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
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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