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

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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.

Related facts

Opposite facts

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.

References