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Does every group with a finite classifying space have finitely generated center?

Remarks:

  1. If $G$ is a finitely generated group with infinitely generated center $Z(G)$, then the quotient $G/Z(G)$ is not finitely presented (as follows from a result of B.H Newmann).

  2. Finite classifying space means that the group is the fundamental group of a finite aspherical cell complex.

  3. I suspect the above question is a well-known open problem, but cannot find it stated in the literature, so a reference would be appreciated.

  4. Alperin-Shalen (Inventiones, 1982) showed that the answer is yes for every subgroup of $GL_n(K)$ where $n>0$ and $K$ is a field of characteristic zero.

  5. The answer is also yes for elementary amenable groups. (I know a proof, but have no reference).

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@Igor, check Geoff Mess' paper on Seifert conjecture, he had some comments on this question (I forgot what he said though) and I do not have access to his preprint right now. –  Misha Aug 5 '12 at 4:19
    
@Misha, I was unable to find a copy online (and it might not exist given that it was written in 1988). –  Igor Belegradek Aug 5 '12 at 11:33
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btw, is there a known $F_\infty$ group with infinitely generated center? Recall that $F_\infty$ means there's a $K(G,1)$ finite in each dimension, so it is weaker the existence of a finite $K(G,1)$. (Abels-Brown found $F_n$ groups with infinitely generated center for all $n$, where $F_n$ means: there exists $K(G,1)$ finite up to dimension $n$.) –  YCor Aug 5 '12 at 11:52
    
@Yves, I don't know such examples. The above results in 4-5 hold for groups of type FP, so assuming type F (like I do) is an overkill, but type F is what I need for an application. –  Igor Belegradek Aug 5 '12 at 12:28
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1 Answer

The answer to another question implies that there is a finitely presented group which has $\mathbb{Q}$ as its center (or any recursively presentable abelian group). However, you'd have to go through the paper to see if Houcine's proof could produce a group with a finite complex (seems unlikely).

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Thanks for pointing this out! I missed the new answer to that other question. It does seem unlikely that you get a group of finite cohomological dimension, let alone of type F, but I will look into the matter. –  Igor Belegradek Sep 19 '12 at 10:25
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