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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.

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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?

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    $\begingroup$ Yes, of course, this is a mistake running through the whole thing, where it should be $<$ not $\leq$ $\endgroup$ – Stanley Jun 1 '15 at 0:22

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