TITLE: Homology groups of divisible and powered (nilpotent) groups
QUESTION [5 upvotes]: (1) Suppose $\pi$ is a set of primes and $G$ is a $\pi$-divisible nilpotent group, i.e., for any $g \in G$ and $p \in \pi$, there exists $x \in G$ such that $x^p = g$. Is it necessary that all the homology groups of $G$ for the trivial group action with coefficients in $\mathbb{Z}$ are also $\pi$-divisible groups? I am in particular interested in the second homology group $H_2(G;\mathbb{Z})$, which is the Schur multiplier of $G$. Note that the result is true for $H_1(G;\mathbb{Z})$, which is the abelianization of $G$.
I believe that the following example shows that the result is false for $H_2(G;\mathbb{Z})$, but I'm not sure whether I am applying results correctly. The example group $G$ that I have in mind is defined as follows. Let $UT(3,\mathbb{Q})$ denote the group of $3 \times 3$ upper triangular matrices over the rational numbers that have 1s on the diagonal. Let $Z$ be an infinite cyclic subgroup of $UT(3,\mathbb{Q})$ generated by any non-identity element in its center. Let $G = UT(3,\mathbb{Q})/Z$. Then, $G$ is divisible by all primes, on account of being a quotient of $UT(3,\mathbb{Q})$, which is divisible by all primes. $G$ admits $UT(3,\mathbb{Q})$ as a stem extension group and the base of the extension is $Z$, isomorphic to $\mathbb{Z}$, which is not divisible by any prime. The base of any stem extension is a quotient of the Schur multiplier, so the Schur multiplier (which I haven't computed precisely) should not be divisible by any prime.
(2) Suppose $\pi$ is a set of primes and $G$ is a $\pi$-powered nilpotent group, i.e., for any $g \in G$ and $p \in \pi$, there exists unique $x \in G$ such that $x^p = g$. Is it necessary that all the homology groups of $G$ for the trivial group action with coefficients in $\mathbb{Z}$ are also $\pi$-powered groups? I am in particular interested in the second homology group $H_2(G;\mathbb{Z})$, which is the Schur multiplier of $G$. Note that the result is true for $H_1(G;\mathbb{Z})$, which is the abelianization of $G$ (the proof of this uses that $G$ is nilpotent).
Can anything be said in general, in the non-nilpotent case?

REPLY [3 votes]: For $G=UT(3,\mathbb{Q})/Z$, I computed $H_2(G,\mathbb{Z})=\mathbb{Z}$, which confirms that $G$ is a counterexample to (1). (This doesn't contradict Adam's answer, since $G\supseteq  \mathbb{Q}/\mathbb{Z}$ isn't uniquely divisible)
The essential steps in my computation are (coefficients are always $\mathbb{Z}$ which are suppressed): 

The LHS spectral sequence of the central extension $\mathbb{Q}/\mathbb{Z} \hookrightarrow G \twoheadrightarrow \mathbb{Q}^2$ yields $H_3(G)=0$ and $H_2(G) = E^3_{2,0} \le H_2(\mathbb{Q}^2)=\mathbb{Q}$. 
The LHS of the central extension $\mathbb{Q} \hookrightarrow UT(3,\mathbb{Q}) \twoheadrightarrow \mathbb{Q}^2$ yields $H_2(UT(3,\mathbb{Q}))=\mathbb{Q}^2$ 
The LHS of the central extension $\mathbb{Z} \hookrightarrow UT(3,\mathbb{Q}) \twoheadrightarrow G$ has 
$$E^2=\begin{array}{cccc}
|0 & & & & \newline 
|\mathbb{Z} & \mathbb{Q}^2 & & \newline 
|\underline{\mathbb{Z}} & \underline{\mathbb{Q}^2} & \underline{H_2(G)} & \underline{0} \newline 
\end{array}$$ 
Since $UT(3,\mathbb{Q}) \twoheadrightarrow G$ induces an isomorphism on the abelianization, we have $E^3_{01}=0$. Hence $d^2: H_2(G)\to \mathbb{Z}$ is epi. For $K = \ker d^2$ there is a s.e. sequence $\mathbb{Q}^2 \hookrightarrow H_2(UT(3,\mathbb{Q})) \twoheadrightarrow K$. Tensoring yields $K\otimes \mathbb{Q}=0$. If $K \neq 0$ there is $\mathbb{Z} \hookrightarrow K$ (since $K\le H_2(G)\le \mathbb{Q})$ and tensoring gives the contradiction $\mathbb{Q} \hookrightarrow K\otimes \mathbb{Q}=0$. Hence $K=0$ and $H_2(G)\cong \mathbb{Z}$ as claimed.