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I'm having trouble to prove that if $G$ is a finite simple group and $H$ is a subgroup of finite index, then $|G|$ divides $\frac{k!}{2}$, where $k$ is the index of H.

I've prove that in this conditions $G$ is isomorphic to a subgroup $F$ of the symmetric group of $k$, $S_k$. As we are trying to prove that $|G| = |F|$ divides $\frac{k!}{2}$ I think that I need to show that $F \leq A_k$ where $A_k$ is the alternating group, but I don't get to see clearly why this is true.

Also, I saw a similar question here but don't completely understand the answer. I would appreciate any help.

Shaun
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alucard hellsing
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2 Answers2

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You need to assume $H\neq G$, of course... Also, the statement is false for $G=C_2$ and $H=\{e\}$. Then $k=2$, but $|G|$ does not divide $\frac{2!}{2}=1$. So assume $G\neq C_2$.

Let $G$ act on the cosets of $H$, which gives us a homomorphism $\varphi\colon G\to S_k$; the kernel is normal in $G$, but cannot equal $G$ because $H\neq G$ and the action is transitive. Hence, $G$ is isomorphic to a subgroup of $S_k$.

Now, $\varphi(G)\cap A_k$ is normal in $\varphi(G)$, since $A_k$ is normal in $S_k$. Thus, either $\varphi(G)\cap A_k=\varphi(G)$, or else $\varphi(G)\cap A_k=\{e\}$.

In the latter case, let $g\in G$, $g\neq e$. Then $\varphi(g^2)=e$, since $\varphi(g)^2\in A_k\cap \varphi(G)$. Thus, $\varphi(G)$ has exponent $2$ and is abelian, hence being simple you must have $G$ cyclic of order $2$, contradicting our working assumption that $G\neq C_2$. Thus, $\varphi(G)\cap A_k$ cannot be trivial, and so $\varphi(G)\cap A_k=\varphi(G)$, proving that $\varphi(G)\subseteq A_k$. Thus, $|G|=|\varphi(G)|$ divides $|A_k| = \frac{k!}{2}$, as claimed.

Arturo Magidin
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Hint: Assume $|G|\gt2$ and $k\gt1$. Let $G$ act by left multiplication on the set of cosets $G/H$. Then we get a morphism $\varphi:G\to S_k$. Furthermore, $\varphi(G)\cong G$, since the kernel has to be trivial. Now $A_k\cap\varphi(G)$ is normal in $\varphi(G)$. From there it takes a little more work to get that $\varphi(G)\leqslant A_k$.