From: Allan Adler Subject: Re: Sylow Thm Application Date: 17 Dec 1999 03:38:36 -0500 Newsgroups: sci.math Keywords: cyclic 2-subgroups have normal complements jprognes@inst.augie.edu writes: > I'm trying to prove the following: If |G|=(2^n)m where m is odd and G > has a cyclic Sylow 2-subgroup, then G has a normal subgroup of order m. > I've been preceding by induction on n. Anchoring it isn't a problem, > and then if I assume it for all groups of order (2^n)m, and then > consider G of order 2^(n+1)m, I can do the following: G has a subgroup N > of order 2, I think you mean "index 2". > hence N is normal and has order 2^(n)m. Also, N's Sylow > 2-subgroup is cyclic, so by assumption N has a normal subgroup M of > order m. The problem, of course, is that M isn't necessarily normal in > G. I've been playing around with everything from counting elements in > conjugacy classes to composition series and automorphisms to either > prove that M is characteristic in N, or to prove directly that M is > normal in G, but nothing has given me a clear solution. Use Theorem 7.4.4 on p.253 of Gorenstein's book Finite Groups to prove that there is a nontrivial homomorphism from G onto a cyclic group of order 2. Your subgroup of index 2 is characteristic because the group of characters of G of order a power of 2 form a nontrivial cyclic 2-group which therefore has a unique element of order 2. This gives you the induction hypothesis you need to climb down to a normal subgroup of G of order m. Allan Adler ara@altdorf.ai.mit.edu **************************************************************************** * * * Disclaimer: I am a guest and *not* a member of the MIT Artificial * * Intelligence Lab. My actions and comments do not reflect * * in any way on MIT. Morever, I am nowhere near the Boston * * metropolitan area. * * * **************************************************************************** ============================================================================== From: mareg@mimosa.csv.warwick.ac.uk () Subject: Re: Sylow Thm Application Date: 20 Dec 1999 18:08:16 GMT Newsgroups: sci.math In article <3859A8B7.D88@inst.augie.edu>, jprognes@inst.augie.edu writes: >I'm trying to prove the following: If |G|=(2^n)m where m is odd and G >has a cyclic Sylow 2-subgroup, then G has a normal subgroup of order m. >I've been preceding by induction on n. Anchoring it isn't a problem, >and then if I assume it for all groups of order (2^n)m, and then >consider G of order 2^(n+1)m, I can do the following: G has a subgroup N >of order 2, hence N is normal and has order 2^(n)m. You seem to mean index 2, not order 2. But how do you know that G has a subgroup N of index 2? Proving that is the hardest part of the problem. >Also, N's Sylow >2-subgroup is cyclic, so by assumption N has a normal subgroup M of >order m. The problem, of course, is that M isn't necessarily normal in >G. Yes it is! The point is that M must be the unique normal subgroup of M of order m, because if there were another one, M', then MM' would be a subgroup of M of odd order greater than m, which is impossible. Since, for g in G, g^-1 M g is a normal subgroup of N of order N, it follows by uniqueness that g^-1 M g = M,so M is normal in G. (Another way of saying all this is that M is characteristic in N, which implies M normal in G.) But you still have to do the hard part, which is to show that G has a subgroup of index 2. There happens to be a quick and easy way of doing this, but it is not one that most people would think of without a hint. By Cayley's theorem, G permutes its own elements by multiplication, and so is a subgroup of the group of permutations of |G| points, with the property that no non-1 element of G fixes any point. An element of order 2^n must consist of a product of m cycles of length 2^n, and so it is an odd permutation. Now intersect with the alternating group to get a subgroup of G of index 2. Derek Holt.