From: Dr D F Holt
Subject: Re: learning group theory
Date: Mon, 1 Feb 1999 10:12:59 GMT
Newsgroups: [missing]
To: rusin@math.niu.edu
Keywords: Converse of Lagrange's Theorem implies solvable
> >I had not realized before that only solvable groups can ever satisfy
> >the converse of Lagrange's Theorem
>
> I must have missed that thread. Is it obvious? Clearly non-solvable groups
> with a simple factor near the top will violate the condition, since they
> can't contain a subgroup of small index unless the simple factors are
> themselves small, but that doesn't get us very far with non-simple non-solvable
> groups. Is there a quick argument to use here?
I thought I had a really quick argument, but it didn't quite work!
However, the result is fairly well-known as it turns out.
Looking through my old copy of Marshall Hall, "The Theory of Groups".
Theorem 9.3.3 on page 144 states "If G has a p-complement for all primes
p dividing its order, then G is solvable".
A p-complement is a subgroup of order m where |G| = p^a m, and p does not
divide m.
The proof is a fairly elementary reduction to the case where only two
primes divide |G|, which is settled by Burnside's p^a q^b theorem.
Derek.