Associated Topics || Dr. Math Home || Search Dr. Math

### Proof That G Is Abelian

```Date: 03/05/2003 at 19:53:01
From: Jim
Subject: Abelian Groups

Let H be a subgroup of G that is different from G and let x*y=y*x
for all x and y in G minus H. Prove that G is Abelian.
```

```
Date: 03/06/2003 at 04:46:58
From: Doctor Jacques
Subject: Re: Abelian Groups

Hi Jim,

As H is a proper subgroup, there exists an element c in (G\H). This
element will be kept fixed in what follows.

We will first prove that any element of H commutes with c.

Let a be an element of H.

As H is a subgroup, and c is not in H, neither a*c nor c^(-1) are in
H.

By hypothesis, we have:

(a*c)*c^(-1) = c^(-1)*(a*c)
a = c^(-1)*(a*c)

and you should be able to conclude that c*a = a*c.

It remains to prove that any two elements of H commute. Let a and b
be two elements of H. We can write:

a*b = (a*c)*(c^(-1)*b)

and neither (a*c) not (c^(-1)*b) belong to H.

This should allow you to prove that a*b = b*a, using the previous
result and a similar technique.

As this exhausts all possibilities, it proves the proposition.

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: 03/10/2003 at 12:38:13
From: Jim
Subject: Thank you (Abelian Groups)

Thank you, Dr. Jacques, for your time and quick response.
The help was right on.
Jim
```
Associated Topics:
College Modern Algebra

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

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/