From: rchapman@mpce.mq.edu.au Subject: Re: Groups isomorphism Date: Mon, 25 Jan 1999 03:25:34 GMT Newsgroups: sci.math To: jean-pierre.merx@wanadoo.fr Keywords: G/Z(G) is a square if Z cyclic, contains G' In article <36AB87CE.2BF9@wanadoo.fr>, Jean-Pierre.Merx@wanadoo.fr wrote: > Can someone help me to prove that: > > If the center Z(G) of a group G is cyclic and G/Z(G) is commutative then > there exists a group H such that HxH is isomorphic to G/Z(G). This is quite am interesting question. Consider the map f from G x G to G given by f(x, y) = xyx^{-1}y^{-1}. Then since G/Z(G) is commutative this takes values in Z = Z(G). Also as Z is the centre of G, then f induces a map from G/Z to G/Z to Z which is symmetric and bilinear. Also f is symplectic: f(x,x) is always the identity. Again it follows from the fact that Z is the centre of G that f is non-degenerate on G/Z, that is if f(x,y) is the identity for all y in G/Z then x is the identity of G/Z. Now we use standard theory of bilinear forms to show that G/Z is of the form H x H. Indeed one can take H to be a maximal isotropic subgroup of G/Z, one for which f(H,H) is trivial. Robin Chapman + "Going to the chemist in Australia Department of Mathematics, DICS - can be more exciting than going Macquarie University + to a nightclub in Wales." NSW 2109, Australia - rchapman@mpce.mq.edu.au + Howard Jacobson, http://www.maths.ex.ac.uk/~rjc/rjc.html - In the Land of Oz -----------== Posted via Deja News, The Discussion Network ==---------- http://www.dejanews.com/ Search, Read, Discuss, or Start Your Own ============================================================================== From: rchapman@mpce.mq.edu.au Subject: Re: Groups isomorphism Date: Tue, 26 Jan 1999 22:27:40 GMT Newsgroups: sci.math In article <78il1r$aa6$1@platane.wanadoo.fr>, "Jean-Pierre Merx" wrote: > Hello Robin, > > Thanks for your answer. > > I have somme difficulty to understand your message. Being french is not a > help to understand english math wording ! > > This is quite OK for me: > > >This is quite am interesting question. Consider the map f from G x G to G > >given by f(x, y) = xyx^{-1}y^{-1}. Then since G/Z(G) is commutative this > >takes values in Z = Z(G). > > I just presume that a map is just a function, and not a morphism of the > group GxG to the group Z ? Yes. > >Also as Z is the centre of G, then f induces > >a map from G/Z to G/Z to Z which is symmetric and bilinear. > > I don't know what means bilinear for groups ? We're dealing with abelian groups here. I mean f(ab,c) = f(a,c)f(b,c) and f(a,cd) = f(a,c) f(a,d). In additive notation, this terminology would be an obvious choice. > >Also f is symplectic: f(x,x) is always the identity. Again > >it follows from the fact that Z is the centre of G that f is > >non-degenerate on G/Z, that is if f(x,y) is the identity > >for all y in G/Z then x is the identity of G/Z. Now we use standard > >theory of bilinear forms to show that G/Z is of the form H x H. Indeed one > can > >take H to be a maximal isotropic subgroup of G/Z, one for which f(H,H) is > >trivial. > > I am surprised also by the fact that you do not use the assomption that Z is > cyclic. I'm summarizing a non-trivial argument here. One can prove the existence of H be mimicking the standard arguments for vector spaces while attending to some technical details not arising there. In particular one needs that the image of the map f is cyclic. This allows us to use duality arguments, e.g., the fact that G/Z and Hom (G/Z,Z) are isomorphic. Robin Chapman + "Going to the chemist in Australia Department of Mathematics, DICS - can be more exciting than going Macquarie University + to a nightclub in Wales." NSW 2109, Australia - rchapman@mpce.mq.edu.au + Howard Jacobson, http://www.maths.ex.ac.uk/~rjc/rjc.html - In the Land of Oz -----------== Posted via Deja News, The Discussion Network ==---------- http://www.dejanews.com/ Search, Read, Discuss, or Start Your Own