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 $k!$.


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 $\,|G|\,\mid\,[G:H]!\,$ ...

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    $\begingroup$ I think I saw the similar approach in Dixon's book. :+) $\endgroup$ – mrs Aug 15 '13 at 13:30
  • $\begingroup$ Is this reasoning of the fact that the kernel is contained in $H$ correct? If $x$ lies in the kernel, then $xgH=gH$ for all $g\in G$. But then $g^{-1}xg\in H$ or equivalently $x\in gHg^{-1}$ for all $g\in G$. In particular this holds for $g=1$, so $x\in H$. The kernel itself is the set $\{ghg^{-1}:g\in G, h\in H\}$. Also, why do we need that the kernel is contained in $H$? In your solution you only use simplicity, first iso theorem, and Lagrange's theorem as far as I can see. $\endgroup$ – user437309 Jul 19 '18 at 3:58

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