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Topic: Given subgroup H maximal nonnormal in some group? (Read 763 times) 

ecoist
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Given subgroup H maximal nonnormal in some group?
« on: Apr 12^{th}, 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, nonnormal subgroup of G.


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Eigenray
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Re: Given subgroup H maximal nonnormal in some gr
« Reply #1 on: Apr 12^{th}, 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 = Z_{p} for some prime p. Pick a prime q =1 mod p, and let G = Z_{q} H, using H > S > Aut(Z_{q}), where the first map is projection, and the second takes S to the unique subgroup of Aut(Z_{q}) = Z_{q}^{*} of order p. Then the composition is nontrival, so H is nonnormal, and since it has prime index, it's maximal. (2) If S is nonabelian, 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 nonabelian, this composition is nontrivial, so H is nonnormal. 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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