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.