From: Robin Chapman <rjchapman@my-deja.com>
Subject: Re: Commutators in GL_n(K)
Date: Thu, 08 Feb 2001 14:07:36 GMT
Newsgroups: sci.math
Summary: Commutator subgroup, and set of all commutators, in GL_n(K)

In article <95u81e$kpq$1@nnrp1.deja.com>,
  azmi_tamid@my-deja.com wrote:
> Let K be a field and G=GL_n(K) the group of invertible n by n matrices
> over K.
>
> If g in G is a commutator i.e g=aba^(-1)b^(-1) then det(g)=1 so g is in
> SL_n(K).
>
> Conversely if det(g) = 1 then can we write g as a commutator in G ?

Not if n = 2 and |K| = 2. But I can't see any other obvious
exceptions.....

--
Robin Chapman
 http://www.maths.ex.ac.uk/~rjc/rjc.html
  "His mind has been corrupted by colours, sounds and shapes."
    The League of Gentlemen


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==============================================================================

From: mareg@mimosa.csv.warwick.ac.uk ()
Subject: Re: Commutators in GL_n(K)
Date: 9 Feb 2001 19:09:12 GMT
Newsgroups: sci.math

In article <960ena$hnk$1@nnrp1.deja.com>,
	tim_brooks@my-deja.com writes:
>Is there a simple proof that SL_n(K) is equal to the
>commutator subgroup of GL_n(K), (except when n=|K|=2) or a reference
>for such proof ?

There is a proof in Rotman's book "An introduction to the Theory of Groups".
(It is Theorem 8.24 in the Third Edition.)

The idea is as follows. Define a transvection to be an element of SL_n(K)
that fixes a hyperplane pointwise. Show that all transvections are conjugate
in GL_n(K) - that is not hard. Then show that SL_n(K) is generated by
transvections - again not too hard. Finally express some trnasvection as
a commutator, and you are done, because the commutator subgroup us normal,
and so if it contains one transvection then it contains them all.

Derek Holt.
==============================================================================

From: mareg@mimosa.csv.warwick.ac.uk ()
Subject: Re: Commutators in GL_n(K)
Date: 10 Feb 2001 16:09:38 GMT
Newsgroups: sci.math

In article <3A8416A9.5F25C3D0@jps.net>,
	alperin@jps.net writes:
>Questions of this kind are considered in a recent preprint by U. Rehmann
>and N. Gordeeva, available from the preprint server at Bielefeld
>University, Germany.
>
>Roger  Alperin

I just looked this up. In fact the answer to the original question, of
whether every element of SL_n(K) is a commutator in GL_n(K) is yes, except
when n=|K|=2. This was proved by R.C.Thomson in 1961, in
"Commutators in the special and general linear groups", Trans. Amer. Math.
Soc. 101 (1961), 16-23. There are some cases however when certain
scalar matrices are not commutators within the subgroup SL_n(K).

For simple groups, it is known that every element is a commutator in the
alternating groups, the sporadic groups and in finite simple Lie groups
over fields with at least 9 elements.

Derek Holt.