# 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.

## Proof

Further information: quotient of UT(3,Q) by a central Z

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.