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Let $\Gamma$ be a finite graph, then $H^1(\Gamma,\mathbb{Z})\cong \mathbb{Z}^{g(\Gamma)}$ can be viewed as a $\mathrm{Aut}(\Gamma)$ module.

Conversely, given a finite group $G$, and a $G$-module $\mathbb{Z}^n$, does there always exist a finite graph $\Gamma$ such that the $G$ module $\mathbb{Z}^n$ arises as $G\overset{f}{\to}\mathrm{Aut}(\Gamma)\curvearrowright H^1(\Gamma,\mathbb{Z})$ for some $f\colon G\to \mathrm{Aut}(\Gamma)$?

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No. An action of a group $G$ on a graph $\Gamma$ induces a homomorphism $G\to \mathrm{Out}(\pi_1(\Gamma))=\mathrm{Out}(F_n)$. So a representation $G\to \mathrm{GL}_n({\mathbb Z})$ can come from an action on a graph only if it lifts to a homomorphism $G\to \mathrm{Out}(F_n)$ over the quotient homomorphism $\mathrm{Out}(F_n)\to \mathrm{GL}_n({\mathbb Z})$. This paper of Zimmermann gives an example of a finite cyclic subgroup of $\mathrm{GL}_n({\mathbb Z})$ that does not lift.

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  • $\begingroup$ How do you define "an action by outer automorphisms" for a group action on another group? $\endgroup$
    – YCor
    Commented Jul 13, 2018 at 18:48
  • $\begingroup$ What you mean is correct; it's just that it sounded weird to me to call this an "action by outer automorphisms" (note that what you say in your comment does not make use of the group structure of $G$). Possibly by outer action of $G$ on a group $H$ one can just mean a homomorphism $G\to\mathrm{Out}(H)$. $\endgroup$
    – YCor
    Commented Jul 13, 2018 at 18:56
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    $\begingroup$ I'll just throw in that this terminology, however imprecise, is nonetheless pretty well established in group theory. $\endgroup$
    – Lee Mosher
    Commented Jul 13, 2018 at 18:59
  • $\begingroup$ Anyhow, I edited. $\endgroup$
    – Gregory Arone
    Commented Jul 13, 2018 at 19:01
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    $\begingroup$ @YCor: It's a theorem, proved independently by Culler, by Zimmerman, and by Khramstov, that every finite subgroup of $\text{Out}(F_n)$ can be realized by an action on a graph. For example the order 3 cyclic group acts on the rank 2 theta graph. $\endgroup$
    – Lee Mosher
    Commented Jul 14, 2018 at 14:06

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