Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Only Two Abelian Groups

Date: 02/25/2003 at 18:56:08
From: Anonymous
Subject: College Algebra

Show that any group with order p^2, where p is a prime, is Abelian.
Show that up to isomorphism that only two such groups exist.

Let G be a group with C the center of G. C is a subgroup of G. By 
Lagrange's theorem |C| divides |G|. Thus |C|=1,p,p^2. If |C|=p^2, 
then C=G and thus G is Abelian. However I am stumped when it comes to 
ruling out |C|=1 and |C|=p and I have no clue how to find the two 
groups that are supposed to exist.

Any finite group G of prime power order has center C/=1. The proof is 
unclear to me. Therefore I am not going to rule out |C|=1 just yet. 
Even so, I still have the case of |C|=p. Any thoughts will help!


Date: 02/26/2003 at 04:09:19
From: Doctor Jacques
Subject: Re: College Algebra

Hi,

The crucial point is indeed that the center of a group of prime power 
order is not trivial. This is usually proven by using the concepts of 
group action on a set and conjugation.

Let us assume for the moment that we know that |C| <> 1, which 
implies that |C| = p.

As the center is a normal subgroup, it defines a quotient group G/C, 
which is of order p, and therefore cyclic. Let gC be a generator of 
that cyclic quotient group.

Every element x of G can be written as (g^i)*c, for some c in C and 
some integer i.

Consider two elements x and y of G.

We have:

  x = (g^i)*c_1
  y = (g^j)*c_2

with c_1 and c_2 in C.

This yields:

  xy = (g^i)*c_1*(g^j)*c_2
     = (g^(i+j))*c_1*c_2

because c_1 and c_2 are in C.

If you compute yx in the same way, you will find that yx = xy, i.e. 
any two elements commute, and C = G after all.

This shows that G is abelian; in fact any group G with center C such 
that G/C is cyclic is abelian.

There are two cases to consider:

Case 1
------
G has an element of order p^2. In this case, G is cyclic of order p^2

Case 2
------
G has no element of order p^2. In this case, all non-identity 
elements of G are of order p.

Consider one such element a, and let H = <a> be the cyclic group (of 
order p) generated by a. Let b be an element not in H, and K = <b> be 
the cyclic subgroup (of order p) generated by b. We claim that G is 
isomorphic to H x K (the direct product).

As G is abelian, we need only to prove:

(1) (H intersect K) is the trivial subgroup.
(2) Every element of G can be written as hk, with h in H and k in K.

To prove (1), assume that x <> e belongs to H and K.

As every non-identity element of a cyclic group of prime order is a 
generator, this means that x is a generator of both H and K, and that 
these subgroups are identical, which contradicts the choice of b.

To prove (2), consider the quotient G/H (as G is Abelian, we are free 
to take quotients). G/H being of order p, it is cyclic. As b is not 
in H, bH is a generator of G/H. We can therefore write any element x 
of G as:

  x = (b^i)*h     with h in H

As b^i is in K, this proves (2).

To summarize, G can only be cyclic of order p^2, or isomorphic to the 
direct product C_p x C_p, where C_p is the cyclic group of order p.

Does this help? Please feel free to write back if you want to 
discuss this further.

- Doctor Jacques, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
College Modern Algebra
College Number Theory

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/