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

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 ?

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