Is there an example of a finitely generated subgroup $\Gamma \subset \mathrm{GL}_n(\mathbb{C})$ such that the group of outer automorphisms $\mathrm{Out}(\Gamma)$ contains finite subgroups of unbounded orders?


1 Answer 1


Yes. Furthermore, every recursively presented countable group embeds in such a group. Indeed, first Higman's embedding reduces to proving that every finitely presented group embeds into such a group (if you want all finite groups, it is enough to use a single known f.p. group with all finite groups in it, e.g., Thompson's group $V$).

Now the Rips construction says that for every f.p. group $Q$, there exists a $C'(1/6)$ f.p. small cancelation group $G$ and a finitely generated and normal subgroup $N$ such that $G/N\simeq Q$. Hence $Q$ embeds into $\mathrm{Out}(N)$.

Finally, we use Haglund-Wise-Agol's results, which ensure that every $C'(1/6)$ f.p. small cancelation group is linear over $\mathbf{Z}$.

Edit: as suggested by HJWR in a comment, it is more direct to directly use the Haglund-Wise "Rips construction" (GAFA 2008-other link), which directly provides a group that is linear over $\mathbf{Z}$, without need of the Wise machinery plus Agol.

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    $\begingroup$ For the record: to get this for all $C'(1/6)$ groups uses the full strength of the Agol--Wise machinery, which is pretty challenging. But Haglund--Wise gave a virtually special (in particular, linear over $\mathbb{Z}$) version of the Rips construction in their GAFA paper `Special cube complexes'. This is much easier. $\endgroup$
    – HJRW
    May 30 at 9:33

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