Mixing
Kernel of a homomorphism from $S_4$ to $mathbb R^*$
$begingroup$
Let $S_4$ be the symmetric group on $4$ symbols. Let $f: S_4 to Bbb R^*$ be a homomorphism, where $Bbb R^*$ denotes the non-zero real numbers. Then the number of elements in the set ${ xin S_4: f(x) =1} $ is ....?
abstract-algebra group-theory permutations
edited Jan 27 at 19:38
Matt Samuel
39k63769
asked Jan 27 at 1:58
NizamudheenNizamudheen
385
$endgroup$
add a comment |
$begingroup$
Let $S_4$ be the symmetric group on $4$ symbols. Let $f: S_4 to Bbb R^*$ be a homomorphism, where $Bbb R^*$ denotes the non-zero real numbers. Then the number of elements in the set ${ xin S_4: f(x) =1} $ is ....?
abstract-algebra group-theory permutations
edited Jan 27 at 19:38
Matt Samuel
39k63769
asked Jan 27 at 1:58
NizamudheenNizamudheen
385
$endgroup$
2
$begingroup$
Is $f$ injective? The questions is asking for the cardinality of the kernel of $f$, so do we know more about $f$?
$endgroup$
– Dave
Jan 27 at 2:04
$begingroup$
Kernal of $f$ is normal group so possible orders 1,4,12,24 respecteively the order of normal subgroups trivial group, klein 4 group, $A_4$, $S_4$. Concluding which one
$endgroup$
– Nizamudheen
Jan 27 at 5:47
add a comment |
$begingroup$
Let $S_4$ be the symmetric group on $4$ symbols. Let $f: S_4 to Bbb R^*$ be a homomorphism, where $Bbb R^*$ denotes the non-zero real numbers. Then the number of elements in the set ${ xin S_4: f(x) =1} $ is ....?
abstract-algebra group-theory permutations
edited Jan 27 at 19:38
Matt Samuel
39k63769
asked Jan 27 at 1:58
NizamudheenNizamudheen
385
$endgroup$
Let $S_4$ be the symmetric group on $4$ symbols. Let $f: S_4 to Bbb R^*$ be a homomorphism, where $Bbb R^*$ denotes the non-zero real numbers. Then the number of elements in the set ${ xin S_4: f(x) =1} $ is ....?
abstract-algebra group-theory permutations
abstract-algebra group-theory permutations
edited Jan 27 at 19:38
Matt Samuel
39k63769
asked Jan 27 at 1:58
NizamudheenNizamudheen
385
edited Jan 27 at 19:38
Matt Samuel
39k63769
asked Jan 27 at 1:58
NizamudheenNizamudheen
385
edited Jan 27 at 19:38
Matt Samuel
39k63769
edited Jan 27 at 19:38
Matt Samuel
39k63769
edited Jan 27 at 19:38
Matt Samuel
39k63769
39k63769
asked Jan 27 at 1:58
NizamudheenNizamudheen
385
asked Jan 27 at 1:58
NizamudheenNizamudheen
385
asked Jan 27 at 1:58
NizamudheenNizamudheen
385
385
2
$begingroup$
Is $f$ injective? The questions is asking for the cardinality of the kernel of $f$, so do we know more about $f$?
$endgroup$
– Dave
Jan 27 at 2:04
$begingroup$
Kernal of $f$ is normal group so possible orders 1,4,12,24 respecteively the order of normal subgroups trivial group, klein 4 group, $A_4$, $S_4$. Concluding which one
$endgroup$
– Nizamudheen
Jan 27 at 5:47
add a comment |
2
$begingroup$
Is $f$ injective? The questions is asking for the cardinality of the kernel of $f$, so do we know more about $f$?
$endgroup$
– Dave
Jan 27 at 2:04
$begingroup$
Kernal of $f$ is normal group so possible orders 1,4,12,24 respecteively the order of normal subgroups trivial group, klein 4 group, $A_4$, $S_4$. Concluding which one
$endgroup$
– Nizamudheen
Jan 27 at 5:47
2
2
$begingroup$
Is $f$ injective? The questions is asking for the cardinality of the kernel of $f$, so do we know more about $f$?
$endgroup$
– Dave
Jan 27 at 2:04
$begingroup$
Is $f$ injective? The questions is asking for the cardinality of the kernel of $f$, so do we know more about $f$?
$endgroup$
– Dave
Jan 27 at 2:04
$begingroup$
Kernal of $f$ is normal group so possible orders 1,4,12,24 respecteively the order of normal subgroups trivial group, klein 4 group, $A_4$, $S_4$. Concluding which one
$endgroup$
– Nizamudheen
Jan 27 at 5:47
$begingroup$
Kernal of $f$ is normal group so possible orders 1,4,12,24 respecteively the order of normal subgroups trivial group, klein 4 group, $A_4$, $S_4$. Concluding which one
$endgroup$
– Nizamudheen
Jan 27 at 5:47
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There is only one nonidentity element of finite order in $mathbb R^*$, namely $-1$. Thus either the homomorphism is trivial or it has image of order $2$. Both are possible. Therefore there are two possible orders of the kernel, which I believe you can figure out.
answered Jan 27 at 18:06
Matt SamuelMatt Samuel
39k63769
$endgroup$
$begingroup$
The possible kernels are $A_4$ and $S_4$.
$endgroup$
– Nizamudheen
Jan 28 at 19:18
1
$begingroup$
@niz Yes, that's right.
$endgroup$
– Matt Samuel
Jan 28 at 19:28
add a comment |
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1 Answer
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1 Answer
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$begingroup$
There is only one nonidentity element of finite order in $mathbb R^*$, namely $-1$. Thus either the homomorphism is trivial or it has image of order $2$. Both are possible. Therefore there are two possible orders of the kernel, which I believe you can figure out.
answered Jan 27 at 18:06
Matt SamuelMatt Samuel
39k63769
$endgroup$
$begingroup$
The possible kernels are $A_4$ and $S_4$.
$endgroup$
– Nizamudheen
Jan 28 at 19:18
1
$begingroup$
@niz Yes, that's right.
$endgroup$
– Matt Samuel
Jan 28 at 19:28
add a comment |
$begingroup$
There is only one nonidentity element of finite order in $mathbb R^*$, namely $-1$. Thus either the homomorphism is trivial or it has image of order $2$. Both are possible. Therefore there are two possible orders of the kernel, which I believe you can figure out.
answered Jan 27 at 18:06
Matt SamuelMatt Samuel
39k63769
$endgroup$
$begingroup$
The possible kernels are $A_4$ and $S_4$.
$endgroup$
– Nizamudheen
Jan 28 at 19:18
1
$begingroup$
@niz Yes, that's right.
$endgroup$
– Matt Samuel
Jan 28 at 19:28
add a comment |
$begingroup$
There is only one nonidentity element of finite order in $mathbb R^*$, namely $-1$. Thus either the homomorphism is trivial or it has image of order $2$. Both are possible. Therefore there are two possible orders of the kernel, which I believe you can figure out.
answered Jan 27 at 18:06
Matt SamuelMatt Samuel
39k63769
$endgroup$
There is only one nonidentity element of finite order in $mathbb R^*$, namely $-1$. Thus either the homomorphism is trivial or it has image of order $2$. Both are possible. Therefore there are two possible orders of the kernel, which I believe you can figure out.
answered Jan 27 at 18:06
Matt SamuelMatt Samuel
39k63769
answered Jan 27 at 18:06
Matt SamuelMatt Samuel
39k63769
answered Jan 27 at 18:06
Matt SamuelMatt Samuel
39k63769
answered Jan 27 at 18:06
Matt SamuelMatt Samuel
39k63769
39k63769
$begingroup$
The possible kernels are $A_4$ and $S_4$.
$endgroup$
– Nizamudheen
Jan 28 at 19:18
1
$begingroup$
@niz Yes, that's right.
$endgroup$
– Matt Samuel
Jan 28 at 19:28
add a comment |
$begingroup$
The possible kernels are $A_4$ and $S_4$.
$endgroup$
– Nizamudheen
Jan 28 at 19:18
1
$begingroup$
@niz Yes, that's right.
$endgroup$
– Matt Samuel
Jan 28 at 19:28
$begingroup$
The possible kernels are $A_4$ and $S_4$.
$endgroup$
– Nizamudheen
Jan 28 at 19:18
$begingroup$
The possible kernels are $A_4$ and $S_4$.
$endgroup$
– Nizamudheen
Jan 28 at 19:18
1
1
$begingroup$
@niz Yes, that's right.
$endgroup$
– Matt Samuel
Jan 28 at 19:28
$begingroup$
@niz Yes, that's right.
$endgroup$
– Matt Samuel
Jan 28 at 19:28
add a comment |
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$begingroup$
Is $f$ injective? The questions is asking for the cardinality of the kernel of $f$, so do we know more about $f$?
$endgroup$
– Dave
Jan 27 at 2:04
$begingroup$
Kernal of $f$ is normal group so possible orders 1,4,12,24 respecteively the order of normal subgroups trivial group, klein 4 group, $A_4$, $S_4$. Concluding which one
$endgroup$
– Nizamudheen
Jan 27 at 5:47