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        riddles >> putnam exam (pure math) >> Given subgroup H maximal non-normal in some group?
        
(Message started by: ecoist on Apr 12^th, 2007, 10:58am)

Title: Given subgroup H maximal non-normal in some group?
Post by ecoist on Apr 12^th, 2007, 10:58am

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.

Title: Re: Given subgroup H maximal non-normal in some gr
Post by Eigenray on Apr 12^th, 2007, 4:29pm

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 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/rtimes.gif 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 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 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/rtimes.gif H, where H acts on S=H/N by conjugation, i.e., using

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/phi.gif: H -> S -> Inn(S) -> Aut(S).

Since S is non-abelian, this composition is non-trivial, so H is non-normal. Suppose H http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif K http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif 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 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/phi.gif(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.