# simple group with finite index, deriving a precise estimate

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!

• 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... – DonAntonio Feb 17 '19 at 17:45

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 $$|G|\leq\frac{n!}{2}$$.