:1mapping or 60:1 mapping. Consider (12)(34) which belongs to A5. It's image can be element of order 2 or identity.there are 15 elements of order 2 . suppose these 15 elements are mapped to some element 'g' of order 2 of S4, you need another 15 elements to get mapped to 'g' to have 30 :1 mapping. Other type of elements left in A5 is of order 3. None of them can be mapped to g. hence 15 elements of order 2 should be mapped to identity .. so , (24+15=39) elements mapped to identity.As mentioned earlier it should be 30 :1 or 60:1 mapping. So it must be 60:1 mapping.Hence a trivial homomorphism. Answer is 1.
I wanted to know is there any other technique which can be used to find number of homomorphism in the above question ?
In general, how to find number of homomorphism between any two arbitrary groups ?
abstract-algebra group-theory finite-groups permutation-cycles
abstract-algebra group-theory finite-groups permutation-cycles
asked Dec 20 '18 at 18:23
Bharadwaj RcrBharadwaj Rcr
315
asked Dec 20 '18 at 18:23
Bharadwaj RcrBharadwaj Rcr
315
asked Dec 20 '18 at 18:23
Bharadwaj RcrBharadwaj Rcr
315
asked Dec 20 '18 at 18:23
Bharadwaj RcrBharadwaj Rcr
315
asked Dec 20 '18 at 18:23
Bharadwaj RcrBharadwaj Rcr
315
315
4
$begingroup$
$A_5$ is simple.
$endgroup$
– Lord Shark the Unknown
Dec 20 '18 at 18:26
3
$begingroup$
So $A_5$ is simple and since order of $A_5$ is greater than order of $S_4$ , It cannot be one-one mapping. So trivial
$endgroup$
– Bharadwaj Rcr
Dec 20 '18 at 18:40
1
$begingroup$
The question has been answered here already.
$endgroup$
– Dietrich Burde
Dec 20 '18 at 19:15
1
$begingroup$
List all the functions between the two sets of group elements and count which ones are group homomorphisms.
$endgroup$
– Somos
Dec 20 '18 at 19:26
1
$begingroup$
The exact number of homomorphisms between two arbitrary groups is rarely of much interest, although for some reason it seems to be a popular homework problem. It is of course interesting to know hwether two groups are isomorphic, or whether one is isomorphic to a subgroup or quotient of the other, but the exact number of homomorphisms has no particular significance.
$endgroup$
– Derek Holt
Dec 21 '18 at 8:31
add a comment |
4
$begingroup$
$A_5$ is simple.
$endgroup$
– Lord Shark the Unknown
Dec 20 '18 at 18:26
3
$begingroup$
So $A_5$ is simple and since order of $A_5$ is greater than order of $S_4$ , It cannot be one-one mapping. So trivial
$endgroup$
– Bharadwaj Rcr
Dec 20 '18 at 18:40
1
$begingroup$
The question has been answered here already.
$endgroup$
– Dietrich Burde
Dec 20 '18 at 19:15
1
$begingroup$
List all the functions between the two sets of group elements and count which ones are group homomorphisms.
$endgroup$
– Somos
Dec 20 '18 at 19:26
1
$begingroup$
The exact number of homomorphisms between two arbitrary groups is rarely of much interest, although for some reason it seems to be a popular homework problem. It is of course interesting to know hwether two groups are isomorphic, or whether one is isomorphic to a subgroup or quotient of the other, but the exact number of homomorphisms has no particular significance.
$endgroup$
– Derek Holt
Dec 21 '18 at 8:31
4
4
$begingroup$
$A_5$ is simple.
$endgroup$
– Lord Shark the Unknown
Dec 20 '18 at 18:26
$begingroup$
$A_5$ is simple.
$endgroup$
– Lord Shark the Unknown
Dec 20 '18 at 18:26
3
3
$begingroup$
So $A_5$ is simple and since order of $A_5$ is greater than order of $S_4$ , It cannot be one-one mapping. So trivial
$endgroup$
– Bharadwaj Rcr
Dec 20 '18 at 18:40
$begingroup$
So $A_5$ is simple and since order of $A_5$ is greater than order of $S_4$ , It cannot be one-one mapping. So trivial
$endgroup$
– Bharadwaj Rcr
Dec 20 '18 at 18:40
1
1
$begingroup$
The question has been answered here already.
$endgroup$
– Dietrich Burde
Dec 20 '18 at 19:15
$begingroup$
The question has been answered here already.
$endgroup$
– Dietrich Burde
Dec 20 '18 at 19:15
1
1
$begingroup$
List all the functions between the two sets of group elements and count which ones are group homomorphisms.
$endgroup$
– Somos
Dec 20 '18 at 19:26
$begingroup$
List all the functions between the two sets of group elements and count which ones are group homomorphisms.
$endgroup$
– Somos
Dec 20 '18 at 19:26
1
1
$begingroup$
The exact number of homomorphisms between two arbitrary groups is rarely of much interest, although for some reason it seems to be a popular homework problem. It is of course interesting to know hwether two groups are isomorphic, or whether one is isomorphic to a subgroup or quotient of the other, but the exact number of homomorphisms has no particular significance.
$endgroup$
– Derek Holt
Dec 21 '18 at 8:31
$begingroup$
The exact number of homomorphisms between two arbitrary groups is rarely of much interest, although for some reason it seems to be a popular homework problem. It is of course interesting to know hwether two groups are isomorphic, or whether one is isomorphic to a subgroup or quotient of the other, but the exact number of homomorphisms has no particular significance.
$endgroup$
– Derek Holt
Dec 21 '18 at 8:31
add a comment |
1 Answer
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$begingroup$
Suppose $f:A_5 to S_4$ be a homomorphism. Then $ker f$ is a normal subgroup of $A_5$. But $A_5$ is simple, so $$ker f in Big{ {e},A_5Big}$$
$ker f={e}$ implies $$A_5/{e} sim f(A_5)$$ and so $f(A_5)$ is a subgroup of order $60$ in $S_4$, which is not possible in $S_4$.
$ker f=A_5$ implies $f$ is trivial
Hence $$Bigvert{f ;vert ;f:A_5 to S_4 ;text{is a homomorphism} }Bigvert=1$$
For finding homomorphism $f$ for arbitraay two groups, use the following facts:
$vert f(g) vert$ divides $vert g vert$ where $g$ belong to the domain with $vert g vert < infty$ [this is useful for finite groups]- $f(g^n)=[f(g)]^n$
- List all normal subgroups of domain and use first isomorphism theorem
answered Dec 21 '18 at 4:13
Chinnapparaj RChinnapparaj R
5,8082928
$endgroup$
add a comment |
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$begingroup$
Suppose $f:A_5 to S_4$ be a homomorphism. Then $ker f$ is a normal subgroup of $A_5$. But $A_5$ is simple, so $$ker f in Big{ {e},A_5Big}$$
$ker f={e}$ implies $$A_5/{e} sim f(A_5)$$ and so $f(A_5)$ is a subgroup of order $60$ in $S_4$, which is not possible in $S_4$.
$ker f=A_5$ implies $f$ is trivial
Hence $$Bigvert{f ;vert ;f:A_5 to S_4 ;text{is a homomorphism} }Bigvert=1$$
For finding homomorphism $f$ for arbitraay two groups, use the following facts:
$vert f(g) vert$ divides $vert g vert$ where $g$ belong to the domain with $vert g vert < infty$ [this is useful for finite groups]- $f(g^n)=[f(g)]^n$
- List all normal subgroups of domain and use first isomorphism theorem
answered Dec 21 '18 at 4:13
Chinnapparaj RChinnapparaj R
5,8082928
$endgroup$
add a comment |
$begingroup$
Suppose $f:A_5 to S_4$ be a homomorphism. Then $ker f$ is a normal subgroup of $A_5$. But $A_5$ is simple, so $$ker f in Big{ {e},A_5Big}$$
$ker f={e}$ implies $$A_5/{e} sim f(A_5)$$ and so $f(A_5)$ is a subgroup of order $60$ in $S_4$, which is not possible in $S_4$.
$ker f=A_5$ implies $f$ is trivial
Hence $$Bigvert{f ;vert ;f:A_5 to S_4 ;text{is a homomorphism} }Bigvert=1$$
For finding homomorphism $f$ for arbitraay two groups, use the following facts:
$vert f(g) vert$ divides $vert g vert$ where $g$ belong to the domain with $vert g vert < infty$ [this is useful for finite groups]- $f(g^n)=[f(g)]^n$
- List all normal subgroups of domain and use first isomorphism theorem
answered Dec 21 '18 at 4:13
Chinnapparaj RChinnapparaj R
5,8082928
$endgroup$
add a comment |
$begingroup$
Suppose $f:A_5 to S_4$ be a homomorphism. Then $ker f$ is a normal subgroup of $A_5$. But $A_5$ is simple, so $$ker f in Big{ {e},A_5Big}$$
$ker f={e}$ implies $$A_5/{e} sim f(A_5)$$ and so $f(A_5)$ is a subgroup of order $60$ in $S_4$, which is not possible in $S_4$.
$ker f=A_5$ implies $f$ is trivial
Hence $$Bigvert{f ;vert ;f:A_5 to S_4 ;text{is a homomorphism} }Bigvert=1$$
For finding homomorphism $f$ for arbitraay two groups, use the following facts:
$vert f(g) vert$ divides $vert g vert$ where $g$ belong to the domain with $vert g vert < infty$ [this is useful for finite groups]- $f(g^n)=[f(g)]^n$
- List all normal subgroups of domain and use first isomorphism theorem
answered Dec 21 '18 at 4:13
Chinnapparaj RChinnapparaj R
5,8082928
$endgroup$
Suppose $f:A_5 to S_4$ be a homomorphism. Then $ker f$ is a normal subgroup of $A_5$. But $A_5$ is simple, so $$ker f in Big{ {e},A_5Big}$$
$ker f={e}$ implies $$A_5/{e} sim f(A_5)$$ and so $f(A_5)$ is a subgroup of order $60$ in $S_4$, which is not possible in $S_4$.
$ker f=A_5$ implies $f$ is trivial
Hence $$Bigvert{f ;vert ;f:A_5 to S_4 ;text{is a homomorphism} }Bigvert=1$$
For finding homomorphism $f$ for arbitraay two groups, use the following facts:
$vert f(g) vert$ divides $vert g vert$ where $g$ belong to the domain with $vert g vert < infty$ [this is useful for finite groups]- $f(g^n)=[f(g)]^n$
- List all normal subgroups of domain and use first isomorphism theorem
answered Dec 21 '18 at 4:13
Chinnapparaj RChinnapparaj R
5,8082928
edited Dec 21 '18 at 6:48
answered Dec 21 '18 at 4:13
Chinnapparaj RChinnapparaj R
5,8082928
answered Dec 21 '18 at 4:13
Chinnapparaj RChinnapparaj R
5,8082928
answered Dec 21 '18 at 4:13
Chinnapparaj RChinnapparaj R
5,8082928
5,8082928
add a comment |
add a comment |
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4
$begingroup$
$A_5$ is simple.
$endgroup$
– Lord Shark the Unknown
Dec 20 '18 at 18:26
3
$begingroup$
So $A_5$ is simple and since order of $A_5$ is greater than order of $S_4$ , It cannot be one-one mapping. So trivial
$endgroup$
– Bharadwaj Rcr
Dec 20 '18 at 18:40
1
$begingroup$
The question has been answered here already.
$endgroup$
– Dietrich Burde
Dec 20 '18 at 19:15
1
$begingroup$
List all the functions between the two sets of group elements and count which ones are group homomorphisms.
$endgroup$
– Somos
Dec 20 '18 at 19:26
1
$begingroup$
The exact number of homomorphisms between two arbitrary groups is rarely of much interest, although for some reason it seems to be a popular homework problem. It is of course interesting to know hwether two groups are isomorphic, or whether one is isomorphic to a subgroup or quotient of the other, but the exact number of homomorphisms has no particular significance.
$endgroup$
– Derek Holt
Dec 21 '18 at 8:31