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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?

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Richard Kent had an answer that he deleted, but I think is correct. Take the $BS(1,2)=\langle a,b|bab^{−1}=a^2\rangle$, This has a homomorphism $\phi:BS(1,2)→\mathbb{Z}$ sending $a\mapsto 0,b\mapsto 1$. The kernel has $H_1(ker(\phi),\mathbb{Z})=ker(\phi)=\mathbb{Z}[1/2]$, the dyadic integers. Crossing with $\mathbb{Z}$, one gets a group with $H_2(ker(\phi)\times\mathbb{Z},\mathbb{Z})=\mathbb{Z}[1/2]$ via the Kunneth formula. – Ian Agol Aug 18 '11 at 23:15
The problem with my answer was that my subgroup $H$ isn't finitely generated (which is what Mustafa wants). I would expect some variant would do the trick, though. – Richard Kent Aug 19 '11 at 0:45
Thanks Agol and Richard. I think I can show that if $G\H$ is $\mathbb{Z}$ (maybe also if the quotient is free) and $H$ is finitely generated then the multiplier is always a direct product of f.g. abelian groups. SO one has to look for other type of examples. – Mustafa Gokhan Benli Aug 19 '11 at 1:15
Good point, Richard - I missed the finitely generated hypothesis as well. – Ian Agol Aug 19 '11 at 2:12
up vote 6 down vote accepted

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

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That's also good. – Richard Kent Aug 19 '11 at 3:43
@Ben. I am sorry but I have difficulty following the argument . Can you explicitly tell me what is the group and what is the subgroup? Thanks alot. – Mustafa Gokhan Benli Aug 19 '11 at 17:55
The group is $\mathbb Z^2\ltimes \mathbb Z[\frac12]^2$. The subgroup is $\mathbb Z\ltimes \mathbb Z[\frac12]^2$. The subgroup consists of those elements whose image in $\mathbb Z^2$ is anti-diagonal: $(n,-n)$. The group is fp because it's the square of the fp $BS(1,2)$. The subgroup is fg by 3 elements, the $1\in \mathbb Z$ and the two $1\in\mathbb Z[\frac12]$. – Ben Wieland Aug 20 '11 at 2:17

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 $H_2(Dub(BS(1,2),ker(\phi)), \mathbb{Z})\cong \mathbb{Z}[\frac12]$.

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Oh good. That's nice. – Richard Kent Aug 19 '11 at 3:37
Agol, why is the multiplier not a direct product of finitely generated abelian groups? – Mustafa Gokhan Benli Aug 19 '11 at 17:53
@Mustafa: I guess the argument I gave was incomplete. I think I've fixed it, adding some more exposition (an extra term in the Mayer-Vietoris sequence is needed). I think Ben Wieland's answer is simpler and probably easier to understand though. – Ian Agol Aug 19 '11 at 19:17

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