[FOM] correction

[Andreas Blass]

Rob Arthan has pointed out that I erred in writing (on August 1):

        Something similar, but not quite so bad, happens with (Q-{0},*),
which is the product of a cyclic group of order 2 and a
countable-dimensional vector space over the rationals.

In fact, this group is the direct sum of the cyclic group of order 2
generated by -1 and countably many cyclic groups of infinite order, each
generated by a prime.  Ignoring the factor of order 2, and thus working
with the multiplicative group of positive rationals, we just have a free
abelian group on a countable infinity of generators.  So a homomorphism of
this group into itself is obtained by assigning arbitrary images to the
primes and then extending to all positive rationals multiplicatively.
That doesn't yet take *almost* homomorphisms into account.  Fortunately,
the rest of what I wrote about this case was vague enough to be correct
despite my error:

You do get something of cardinality only the continuum, but it doesn't
look much like the reals.

In addition to correcting my error, Rob has suggested a further
generalization of Schanuel's construction and has provided a calculation
(which I haven't yet had time to study carefully) of the resulting groups
when you start with finitely generated abelian groups.

Andreas Blass