# Proof that $G$, a simple group is isomorphic to a subgroup of ${S_X}$

Let $G$ be a group acting on a set $X$. Let $g\in G$. A map $\theta_{g}:X \rightarrow X$ is defined by setting $\theta_{g}(x)=g\cdot x$ for all $x\in X$. Let $S_x{}$ be the group of permutations of $X$

The map $f:G \rightarrow S_{x}$ given by $f(g)=\theta_{g}$ is a homomorphism.

Part 1) asks: Suppose now that $G$ is simple. Show that either $g\cdot x =x$ for all $X\in X$, or that $G$ is isomorphic to a subgroup of $S_X$. - This part I believe I have done fine using the homomorphism theorem. It is part 2) that I need help with.

2) Now let $G$ be a simple group, $H$ a subgroup of index $n>1$ in $G$. By applying part 1) to the action of $G$ by left multiplication on the set $X$ of left $H$-cosets, prove that $G$ is isomorphic to a subgroup of $S_{X}$ and hence that $|G|$ divides $n!$.

I can make observations such as from the Homomorphism theorem and Lagrange, |im$\theta_g |$ divides |H| and $|G|=$|im$\theta_g|$ since ther kernel is trivial. And $|X|=n$ but don't quite now how to put it all together for a proof involving $X$ and left cosets.

$G$ is simple and for $g\in G\setminus H$ you have $gH\ne H$. By the first part $G$ is isomorphic to a subgroup of $S_X$. Therefore $|G|$ divides $|S_X|=|X|!=n!$
• Thanks for your help! How does '$G$ is simple and for $g∈G∖H$ you have $gH≠H$' relate to '$G$ is isomorphic to a subgroup of $S_X$'?. I am told to apply part 1) specifically to the action of $G$ by left multiplication on the set $X$ of left $H$ cosets to prove that $G$ is isomorphic to a subgroup of $S_X$ but I can't see how you did that in your answer, could you please explain? – harry55 Feb 13 '17 at 16:53
• In the first part you show that that if $G$ is simple then either $gx=x$ for all $x\in X$ or $G$ is isomorphic to a subgroup in $S_X$. The fact $gH\ne H$ rules out '$gx=x$ for all $x\in X$' so we are left with $G$ is isomorphic to a subgroup in $S_X$. – Robert Chamberlain Feb 13 '17 at 18:50