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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.

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

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