TITLE: If $H \leq G$ and $[G:H]! \leq |G|$ then $G$ is not simple
QUESTION [6 upvotes]: I'm looking for verification:
My claim: If $G$ is a finite group and $H$ is a (proper)subgroup of index $k>1$, where $k! \leq |G|$, then $G$ is not simple.
Proof:
Consider the set of left cosets of $G$ by $H$:
$$G/H = \{H,g_1H,\dots,g_kH\}$$
Then $\phi: G \rightarrow S_{G/H}$ is a homomorphism where we define:
$$\phi(g)[g_iH] = g g_iH$$
($S_{G/H}$ is the group of permutations on elements of $G/H$ and of course $S_{G/H} \cong S_k$.)
Also $\ker \phi \neq G$ because $\phi(g_1)$ is clearly not the identity in $S_{G/H}$.
Because $\phi$ is a homomorphism,
$$ G/\ker\phi \cong \text{im}\phi$$
Then $|\text{im} \phi| \leq |S_k| = k!$ and if $k! \leq |G|$ then $|\ker\phi| > 1$ and $\ker\phi$ is a non-trivial normal subgroup of $G$.
Example: If a group of order $36$ has a subgroup of index $3$ or $4$ then the group is not simple.

REPLY [2 votes]: It's almost all fine, except this line:

Then $|\text{im} \phi| \leq |S_k| = k!$ and if $k! \leq |G|$ then $|\ker\phi| > 1$ and $\ker\phi$ is a non-trivial normal subgroup of $G$.

The inequalities tell you that $|\operatorname{im} \phi| \leq |G|$, and you claim that therefore $|\ker\phi| > 1$. This isn't necessarily true - if $|\operatorname{im} \phi| = |G|$ then $|\ker\phi| = 1$. However, this situation is easily fixed - can  you see how?