8
$\begingroup$

I'm interested in the homology of infinite groups and especially in low-dimension integral homology.

If $G$ is a locally finite group of finite exponent, one has that also $H_*(G;\mathbb{Z})$ has finite exponent. This follows from the fact that the integral homology of a finte group is bounded by the exponent of the group itself.

Is it true that, in general, a group of finite exponent has finite exponent integral homology? Are there counterexamples?

$\endgroup$

1 Answer 1

8
$\begingroup$

The answer is no. It is shown in Olshanskii's book that the second integral homology of a free Burnside group of odd exponent sufficiently large is free abelian of infinite rank. See Corollary 31.2 of Geometry of Defining Relations in Groups page 336.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .