$G$ be a finite group of order $n$ , $H$ be a proper subgroup of order $m$ such that $(n/m)!<2n$ ; $G$ is not simple Let $G$ be a finite group of order $n$ , $H$ be a proper subgroup of order $m$ such that $(n/m)!<2n$ ; then how to show that $G$ is not simple ? I have proceeded by Cayley's theorem , $\ker f$ is normal and contained in $H$ ; so it is a proper normal subgroup of $G$ , we need moreover a non-singleton normal subgroup , which I am not able to get . Please help . Thanks in advance . 
 A: Assume that $G$ is simple. So the action $f$ of $G$ on the left cosets of $H$ has trivial kernel. That is, $f$ is an injective homomorphism $f:G \to S_{n/m}$ to the symmetric group of degree $n/m$. Hence $|G| = n$ divides $(n/m)!$. But then $(n/m)! < 2n$ implies $n = (n/m)!$, so $f$ must be an isomorphism, and $G \cong S_{n/m}$, which is not simple, contradiction.
A: Assume by contradiction that $G$ is simple. Consider the action on the coset space $G/H$ by left multiplication. This induces a homomorphism $\rho:$ $G \to S_{\frac{m}{m}}$ with $\ker \rho \triangleleft H$. Since $H$ proper, and $G$ simple, then $\ker \rho=\lbrace 1 \rbrace $. So $\rho $ injective. Lagrange's theorem implies that $n$ divides $(\frac{n}{m})!$ and since $(\frac{n}{m})!<2n$, then we mut have $n=(\frac{n}{m})!$, so $G \cong S_{\frac{n}{m}}$. Now since $G$ is simple and $H$ is proper, we must have $n>2m$ (any subgroup of index 2 is normal!)and so $\frac{n}{m} \geq 3$, but $S_{k}$ for $k \geq 3$ is not simple. Contradiction.
