Let $G$ be a simple group of order $n$. Let $H$ be a subgroup of $G$ of index $k$ with $H\ne G$. Show that $n$ divides $k!$.
[Math] Let $G$ be a simple group of order $n$. Let $H$ be a subgroup of $G$ of index $k$. Show that $n$ divides $k!$
abstract-algebrasimple-groups
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Best Answer
A pity such a question will remain unanswered. Define a group action
$$X:=\left\{\;gH\;;\;g\in G\;\right\}\;,\;\;G\times X\to X\;,\;\;x(gH):=(xg)H$$
Prove the above indeed is an action of $\,G\,$ on $\,X\,$ , and thus we get the induced homomorphism $\,\phi:G\to \text{Sym}(X)\,$ . The kernel of this homomorphism, also known as the core of $\,H\,$ in $\,G\,$ , is characterized as the largest normal subgroup of $\,G\,$ contained in $\,H\,$ .
But since $\,G\,$ is simple we get then that $\,\ker\phi=1\,$ , and this means we can embed $\,G\,$ into $\,\text{Sym}(X)\,$ , and this means, by Lagrange's Theorem, that $\,|G|\,\mid\,[G:H]!\,$ ...