Group cohomology of Q/Z 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 $0 \to \mathbb{Z} \to \mathbb{Q} \to \mathbb{Q}/\mathbb{Z} \to 0$?
 A: 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.
A: Short answer:
$$
H^i(\mathbb{Q}/\mathbb{Z}) = 
\begin{cases}
\mathbb{Z}, & i = 0,\\
0, & i \equiv 1 \mod 2,\\
A, & i\equiv 0 \mod 2, i>1,
\end{cases}$$
where $A$ fits into a short exact sequence $\mathbb{Z}\to A\to H^2(\mathbb{Q})$.
Explanation:
We can use the short exact sequence to compute the cohomology of $\mathbb{Q}$ from the cohomology of $\mathbb{Q}/\mathbb{Z}$ and $\mathbb{Z}$.
If we look at the spectral sequence we see that only the first two rows are non-zero since the $(p, q)$ entry is $H^p(\mathbb{Q}/\mathbb{Z}; H^q(\mathbb{Z}; \mathbb{Z}))$ which is non-zero only if $H^q(\mathbb{Z})\neq 0$ only if $q = 0, 1$.
Thus we end up with a long exact sequence of cohomology groups
$$0 \to H^1(\mathbb{Q}/\mathbb{Z})\to H^1(\mathbb{Q})\to H^0(\mathbb{Q}/\mathbb{Z})\to \dotsb \to H^i(\mathbb{Q})\to H^{i-1}(\mathbb{Q}/\mathbb{Z}) \to H^{i+1}(\mathbb{Q}/\mathbb{Z})\to H^{i+1}(\mathbb{Q})\to\dotsb.$$
Now $\mathbb{Q}$ has cohomological dimension 2. So we obtain $H^i(\mathbb{Q}/\mathbb{Z}) = H^{i+2}(\mathbb{Q}/\mathbb{Z})$ for $i>2$.
For the low dimensional cases since $\mathbb{Q}/\mathbb{Z}$ is torsion we see $H^1(\mathbb{Q}/\mathbb{Z})$ vanishes.
Also $H^1(\mathbb{Q})$ vanishes so we get a short exact sequence $0\to H^0(\mathbb{Q}/\mathbb{Z})\to H^2(\mathbb{Q}/\mathbb{Z})\to H^2(\mathbb{Q})\to 0$. Hence the even terms.
We also see that a portion of the long exact sequence is:
$$\dotsb\to H^{1}(\mathbb{Q}/\mathbb{Z}) \to H^{3}(\mathbb{Q}/\mathbb{Z})\to H^{3}(\mathbb{Q})\to \dotsb.$$
So $H^3(\mathbb{Q}/\mathbb{Z})$ vanishes.
Unfortunately, I do not understand $H^2(\mathbb{Q})$, let alone an extension of it by $\mathbb{Z}$. I believe the universal coefficient theorem shows that $H^2(\mathbb{Q})\cong \operatorname{Ext}(\mathbb{Q}, \mathbb{Z})$, which is described in some detail on page 5 of the notes Boardman - Some common Tor and Ext groups.
