Proof That G Is AbelianDate: 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 |
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