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Math Help - Simple group's order divides n!/2

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

    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!
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  2. #2
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    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}.
    Last edited by Giraffro; March 21st 2010 at 05:00 PM. Reason: Proof was wrong
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  3. #3
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    Quote Originally Posted by Giraffro View Post
    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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