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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 ?

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    $\begingroup$ Your action is not trivial, so $\phi$ is not trivial and $K\neq G$. $\endgroup$ – GreginGre Oct 23 '19 at 8:36
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    $\begingroup$ If $K=G$, then there is only one coset in $[G:H]$, namely $H$ itself. So $H = G$. $\endgroup$ – Hongyi Huang Oct 23 '19 at 8:39
  • $\begingroup$ See the proof here and here. $\endgroup$ – Dietrich Burde Oct 23 '19 at 8:40
  • $\begingroup$ Thanks i got it :) $\endgroup$ – Sabhrant Oct 23 '19 at 8:46
  • $\begingroup$ See also math.stackexchange.com/questions/88719/… $\endgroup$ – lhf Oct 23 '19 at 10:00