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Finite Groups and Normal Subgroups

Date: 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/ 
Associated Topics:
College Modern Algebra

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