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$\DeclareMathOperator\vcd{vcd}\DeclareMathOperator\Aut{Aut}\DeclareMathOperator\Inn{Inn}\DeclareMathOperator\Out{Out}$Here I mean $\vcd(G)$ to be the virtual cohomological dimension of $G$. Some possible examples of what I mean by $H$ are $H=\Aut(G)$, $\Inn(G)$ or $\Out(G)$. I could extend this to other natural, more topological examples as well.

I've wondered about this and wondered if anyone actually speculated about this since the work in the 70s and 80s studying this problem where G is either a surface group (Harer, Penner, Thurston, etc.) or a free group (Culler, Vogtmann, Charney?, Bestvina, etc.). Of course we have the work of Borel, Serre and others for arithmetic groups, where G could be free abelian, or probably polycyclic-by-finite.

I would reframe and expand this question as follows: Let $G$ have $\vcd(G)$ finite so we have a finite dimensional model $X$ of $K(G,1)$. How can we construct a suitably good finite dimensional model $Y$ of $K(\Aut(G),1)$, for example?

As a side question, are there groups $G$ with $\vcd(G)$ finite but infinitely generated center $Z(G)$? What happens with $\Inn(G)=G/Z(G)$ then?

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  • $\begingroup$ I just realized that, to simplify things regarding torsion and finite index subgroups, we may as well assume G is torsion free to start off here. $\endgroup$
    – Mike
    Commented Feb 25, 2023 at 23:28
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    $\begingroup$ For the last question, i.e., are there finitely generated groups with finite vcd but non-finitely generated center, the answer is yes. You can take "Abels's groups" Ab_n, which are the groups of n-by-n upper triangular matrices over Z[1/p] (for any prime p) whose first and last diagonal entries are 1. For n>2, Ab_n is finitely generated, but the center is the copy of Z[1/p] "in the top right" so it's not fin. gen. $\endgroup$ Commented Feb 26, 2023 at 11:39
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    $\begingroup$ The related question of whether a group $G$ of type F can have a $\mathrm{Out}(G)$ (or $\mathrm{Aut}(G)$) with infinite vcd seems harder. The examples given in Moishe Kohan's answer are finite dimensional, but never finitely presented. It may be possible to use fibre products to get finitely presented examples, but higher finiteness properties seem out of reach to these techniques. Maybe there's some other kind of example known? $\endgroup$
    – HJRW
    Commented Feb 26, 2023 at 20:42
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    $\begingroup$ @HJRW: I am curious, what about Baumslag-Solitar groups? Some of them have their automorphism groups not finitely generated. Are they sufficiently big to have an infinite vcd? $\endgroup$
    – AGenevois
    Commented Feb 27, 2023 at 7:27
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    $\begingroup$ @Mike The standard example of a Baumslag-Solitar group with non-finitely generated (outer) automorphism group is $BS(2, 4)$, due to Collins and Levine in the 1980s ("Automorphisms and Hopficity of certain Baumslag-Solitar groups" Arch. Math. (1983)); Levitt gave a geometric reason for this non-finite generation in a 2007 G&T paper (link). Theorem 5.2 of this gives a positive answer to your question in the "nice" case of (Generalised) Baumslag-Solitar groups with no $BS(1, n)$ subgroups, $n>1$. $\endgroup$
    – ADL
    Commented Feb 27, 2023 at 15:36

1 Answer 1

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Sorry, it is Rips' and not Mikhailova's construction: I should not comment when I am half-asleep.

Let me start with the Rips construction.

Let $Q$ be a finitely presented group. Rips in [1] constructed $C'(1/\lambda)$-small cancellation groups $G$ (with arbibtarily large $\lambda$) and normal finitely generated subgroups $N< G$ such that $G/N\cong Q$. For $\lambda\ge 7$ the group $G$ will be hyperbolic.

A nice exposition of the Rips construction and its generalizations can be found in these two blog-posts: here and here. Actually, the Rips construction is quite flexible and one can make choices so that no defining relator of $G$ is a proper power; hence, the presentation complex of $G$ is aspherical. In particular, $G$ is torsion-free and is 2-dimensional. The subgroup $N$, therefore, is also 2-dimensional. However, the group $Q$ can be taken to have infinite virtual cohomological dimension.

We, thus, obtain a finitely generated 2-dimensional group $N$ such that $Out(N)$ contains $Q$ and thus has infinite vcd.

I am not sure how to find examples where $Aut$ has infinite vcd.

[1] Rips, E., Subgroups of small cancellation groups, Bull. Lond. Math. Soc. 14, 45-47 (1982). ZBL0481.20020.

Edit. A modification of the Rips construction in [2] yields examples where $Out(N)\cong Q$ even though it is not need for your question.

[2] Bumagin, Inna; Wise, Daniel T., Every group is an outer automorphism group of a finitely generated group., J. Pure Appl. Algebra 200, No. 1-2, 137-147 (2005). ZBL1082.20021.

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  • $\begingroup$ Thanks. How do you get to Out(N)≅Q, though? $\endgroup$
    – Mike
    Commented Feb 26, 2023 at 16:23
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    $\begingroup$ @Mike: no nontrivial element of $G$ commutes with all elements of $N$ (by hyperbolicity). $\endgroup$ Commented Feb 26, 2023 at 16:25
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    $\begingroup$ I see why the map $G\to Aut(N)$ induces an injective map $Q\to Out(N)$, but is it really always surjective? In any case, injectivity is enough to ensure $Out(N)$ has infinite vcd once $Q$ does, as desired (and maybe $Q$ really is isomorphic to $Out(N)$ and I'm just being dense). $\endgroup$ Commented Feb 26, 2023 at 17:11
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    $\begingroup$ Oh, this should be corrected. I only claim injectivity which suffices. $\endgroup$ Commented Feb 26, 2023 at 17:12

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