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Topic: Given subgroup H maximal non-normal in some group? (Read 759 times) |
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ecoist
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Given subgroup H maximal non-normal in some group?
« on: Apr 12th, 2007, 10:58am » |
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Let H be a finite group of order greater than 1. Show that there exists a finite group G in which H ia a maximal, non-normal subgroup of G.
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Eigenray
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Re: Given subgroup H maximal non-normal in some gr
« Reply #1 on: Apr 12th, 2007, 4:29pm » |
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Let N be a maximal normal subgroup of H (possibly trivial). Then S = H/N is simple.
(1) If S is abelian, say S = Zp for some prime p. Pick a prime q =1 mod p, and let G = Zq  H, using
H -> S -> Aut(Zq),
where the first map is projection, and the second takes S to the unique subgroup of Aut(Zq) = Zq* of order p. Then the composition is non-trival, so H is non-normal, and since it has prime index, it's maximal.
(2) If S is non-abelian, then let G = S H, where H acts on S=H/N by conjugation, i.e., using
 : H -> S -> Inn(S) -> Aut(S).
Since S is non-abelian, this composition is non-trivial, so H is non-normal. Suppose H  K G.
Since K contains H, we have K = T x H as sets, for some subset T of S. But then T is actually a subgroup of S, and in fact normal: K is normalized by H, so T is invariant under (H) = Inn(S). Since S is simple, we must have T=1 or S, i.e., K=H or G, and so H is maximal.
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