# Thread: Simple group's order divides n!/2

1. ## Simple group's order divides n!/2

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

What I have so far:

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

Any help would be greatly appreciated. Thanks!

2. Never mind, found a solution myself.

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

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

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

What I have so far:

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

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

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

Tonio