Let $H$ be a subgroup of $G$ with $[G:H]=n$ and $G$ finite. Let $G$ act on left cosets of $H$ by left multiplication. Let $N$ be the kernel of the action. Show that $[G:N]$ divides $n!$.
Ok so I have a theorem that says that if $G$ is simple then it embeds into $S_n$. So if $G$ is simple than I'm done because $G$ divides $n!$ and thus $N$ divides $n!$ and hence $[G:N]$ divides $n!$.
Now I also know that $N$ is the largest normal subgroup of $G$ contained in $H$. Thus if $G$ isn't simple then it contains some non-trivial normal subgroup $K$ and thus $K\cap H$ is normal in $H$ and by the 2nd Iso Theorem $H/K\cap H\cong KH/H$…
As you can see I'm not sure where to go from here. I was hoping to prove that $K\cap H$ wasn't trivial and thus that $N$ couldn't be trivial since it must contain $K\cap H$, but I'm not sure I can, and even if I could I'm not sure that would help me in showing that $[G:N]$ is small enough to divide $n!$.
Can anyone give some guidance? Thanks.
Best Answer
Let $S_{G:H}$ be the set of permutations of $G:H$. Consider $\phi:(G:N)\rightarrow S_{G:H}$ that sends $xN$ to $T_x$ (Where $T_x$ is the permutation of $G:H$ that sends $aH$ to $xaH$ ). It is easy to verify that $\phi$ is an injective group homomorphism. Thus, $G:N$ is isomorphic to a subgrop of $S_{G:H}$. Using Lagrange's theorem, we deduce that $|G:N|$ divides $|S_{G:H}|=n!$