$G$ be a finite group, if $H\lneq G$ and $|G| \nmid [G:H]!$. Prove $G$ is not simple

Asked Oct 23 '19 at 08:28

Active Oct 23 '19 at 08:28

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Let G be a finite group, if $H\lneq G$ and $|G| \nmid ( \, [G:H]! \, ) $ then prove $G$ is not simple.

I used contrapositive argument. Suppose $G$ is simple then we need to prove that $ |G| \mid ( \, [G:H]! \, ) $. Now consider the group action $$f:G\times G/H \rightarrow G/H \\ (g,xH) \mapsto (gxH)$$

Note that it is not necessary for $H$ to be a normal subgroup of $G$. This group action is equivalent to a homomorphism $\phi : G \rightarrow S(G/H)$. Then $K=Ker(\phi) \trianglelefteq G$. If $K={1}$ then $|G|$ divides $|S(G/H)|=[G:H]!$ . Now i am not sure how i can exclude the case of $K=G$. Any hints ?

asked Oct 23 '19 at 08:28

Sam

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1

Your action is not trivial, so $\phi$ is not trivial and $K\neq G$. – GreginGre Oct 23 '19 at 08:36
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If $K=G$, then there is only one coset in $[G:H]$, namely $H$ itself. So $H = G$. – Groups Oct 23 '19 at 08:39
See the proof here and here. – Dietrich Burde Oct 23 '19 at 08:40
Thanks i got it :) – Sam Oct 23 '19 at 08:46
See also https://math.stackexchange.com/questions/88719/a-group-g-with-a-subgroup-h-of-index-n-has-a-normal-subgroup-k-subset-h – lhf Oct 23 '19 at 10:00

$G$ be a finite group, if $H\lneq G$ and $|G| \nmid [G:H]!$. Prove $G$ is not simple

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