Nontrivial conjugacy class and simplicity
Prove that if $G$ has a nontrivial conjugacy class $K$ such that $|G|$ does not divide $|K|!$, then $G$ is not simple.
How can I prove this statement? I think that with the condition, I could find nontrivial normal subgroup. But I failed. Help me!
abstract-algebra group-theory
edited Nov 30 at 13:41
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Nov 30 at 13:36
Pearl
1218
add a comment |
Prove that if $G$ has a nontrivial conjugacy class $K$ such that $|G|$ does not divide $|K|!$, then $G$ is not simple.
How can I prove this statement? I think that with the condition, I could find nontrivial normal subgroup. But I failed. Help me!
abstract-algebra group-theory
edited Nov 30 at 13:41
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Nov 30 at 13:36
Pearl
1218
5
Hint: $G$ acts non-trivially on $K$.
– Tobias Kildetoft
Nov 30 at 13:38
@TobiasKildetoft Thank you. I solve it!
– Pearl
Nov 30 at 13:43
add a comment |
Prove that if $G$ has a nontrivial conjugacy class $K$ such that $|G|$ does not divide $|K|!$, then $G$ is not simple.
How can I prove this statement? I think that with the condition, I could find nontrivial normal subgroup. But I failed. Help me!
abstract-algebra group-theory
edited Nov 30 at 13:41
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Nov 30 at 13:36
Pearl
1218
Prove that if $G$ has a nontrivial conjugacy class $K$ such that $|G|$ does not divide $|K|!$, then $G$ is not simple.
How can I prove this statement? I think that with the condition, I could find nontrivial normal subgroup. But I failed. Help me!
abstract-algebra group-theory
abstract-algebra group-theory
edited Nov 30 at 13:41
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Nov 30 at 13:36
Pearl
1218
edited Nov 30 at 13:41
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Nov 30 at 13:36
Pearl
1218
edited Nov 30 at 13:41
GNUSupporter 8964民主女神 地下教會
12.8k72445
edited Nov 30 at 13:41
GNUSupporter 8964民主女神 地下教會
12.8k72445
edited Nov 30 at 13:41
GNUSupporter 8964民主女神 地下教會
12.8k72445
12.8k72445
asked Nov 30 at 13:36
Pearl
1218
asked Nov 30 at 13:36
Pearl
1218
asked Nov 30 at 13:36
Pearl
1218
1218
5
Hint: $G$ acts non-trivially on $K$.
– Tobias Kildetoft
Nov 30 at 13:38
@TobiasKildetoft Thank you. I solve it!
– Pearl
Nov 30 at 13:43
add a comment |
5
Hint: $G$ acts non-trivially on $K$.
– Tobias Kildetoft
Nov 30 at 13:38
@TobiasKildetoft Thank you. I solve it!
– Pearl
Nov 30 at 13:43
5
5
Hint: $G$ acts non-trivially on $K$.
– Tobias Kildetoft
Nov 30 at 13:38
Hint: $G$ acts non-trivially on $K$.
– Tobias Kildetoft
Nov 30 at 13:38
@TobiasKildetoft Thank you. I solve it!
– Pearl
Nov 30 at 13:43
@TobiasKildetoft Thank you. I solve it!
– Pearl
Nov 30 at 13:43
add a comment |
1 Answer
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Consider the group $G$ act on $K$ via conjugation.
Then by such homomorphism $phi$, with 1st isomorphism theorem, $|G/Kerphi|||K|!$
If $Kerphi={1}$, it contradicts the non-divisibility condition in problem.
If $Kerphi=G$, it contradicts the nontriviality of conjugacy class $K$.
So we get nontrivial normal subgroup $Kerphi$ which shows that G is not simple.
answered Nov 30 at 14:00
user347643
404
add a comment |
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1 Answer
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Consider the group $G$ act on $K$ via conjugation.
Then by such homomorphism $phi$, with 1st isomorphism theorem, $|G/Kerphi|||K|!$
If $Kerphi={1}$, it contradicts the non-divisibility condition in problem.
If $Kerphi=G$, it contradicts the nontriviality of conjugacy class $K$.
So we get nontrivial normal subgroup $Kerphi$ which shows that G is not simple.
answered Nov 30 at 14:00
user347643
404
add a comment |
Consider the group $G$ act on $K$ via conjugation.
Then by such homomorphism $phi$, with 1st isomorphism theorem, $|G/Kerphi|||K|!$
If $Kerphi={1}$, it contradicts the non-divisibility condition in problem.
If $Kerphi=G$, it contradicts the nontriviality of conjugacy class $K$.
So we get nontrivial normal subgroup $Kerphi$ which shows that G is not simple.
answered Nov 30 at 14:00
user347643
404
add a comment |
Consider the group $G$ act on $K$ via conjugation.
Then by such homomorphism $phi$, with 1st isomorphism theorem, $|G/Kerphi|||K|!$
If $Kerphi={1}$, it contradicts the non-divisibility condition in problem.
If $Kerphi=G$, it contradicts the nontriviality of conjugacy class $K$.
So we get nontrivial normal subgroup $Kerphi$ which shows that G is not simple.
answered Nov 30 at 14:00
user347643
404
Consider the group $G$ act on $K$ via conjugation.
Then by such homomorphism $phi$, with 1st isomorphism theorem, $|G/Kerphi|||K|!$
If $Kerphi={1}$, it contradicts the non-divisibility condition in problem.
If $Kerphi=G$, it contradicts the nontriviality of conjugacy class $K$.
So we get nontrivial normal subgroup $Kerphi$ which shows that G is not simple.
answered Nov 30 at 14:00
user347643
404
answered Nov 30 at 14:00
user347643
404
answered Nov 30 at 14:00
user347643
404
answered Nov 30 at 14:00
user347643
404
404
add a comment |
add a comment |
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5
Hint: $G$ acts non-trivially on $K$.
– Tobias Kildetoft
Nov 30 at 13:38
@TobiasKildetoft Thank you. I solve it!
– Pearl
Nov 30 at 13:43