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.