Finite Groups and Normal SubgroupsDate: 10/30/2004 at 15:22:12 From: Emily Subject: Finite groups and normal subgroups Let G be a finite group of order n such that G has a subgroup of order d for every positive integer d dividing n. Prove that G has a proper normal subgroup N such that G/N is Abelian. If n is prime, we're done because G is Abelian. If n is even, there is a subgroup of index 2, which is normal, and G/N has order 2, so will be Abelian. I'm not sure how to start the problem if n is odd. Perhaps I'm on the wrong track in splitting this into cases? Date: 10/31/2004 at 04:25:16 From: Doctor Jacques Subject: Re: Finite groups and normal subgroups Hi Emily, Note that there is a theorem that says that every group of odd order is solvable, and this implies the result. However, the original proof of that theorem has more than 400 pages, so this may not be the simplest way to prove the result. :) Let p be the smallest prime factor of |G| = n. By hypothesis, G contains a subgroup H of order n/p, and we have [G:H] = p. If H is normal in G, then G/H has order p and is cyclic. It is therefore enough to prove that H is normal. Note that, as [G:H] is prime, H is maximal--there are no subgroups between H and G. Let N(H) be the normalizer of H in G. N(H) is a subgroup containing H, and therefore, as H is maximal, N(H) = H or G. If N(H) = G, this means that H is normal, and we are done. Assume now that N(H) = H. Let G act by conjugation on the conjugates of H. The number of such conjugates is [G:N(H)] = [G:H] = p. This defines a homomorphism: f : G -> S_p Let K = ker f be the kernel of this homomorphism. K is the intersection of all the conjugates of H; in particular, K is a subgroup of H, and [H:K] > 1, since the conjugates of H are distinct (because H is not normal). In other words, we have: K < H < G where "<" means "is a proper subgroup of". The image of f, img f, is a subgroup of S_p of order [G:K]. We have: [G:K] = [G:H][H:K] = p[H:K] [1] As S_p, is of order p!, we may write: p[H:K] | p! [H:K] | (p-1)! where "|" means "divides". As [H:K] > 1, there is a prime q that divides [H:K], and we have: q | [H:K] | (p-1)! Now, all the prime factors of (p-1)! are less than p; in particular, q < p. Equation [1] then shows: q | [H:K] | [G:K] | n and this contradicts the fact that p is the smallest prime factor of n. Note that we have in fact proved something stronger: if p is the smallest prime dividing the order of G, then any subgroup of index p is normal in G. Does this help? Write back if you'd like to talk about this some more, or if you have any other questions. - Doctor Jacques, The Math Forum http://mathforum.org/dr.math/ Date: 10/31/2004 at 13:16:50 From: Emily Subject: Finite groups and normal subgroups Thanks so much for your help! I did not think of using solvable groups to solve this problem. I have a theorem in my book (that I can use) that states the following: The finite group G is solvable if and only if for every divisor n of |G| such that (n,|G|/n)=1, G has a subgroup of order n. Since G has a subgroup of order n for EVERY divisor of |G|, can I conclude that G is solvable? I have more information than the hypothesis requires, but it still works, right? Again, thanks for your help! Emily Date: 11/01/2004 at 04:03:28 From: Doctor Jacques Subject: Re: Finite groups and normal subgroups Hi again Emily, I assume you refer to Hall's theorem. If you do know it, the proof is indeed rather obvious, as you saw. (However, the proof of Hall's theorem itself, or at least any proof I know, is far from easy.) - Doctor Jacques, The Math Forum http://mathforum.org/dr.math/ |
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