TITLE: Group cohomology of Q/Z
QUESTION [5 upvotes]: What is the group cohomology $H^{d}(\mathbb{Q}/\mathbb{Z}, \mathbb{Z})$ with trivial action?
Can it be computed succinctly using the short exact sequence $0 \to \mathbb{Z} \to \mathbb{Q} \to \mathbb{Q}/\mathbb{Z} \to 0$?

REPLY [9 votes]: This can be computed using standard tools of algebraic topology.  The idea is first to compute the homology groups  $H_*({\mathbb Q}/{\mathbb Z},{\mathbb Z})$ and then use the universal coefficient theorem to get the cohomology groups $H^*({\mathbb Q}/{\mathbb Z},{\mathbb Z})$.
The group ${\mathbb Q}/{\mathbb Z}$ is the union of an infinite increasing sequence of finite cyclic subgroups $G_n$. For example one can take $G_n$ to be cyclic of order $n!$.  An Eilenberg-MacLane space $K({\mathbb Q}/{\mathbb Z},1)$ can be constructed as an increasing union of $K(G_n,1)$'s  since homotopy groups commute with direct limits. Homology groups also commute with direct limits so $H_i({\mathbb Q}/{\mathbb Z},{\mathbb Z})$ is the direct limit of the groups $H_i(G_n,{\mathbb Z})$.  The latter is $G_n$ for odd $i$ and $0$ for even $i>0$. Thus $H_i({\mathbb Q}/{\mathbb Z},{\mathbb Z})$ is ${\mathbb Q}/{\mathbb Z}$ for odd $i$ and $0$ for even $i>0$.
To apply the universal coefficient theorem we have $\operatorname{Hom}({\mathbb Q}/{\mathbb Z},{\mathbb Z})=0$ and hence $H^i({\mathbb Q}/{\mathbb Z},{\mathbb Z})$ is $\operatorname{Ext}({\mathbb Q}/{\mathbb Z},{\mathbb Z})$ for even $i>0$ and $0$ for odd $i$.
Computing $\operatorname{Ext}({\mathbb Q}/{\mathbb Z},{\mathbb Z})$ takes a little work so let me just refer to my algebraic topology book where this is done in Section 3F, specifically on page 318 a few lines above Proposition 3F.12. The answer is that $\operatorname{Ext}({\mathbb Q}/{\mathbb Z},{\mathbb Z})$ is the direct product (not direct sum) of the additive groups of $p$-adic integers for all primes $p$.
Comparing this calculation with Robert Kropholler's answer, his group $A$ is the product of the groups of $p$-adic integers and his map ${\mathbb Z}\to A$ is the obvious diagonal map to this product.
In particular the cohomology groups of ${\mathbb Q}/{\mathbb Z}$ are uncountable in each even positive dimension. This is an instance of the general fact that for an abelian group $A$ which is not finitely generated either $\operatorname{Hom}(A,{\mathbb Z})$ or $\operatorname{Ext}(A,{\mathbb Z})$ is uncountable, which is the Proposition 3F.12 mentioned above.