From: Fred W. Helenius <fredh@ix.netcom.com>
Subject: Re: Non-long winded proof that groups of order p^2 are Abelian?
Date: Tue, 08 Aug 2000 23:48:16 -0400
Newsgroups: sci.math
Summary: [missing]

lemma_one@my-deja.com wrote:

>Yesterday the professor covered a proof on the board that all groups of
>order p^2 (p prime) are Abelian. He spent like 35-40 minutes covering it
>on the board.
[snip]
>So is there not too long winded proof that all groups of order p^2 are
>Abelian?

Here's a terse (possibly too terse) proof.  I'll assume you're
familiar with centers, normalizers, and conjugacy classes.


Lemma.  A group of order p^n (p prime, n>0) has a nontrivial center.

Proof.  The size of the conjugacy class of any element x is the order
of the group divided by the order of the normalizer of x.  If x is
not in the center, the normalizer of x is smaller than the group, so
the size of the conjugacy class is a power (greater than 1) of p.  If
the center were trivial, the conjugacy class of the identity would
contain one element and all of the other conjugacy classes would have
size divisible by p.  The total of all the sizes would thus be one
more than a multiple of p.  Since the total is actually a multiple
of p (namely, p^n), it is impossible for the center to be trivial.


Theorem.  A group of order p^2 (p prime) is abelian.

Proof.  From the lemma, the center is nontrivial, and so has order
at least p.  Assume there is some element x not in the center.  The
normalizer of x contains the center and x itself, and so has at
least p+1 elements.  Since the order of the normalizer must divide
p^2, it must be p^2.  That means x is in the center, contradicting
the assumption.  Thus every element is in the center, and the
group is abelian.

-- 
Fred W. Helenius	<fredh@ix.netcom.com>