# Let $G$ be a simple group of order $n$. Let $H$ be a subgroup of $G$ of index $k$. Show that $n$ divides $k!$

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 $k!$.

• Hint: Consider the action of $G$ on the cosets of $H$ by left multiplication. What is the kernel of the corresponding homomorphism? – Tobias Kildetoft May 21 '13 at 15:59
• Try to get the points from Dennis's answer and @Tobias's neat comment. Dennise's makes light on the way. math.stackexchange.com/a/267617/8581 – mrs May 21 '13 at 16:38
• – user437309 Jul 19 '18 at 4:06

$$X:=\left\{\;gH\;;\;g\in G\;\right\}\;,\;\;G\times X\to X\;,\;\;x(gH):=(xg)H$$
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 $\,|G|\,\mid\,[G:H]!\,$ ...
• 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. – user437309 Jul 19 '18 at 3:58