Let $G$ be a finitely presented group and $H$ a finitely generated **normal** subgroup. Is it always true that the Schur Multiplier $H\_2(H,\mathbb{Z})$ is a direct product of finitely generated abelian groups? 



$H_2(H,\mathbb{Z})$ where H is a f.g. normal subgroup of a f.p. group.
$BS(1,2)^2$. I think of $BS(1,2)=\mathbb Z\ltimes \mathbb Z[\frac12]$, where the first component acts by multiplying by $2$. Then $BS(1,2)^2=\mathbb Z^2\ltimes \mathbb Z[\frac12]^2$ has the normal subgroup $\mathbb Z\ltimes \mathbb Z[\frac12]^2$, where the first component acts by $2$ on one factor and by $\frac12$ on the other. Those factors cancel so that it acts trivially on $H\_2(\mathbb Z[\frac12]^2)=\mathbb Z[\frac12]$. The universal central extension can be realized as the matrix group over $\mathbb Z[\frac12]$ of $3\times3$ upper triangular matrices with $1$ on the extreme diagonal entries, but powers of $2$ allowed in the middle entry.


