--- Michael Kleber <michael.kleber@gmail.com> wrote:
Mike Speciner wrote:
Am I missing something? According to mathworld, a free abelian group is an abelian group with no torsion, seemingly meaning that the only element with finite order is the identity (in this case the zero function). Clearly here, no function other than zero has finite order.
Mathword does badly on this one -- its definition is just fine for finitely generated groups, but not in general.
An abelian group G is "free abelian" if there exists a set of generators {g_i, i in I}, for some index set I (not necessarily countable, of course), such that every element can be written as a finite sum of generators and their inverses in exactly one way.
The group P in question is precisely a counterexample to the hypothesis "That just means the group is torsion-free, right?"
--Michael Kleber
Well, not quite. In order to assert that P is a counterexample, it must be shown that P is not free. But here is a simpler counterexample: the additive group Q of rational numbers. Suppose {g1, g2, ... } is a countable set of free generators. Then (1/2)g1 is a finite sum of generators with integer coefficients. But then g1 = (1/2)g1 + (1/2)g1 is an alternative expression for g1 in which all generators appear with even coefficient. So g1 has two different expressions, and Q is not free. Gene __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com