# Automorphisms of virtually abelian groups

Let $G$ be a finitely generated virtually abelian group (i.e., $G$ contains $\mathbb{Z}^n$ with finite index for some $n\ge 2$). Is there anything known about the outer automorphism group $Out(G)$?

For example, is $Out(G)$ finitely presented? Is it linear? Is it residually finite?

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Auslander (sorry, restricted access link) proved that the automorphism group of any polycyclic group is residually finite, linear and finitely presented. As mentioned by Steve in the comments, it extends to virtually polycyclic groups (se the book Polycyclic Groups by Segal).

For $\mathrm{Out}(G)$, we have to mod out by the group of inner automorphisms, which is f.g. virtually abelian. This preserves being finitely presented.

It also follows that $\mathrm{Out}(G)=\mathrm{Aut}(G)/\mathrm{N}$ is residually finite: indeed, let $N$ be a characteristic abelian subgroup of finite index in $G$. The profinite closure $A$ of $\mathrm{Inn}(N)$ is finitely generated abelian: indeed, considering the action of $G$ on $N$, we see that $G$ is an extension of two groups both linear over $\mathbf{Z}$, so has all its abelian subgroups finitely generated. By construction, $G/A$ is residually finite. Since $A/\mathrm{Inn}(N)$ is f.g. and residually finite, it follows that $G/\mathrm{Inn}(N)$ is residually finite. By modding out by a finite subgroup we then get $\mathrm{Out}(G)$, which is residually finite.

I'm not quite sure about linearity, but I think that we could need some slightly more refined result about $\mathrm{Aut}(G)$, saying that it has a linear representation over $\mathbf{Z}$ so that the image of $\mathrm{Inn}(G)$ is virtually unipotent, to pass linearity to the quotient. Or there's a trick to avoid this (and maybe an available reference as well). [One more edit: Steve pointed out that Out of virtually polycyclic groups are indeed linear over $\mathbf{Z}$: Wehrfritz, Two Remarks on Polycyclic Groups, Bull. LMS 26, 543--548]

[Edit: my previous arguments were too flawed to be maintained.]

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@Yves: I do not understand why the kernel of the map $Aut(G) \to Out(G)$ is finite. This kernel is, by definition $Inn(G)\cong G/Z(G)$ so it is finite iff $|G:Z(G)|<\infty$. For example, if $G$ is the fundamental group of the Klein bottle, then $Inn(G)$ is infinite. – Ashot Minasyan Jan 18 '13 at 12:28
Yves - why is linearity preserved by modding out a finite normal subgroup? This must be a silly little thing, but it isn't obvious to me. – HJRW Jan 18 '13 at 13:28
@Henry: this is true for a f.g. linear group by residual finiteness, as it has a finite index subgroup which has trivial intersection with the kernel, and thus embeds into the quotient as a finite index subgroup. – Ashot Minasyan Jan 18 '13 at 13:48
Ashot - of course! Thanks. – HJRW Jan 18 '13 at 13:52
@Yves: Sorry, I couldn't leave this question alone! It is proved in Wehrfritz, Two Remarks on Polycyclic Groups, Bull. LMS 26, 543--548, that $\mathrm{Out}(G)$ is linear for any virtually polycyclic group. Seeing your recent edit, I would add that when I've ben saying linear, I meant as a subgroup of $GL(n,\mathbb{Z})$. – Steve D Jan 18 '13 at 16:58