Nontrivial conjugacy class and simplicity
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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 yesterday
GNUSupporter 8964民主女神 地下教會
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Pearl
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up vote
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down vote
favorite
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 yesterday
GNUSupporter 8964民主女神 地下教會
12.4k72344
asked yesterday
Pearl
1218
4
Hint: $G$ acts non-trivially on $K$.
– Tobias Kildetoft
yesterday
@TobiasKildetoft Thank you. I solve it!
– Pearl
yesterday
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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 yesterday
GNUSupporter 8964民主女神 地下教會
12.4k72344
asked yesterday
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 yesterday
GNUSupporter 8964民主女神 地下教會
12.4k72344
asked yesterday
Pearl
1218
edited yesterday
GNUSupporter 8964民主女神 地下教會
12.4k72344
asked yesterday
Pearl
1218
edited yesterday
GNUSupporter 8964民主女神 地下教會
12.4k72344
edited yesterday
GNUSupporter 8964民主女神 地下教會
12.4k72344
edited yesterday
GNUSupporter 8964民主女神 地下教會
12.4k72344
12.4k72344
asked yesterday
Pearl
1218
asked yesterday
Pearl
1218
asked yesterday
Pearl
1218
1218
4
Hint: $G$ acts non-trivially on $K$.
– Tobias Kildetoft
yesterday
@TobiasKildetoft Thank you. I solve it!
– Pearl
yesterday
add a comment |
4
Hint: $G$ acts non-trivially on $K$.
– Tobias Kildetoft
yesterday
@TobiasKildetoft Thank you. I solve it!
– Pearl
yesterday
4
4
Hint: $G$ acts non-trivially on $K$.
– Tobias Kildetoft
yesterday
Hint: $G$ acts non-trivially on $K$.
– Tobias Kildetoft
yesterday
@TobiasKildetoft Thank you. I solve it!
– Pearl
yesterday
@TobiasKildetoft Thank you. I solve it!
– Pearl
yesterday
add a comment |
1 Answer
1
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votes
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0
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accepted
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 yesterday
user347643
404
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
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 yesterday
user347643
404
add a comment |
up vote
0
down vote
accepted
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 yesterday
user347643
404
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
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 yesterday
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 yesterday
user347643
404
answered yesterday
user347643
404
answered yesterday
user347643
404
answered yesterday
user347643
404
404
add a comment |
add a comment |
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4
Hint: $G$ acts non-trivially on $K$.
– Tobias Kildetoft
yesterday
@TobiasKildetoft Thank you. I solve it!
– Pearl
yesterday