G has 4-Sylow 3-subgroups
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I am working on the following problem:
Prove that if $lvert G rvert = 12$ and $G$ has $4$ Sylow 3-subgroups, then $G equiv A_4$.
If you let $G$ acts by conjugation on the set containing the $4$ Sylow 3-subgroups, then this action induces a homomorphism between $G$ and $S_4$. The issue is that I am having a hard time trying to prove that this homomorphism is injective. Note that this is important as the order of $G$ is $12$. Any thoughts?
abstract-algebra finite-groups symmetric-groups sylow-theory
asked Dec 12 '18 at 6:19
PeterPeter
936
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add a comment |
$begingroup$
I am working on the following problem:
Prove that if $lvert G rvert = 12$ and $G$ has $4$ Sylow 3-subgroups, then $G equiv A_4$.
If you let $G$ acts by conjugation on the set containing the $4$ Sylow 3-subgroups, then this action induces a homomorphism between $G$ and $S_4$. The issue is that I am having a hard time trying to prove that this homomorphism is injective. Note that this is important as the order of $G$ is $12$. Any thoughts?
abstract-algebra finite-groups symmetric-groups sylow-theory
asked Dec 12 '18 at 6:19
PeterPeter
936
$endgroup$
add a comment |
$begingroup$
I am working on the following problem:
Prove that if $lvert G rvert = 12$ and $G$ has $4$ Sylow 3-subgroups, then $G equiv A_4$.
If you let $G$ acts by conjugation on the set containing the $4$ Sylow 3-subgroups, then this action induces a homomorphism between $G$ and $S_4$. The issue is that I am having a hard time trying to prove that this homomorphism is injective. Note that this is important as the order of $G$ is $12$. Any thoughts?
abstract-algebra finite-groups symmetric-groups sylow-theory
asked Dec 12 '18 at 6:19
PeterPeter
936
$endgroup$
I am working on the following problem:
Prove that if $lvert G rvert = 12$ and $G$ has $4$ Sylow 3-subgroups, then $G equiv A_4$.
If you let $G$ acts by conjugation on the set containing the $4$ Sylow 3-subgroups, then this action induces a homomorphism between $G$ and $S_4$. The issue is that I am having a hard time trying to prove that this homomorphism is injective. Note that this is important as the order of $G$ is $12$. Any thoughts?
abstract-algebra finite-groups symmetric-groups sylow-theory
abstract-algebra finite-groups symmetric-groups sylow-theory
asked Dec 12 '18 at 6:19
PeterPeter
936
asked Dec 12 '18 at 6:19
PeterPeter
936
asked Dec 12 '18 at 6:19
PeterPeter
936
asked Dec 12 '18 at 6:19
PeterPeter
936
asked Dec 12 '18 at 6:19
PeterPeter
936
936
add a comment |
add a comment |
1 Answer
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The action is transitive, so the image has order divisible by $4$. If your homomorphism is not injective, then its kernel has order $3$. But that would mean the kernel was a normal Sylow $3$-subgroup, so unique.
answered Dec 12 '18 at 6:54
Lord Shark the UnknownLord Shark the Unknown
105k1160132
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1 Answer
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1 Answer
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$begingroup$
The action is transitive, so the image has order divisible by $4$. If your homomorphism is not injective, then its kernel has order $3$. But that would mean the kernel was a normal Sylow $3$-subgroup, so unique.
answered Dec 12 '18 at 6:54
Lord Shark the UnknownLord Shark the Unknown
105k1160132
$endgroup$
add a comment |
$begingroup$
The action is transitive, so the image has order divisible by $4$. If your homomorphism is not injective, then its kernel has order $3$. But that would mean the kernel was a normal Sylow $3$-subgroup, so unique.
answered Dec 12 '18 at 6:54
Lord Shark the UnknownLord Shark the Unknown
105k1160132
$endgroup$
add a comment |
$begingroup$
The action is transitive, so the image has order divisible by $4$. If your homomorphism is not injective, then its kernel has order $3$. But that would mean the kernel was a normal Sylow $3$-subgroup, so unique.
answered Dec 12 '18 at 6:54
Lord Shark the UnknownLord Shark the Unknown
105k1160132
$endgroup$
The action is transitive, so the image has order divisible by $4$. If your homomorphism is not injective, then its kernel has order $3$. But that would mean the kernel was a normal Sylow $3$-subgroup, so unique.
answered Dec 12 '18 at 6:54
Lord Shark the UnknownLord Shark the Unknown
105k1160132
answered Dec 12 '18 at 6:54
Lord Shark the UnknownLord Shark the Unknown
105k1160132
answered Dec 12 '18 at 6:54
Lord Shark the UnknownLord Shark the Unknown
105k1160132
answered Dec 12 '18 at 6:54
Lord Shark the UnknownLord Shark the Unknown
105k1160132
105k1160132
add a comment |
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