Re: Product of commutators
- From: magidin@xxxxxxxxxxxxxxxxx (Arturo Magidin)
- Date: Fri, 15 Dec 2006 17:27:57 +0000 (UTC)
In article <4582d85f$0$22400$4fafbaef@xxxxxxxxxxxxxxxxxxx>,
bluelabel <bluelabel.invalid@xxxxxxxx> wrote:
Let G be a group.
Could you give me a trivial example of a product of commutators in G which
is not a commutator?
Don't know what you mean by "trivial". The smallest example is a group
G of order 96, according to Rotman's "An Introduction to the Theory of
Groups", 4th edition (pp. 34). Here is a different example he gives,
due to P.J. Cassidy:
Let k[x,y] be the ring of polynomials in two variables over the field
k. Definte G to be the set of all 3 x 3 matrices of the form
( 1 f(x) h(x,y) )
( 0 1 g(y) )
( 0 0 1 )
where f(x) is a polynomial in k[x], g(y) is a polynomial in k[y], and
h(x,y) is a polynomial in k[x,y]. Write such an element as (f(x),g(y),h(x,y)).
Then [G,G] consists of all matrices for which f(x)=g(y)=0.
If (0,0,h) is a commutator, then there are polynomial f1, f2 in k[x]
and g1,g2 in k[y] such that h(x,y) = f1(x)g2(y) - f2(x)g1(y).
However, if h(x,y) = x^2 + xy + y^2, then there is no decomposition as
in the previous paragraph; so (0,0,h) is an element of [G,G] but not a
commutator, proving that there are products of commutators in G that
are not themselves commutators.
--
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"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
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Arturo Magidin
magidin-at-member-ams-org
.
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