(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?