Simple group's order divides n!/2

**Giraffro** · March 21st 2010, 02:14 PM

If $G$ is a finite simple group and $H$ is a proper, non-trivial subgroup, then $|G|$ divides $\frac{[G:H]!}{2}$ .

What I have so far:

Define $\forall g,h \in G, (Hh)(g)\phi := Hhg$ . Then $\phi : G \rightarrow \text{Sym}(\cos(G:H))$ is a well-defined group homomorphism with kernel $\text{core}_G(H) = \{e\}$ . Therefore by the First Isomorphism Theorem, $|G| \, \mid \, |\text{Sym}(\cos(G:H))| = |\cos(G:H)|! = [G:H]!$ . However I'm stuck at this point. I think I need to prove that the image of $\phi$ consists of even permutations on $\cos(G:H)$ , but I can't seem to find a way to prove that. The image must be simple as $\phi$ is injective, but I can't see how that helps.

Any help would be greatly appreciated. Thanks!

**Giraffro** · March 21st 2010, 03:18 PM

Never mind, found a solution myself.

Assume $\Im(\phi) \not \subseteq \text{Alt}(\cos(G:H))$ , so $\exists x \in G$ such that $(x)\phi \not \in \text{Alt}(\cos(G:H))$ . Let $g \in G$ be given such that $(g)\phi \not \in \text{Alt}(\cos(G:H))$ . Then $(g)\phi (x)\phi \in \Im(\phi) \cap \text{Alt}(\cos(G:H))$ so $(g)\phi \in (\Im(\phi) \cap \text{Alt}(\cos(G:H))(x)\phi^{-1}$ . Hence $\Im(\phi) \backslash \text{Alt}(\cos(G:H)) \subseteq (\Im(\phi) \cap \text{Alt}(\cos(G:H)))(x)\phi^{-1}$ and since $(x)\phi \not \in \text{Alt}(\cos(G:H))$ it follows that $[\Im(\phi) : \Im(\phi) \cap \text{Alt}(\cos(G:H))] = 2$ . However as $G$ is simple and $\phi$ is injective, $\Im(\phi)$ is simple. Therefore $\Im(\phi) \cap \text{Alt}(\cos(G:H)) \lhd \Im(\phi)$ , which implies that $|G| = |\Im(\phi)| = 2$ , which is a contradiction as we assumed $G$ has a proper, non-trivial subgroup.

Therefore $\Im(\phi) \subseteq \text{Alt}(\cos(G:H))$ and so $|G| = |\Im(\phi)| \, \mid \, |\text{Alt}(\cos(G:H))| = \frac{|\cos(G:H)|!}{2} = \frac{[G:H]!}{2}$ .

**tonio** · March 21st 2010, 07:09 PM

Originally Posted by Giraffro

If $G$ is a finite simple group and $H$ is a proper, non-trivial subgroup, then $|G|$ divides $\frac{[G:H]!}{2}$ .

What I have so far:

Define $\forall g,h \in G, (Hh)(g)\phi := Hhg$ . Then $\phi : G \rightarrow \text{Sym}(\cos(G:H))$ is a well-defined group homomorphism with kernel $\text{core}_G(H) = \{e\}$ . Therefore by the First Isomorphism Theorem, $|G| \, \mid \, |\text{Sym}(\cos(G:H))| = |\cos(G:H)|! = [G:H]!$ . However I'm stuck at this point. I think I need to prove that the image of $\phi$ consists of even permutations on $\cos(G:H)$ , but I can't seem to find a way to prove that. The image must be simple as $\phi$ is injective, but I can't see how that helps.

Any help would be greatly appreciated. Thanks!

So far so good: you've proved that $G$ is embedded in what you denote $Sym(cos(G:H))\cong S_n\,,\,\,n=[G:H]$ . Now, as any other subgroup of $S_n$ , $G$ (or,

in fact, its isomorphic image in $S_n$ ) is either contained on $A_n$ or else exactly half its elements are odd permutations and the other half are, naturally, even

ones, and thus $|G\cap A_n|=\frac{|G|}{2}$ ... but then $G\cap A_n$ is a subgroup of $G$ of index 2, and this cannot be...

Tonio

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