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Suppose G is a simple group, H its proper subgroup of finite index. The first part of the question was to prove G is finite, which I did by showing it is isomorphic to a subgroup of $S_n$ where $n$ is the index of H in G. I used Cayley's Theorem applied to $G$ acting on the set of left cosets $G/H$ then it becomes clear that the kernel of $f:G \rightarrow Sym(G/H)$ is trivial .

I need to know show that $\lvert G \rvert \leq n!/2$

Obviously we have $\lvert G \rvert \leq n!$, but I'm not sure how to get the stronger inequality, I suspect I have to show $G$ is isomorphic to a subgroup of $A_n$ but I'm really not that confident with this and don't know the implications of $G$ being simple and what not, so any help would be great; thanks!

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    $\begingroup$ Hint: if $\;|G|>\frac{n!}2\;$ then $\;G\cap A_n\;$ is non trivial ( I'm already identifying $\;G\;$ with its isomorphic image in $\;S_n\;$) , so... $\endgroup$ – DonAntonio Feb 17 '19 at 17:45
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So you got an injective map from $G\rightarrow S_n$. If this map is surjective $G\equiv S_n$ which contradicts $G$ is simple. So $|G|<|S_n|=n!$ and hence $|G|\leq\frac{n!}{2}$.

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  • $\begingroup$ ohhh right yeah, thanks $\endgroup$ – Displayname Feb 17 '19 at 17:49

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