Schur multiplier of divisible nilpotent group need not be divisible by any prime

Statement
It is possible to have a divisible nilpotent group $$G$$ (i.e., $$G$$ is a nilpotent group and it is divisible by all primes) such that the  Schur multiplier $$H_2(G;\mathbb{Z})$$ is not a divisible abelian group. In fact, we can choose an example where $$G$$ is a group of nilpotency class two and the Schur multiplier is not divisible by any prime.

Opposite facts

 * Stem extension preserves divisibility for nilpotent groups

Proof
The group $$G$$ described here is a quotient group of unitriangular matrix group:UT(3,Q) by a central subgroup isomorphic to the group of integers, which we can think of as a Z in Q inside the center, which is a copy of $$\mathbb{Q}$$. Explicitly, it is matrices of the form:

$$\{ \begin{pmatrix} 1 & a_{12} & \overline{a_{13}} \\ 0 & 1 & a_{23} \\ 0 & 0 & 1 \\\end{pmatrix} \mid a_{12},a_{23} \in \mathbb{Q}, \overline{a_{13}} \in \mathbb{Q}/\mathbb{Z} \}$$

with the matrix multiplication defined as:

$$\begin{pmatrix} 1 & a_{12} & \overline{a_{13}} \\ 0 & 1 & a_{23} \\ 0 & 0 & 1 \\\end{pmatrix}\begin{pmatrix} 1 & b_{12} & \overline{b_{13}} \\ 0 & 1 & b_{23} \\ 0 & 0 & 1 \\\end{pmatrix} = \begin{pmatrix} 1 & a_{12} + b_{12} & \overline{a_{12}b_{23}} + \overline{a_{13}} + \overline{b_{13}} \\ 0 & 1 & a_{23} + b_{23} \\ 0 & 0 & 1 \\\end{pmatrix}$$

where $$\overline{a_{12}b_{23}}$$ is understood to be the image of $$a_{12}b_{23}$$ under the quotient map $$\mathbb{Q} \to \mathbb{Q}/\mathbb{Z}$$.

The Schur multiplier of $$G$$ turns out to be isomorphic to the group of integers $$\mathbb{Z}$$, which is not divisible by any prime.