Let $G$ a finitely generated group, $\mathrm{Aut}(G)$ is its automorphism group, then it is necessary that $\mathrm{Aut}(G)$ is a finitely generated group? Thanks in advance.

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    $\begingroup$ I am often puzzled by votes to close. It would be helpful if people voting to close would provide some indication of the reason. This seems an interesting and nontrivial question. $\endgroup$
    – Derek Holt
    Commented Jun 8, 2014 at 9:57

2 Answers 2



An example is the Baumslag-Solitar group $BS(2, 4) = \langle a, b \mid a^{-1}b^2a = b^4 \rangle$, which has an infinitely generated automorphism group (as proved here, apparently, though it is not open access). This is just a special case and $BS(m, n) = \langle a, b \mid a^{-1}b^ma = b^n \rangle$ with $|m|,|n| \ne 1$ and $m \mid n$ or $n \mid m$, form an entire family of groups with the same property. See here.

  • $\begingroup$ Very nice example! $\endgroup$ Commented Jun 8, 2014 at 16:09
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    $\begingroup$ The $m\mid n$ etc. condition is both necessary and sufficient (the encyclopedia entry only mentions that these groups have infinitely-generated outer automorphism group). The other direction is proven in the paper Tree actions of automorphism groups by Gilbert, Howie, Metaftsis and Raptis. More generally, one can make similar statements about groups acting on trees where all edge and vertex stabilisers are infinite cyclic (generalised Baumslag-Solitar groups). $\endgroup$
    – user1729
    Commented Jun 9, 2014 at 8:50
  • $\begingroup$ @user1729PhD Thanks for that reference. Especially of interest to me as I study algebraic graph theory. $\endgroup$
    – M. Vinay
    Commented Jun 9, 2014 at 8:52
  • $\begingroup$ If you are interested, there is a good survey article from a few years ago (2005) on GBS groups by Levitt. See here. $\endgroup$
    – user1729
    Commented Jun 9, 2014 at 8:54
  • $\begingroup$ Interesting. Thanks again! $\endgroup$
    – M. Vinay
    Commented Jun 9, 2014 at 9:02

M. Vinay's example is wonderful because the group $G$ is finitely presented rather than just finitely generated. In this answer I will loose the "finitely presented" condition, but maintain "finitely generated". This relaxing of the conditions allow for more groups to appear as $\operatorname{Out}(G)$.

Firstly, $\operatorname{Out}(G)=\operatorname{Aut}(G)/\operatorname{Inn}(G)$ is the outer automorphism group. It is all the automorphisms quotiented out by the obvious ones we always get, and which are boring because they look like the group anyway. Perhaps the best motivation for looking at $\operatorname{Out}(G)$ rather than $\operatorname{Aut}(G)$ is that in topology, certain surfaces have the property that $\operatorname{Out}(\pi_1(S))\cong \operatorname{MCG}(S)$, that is, the fundamental group of the surface is isomorphic to the mapping class group (the group of isotopy-classes of homeomorphisms - homeomorphisms but without wiggling!). Another motivation, and one relevant to here, is that not every group can occur as the automorphism group of a group (for example, the infinite cyclic group does not occur as $\operatorname{Aut}(G)$ for any $G$).

Now, the point of my answer:

Theorem: Every countable group $Q$ occurs as the outer automorphism group of a finitely generated group $G$, $\operatorname{Out}(Q)\cong G$.

This is a result of Bumagin and Wise (see here, third paper from the top). The proof is relatively straight-forward, and relies on a basic working knowledge of small cancellation theory. The groups $G$ they produce are never finitely presented (this is a result intrinsic to the construction they use, called Rips construction - Martin Bridson points out that such groups are never finitely presentable, saying it follows from a result of Bieri). The groups $G$ are always two-generated. If $Q$ is finitely presented then the group $G$ has the strong property of being "residually finite", which means that for every element $g\in G$ there exists some finite object $X$ such that $G$ acts pn $X$ in such a way that the action of $g$ on $X$ is non-trivial.

Now, going back to the question, the result of Bumagin and Wise proves the existence of two-generated groups $G$ with automorphism group (note: not outer automorphism group) possessing one of the following properties.

  • $\operatorname{Aut}(G)$ is not finitely generated.
  • $\operatorname{Aut}(G)$ is finitely generated but not finitely presented.
  • $\operatorname{Aut}(G)$ is finitely generated but not recursively presentable.
  • $\operatorname{Aut}(G)$ is finitely generated and is recursively presentable.
  • $\operatorname{Aut}(G)$ is finitely generated and equal to $\operatorname{Inn}(G)\cong G$ (the groups they produce have trivial centre).

...and so on. This is because we have a short exact sequence $$1\rightarrow\operatorname{Inn}(G)\rightarrow\operatorname{Aut}(G)\rightarrow\operatorname{Out}(G)\rightarrow 1$$ where the kernel $\operatorname{Inn}(G)$ is finitely generated. So if $\operatorname{Out}(G)$ has a property which lifts to pre-images with finitely generated kernel then $\operatorname{Aut}(G)$ also has this property.


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