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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 

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 

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.
Associated Topics:
College Modern Algebra

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