Question

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?

Answer 1

$BS(1,2)^2$. 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.

Answer 2

One may modify Richard's attempt to get an example, using a method of Stallings.

Consider the Baumslag-Solitar group $BS(1,2)=\langle a,b | bab^{-1}=a^2\rangle$, with homomorphism $\phi: BS(1,2) \to H_1(BS(1,2)) = \mathbb{Z}$ given by $b\mapsto 1, a\mapsto 0$. Then $ker(\phi)=\langle b^k a b^{-k} \rangle \cong \mathbb{Z}[\frac12]$, so $ker(\phi)$ is infinitely generated. Take the double of $BS(1,2)$ along $ker(\phi)$, $Dub(BS(1,2),ker(\phi))$. This gives a group with infinite presentation $$Dub(BS(1,2),ker(\phi))=\langle b_1,b_2,a | b_1ab_1^{-1}=a^2=b_2ab_2^{-1}, b_1^kab_1^{-k}=b_2^k a b_2^{-k}\rangle.$$ This group has $H_2(Dub(BS(1,2),ker(\phi)),\mathbb{Z})$ infinitely generated, because of the Mayer-Vietoris sequence (with $\mathbb{Z}$ coefficients) $$H_2(BS(1,2))\oplus H_2(BS(1,2))\to H_2(Dub(BS(1,2),ker(\phi)))$$ $$ \to H_1(ker(\phi)) \to H_1(BS(1,2))\oplus H_1(BS(1,2)).$$ The homomorphism $\mathbb{Z}[\frac12]=H_1(ker(\phi))\to H_1(BS(1,2))\oplus H_1(BS(1,2))=\mathbb{Z}^2$ is trivial, and therefore the homomorphism $H_2(Dub(BS(1,2),ker(\phi)))\to \mathbb{Z}[\frac12]$ is onto. Also, $H_2(BS(1,2))=0$ since it has an aspherical presentation complex with $H_2=0$. This implies that $H_2(Dub(BS(1,2),ker(\phi)))=\mathbb{Z}[\frac12]$, and is therefore infinitely generated.

Now, there is an automorphism of $BS(1,2)$ fixing $b\mapsto b$ and sending $a\mapsto a^2$ (corresponding to multiplication by $2$ on the subgroup $ker(\phi)\cong \mathbb{Z}[\frac12]$). This automorphism extends to $Dub(BS(1,2), ker(\phi))$. Take the mapping torus of this automorphism gives the group presentation $$\langle b_1,b_2,a,x| b_1ab_1^{-1}=a^2=b_2ab_2^{-1}, xb_1=b_1x, xb_2=b_2x, xax^{-1}=a^2\rangle.$$ (One may check this by computing the presentation of the mapping torus in the standard way as an HNN extension, then eliminating all but finitely many of the relators). This group then is finitely presented, whose kernel of homomorphism to $\mathbb{Z}$ given by $x\mapsto 1, a, b_1, b_2\mapsto 0$ is finitely generated with infinitely generated $H_2(Dub(BS(1,2),ker(\phi)), \mathbb{Z})\cong \mathbb{Z}[\frac12]$.