Uniquely divisible groups.

[Tyler Lawson]

2 odd questions about divisible groups.

Say a group G is uniquely divisible (or a Q-group, see [1]) if for
every non-identity element g in G and integer n, there is a unique
element h such that g = h^n.

1) Is the homology H_i(G;Z) of G uniquely divisible for i > 0?

2) Related question. Uniquely divisible groups form a category and
there are _free_ objects, formed by starting with a free group and
iteratedly attaching roots to... well... everything. What are the
homology groups of these free uniquely divisible groups? Are subgroups
of free objects free?


[1] G. Baumslag, Some aspects of groups with unique roots. Acta Math
104, 217--303.

[Mike Hill]

Maybe I am just being stupid, but it seems that the homology should be
at least divisible. If we look at the Hochschild-Serre SS coming from

{1}-> [G,G] -> G -> G_{ab} -> {0},

then isn't the E_2 term already a rational vector space, since G_{ab}
is? I'm not sure how we could then get anything besides a divisible
group (and in fact, it seems that we must get simply divisible ones).

Of course, maybe I am just missing something obvious...

[Tyler Lawson]

The main problem is that the homology of a rational vector space with
coefficients in an arbitrary module isn't necessarily a rational vector
space.

For instance, let G = Q, the rational numbers, and let M be the
G-module of functions from Q to Z/p. M is a product of copies of Z/p
indexed by Q, with Q permuting the product factors. Multiplication by
p is zero on M, and hence on H_*(Q;M).

We can compute the H_1 of Q with coefficients in M as the colimit of
the homologies of (1/n)Z with coefficients in M; it's the colimit of
the (1/n)Z-fixed point sets of M along the norm maps. This colimit is
nonzero.

[Mike Hill]

Something is really odd here. It seems that your module M is
essentially the regular representation of Q reduced mod p. This
shouldn't have any higher homology. It will have some interesting H_0,
but I don't see that it's going to have much beyond that. I guess I'm
implicitly using Q as an abstract group to get around any topological
questions, but still, this is a permutation module. These don't tend to
actually have any higher homologies...

[Tyler Lawson]

It's like the regular representation, but it's a direct product instead
of a direct sum. It behaves more like an injective object than a
projective object.

[Mike Hill]

It's still a permutation representation. Those seem to always have no higher homology...

--
Mais il avait compté sans une force qui, si elle est nourrie d'abord par vanité, vainc le dégoût, le mépris, l'ennui même : c'est l'habitude.