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We are looking for examples of groups $G$ such that $G$ is "big", but $Out(G)$ is trivial. By "big" we mean things like virtually free, or large, or Golod-Shafarevich. However, we would like our groups to be residually-finite.

Edit: Henry, Igor and Lee thank you for all your help. Eventually, we only needed finite $Out(G)$ with $G$ having a trivial center. Thus, $A*B$ was the example we used. It is included in "Large normal subgroup growth and large characteristic subgroup growth" (sorry for the self-promotion, but I thought people might be interested in our motivation and the results of their help).

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  • $\begingroup$ I suspect $\PSL_2(\mathbb{Z})$ has trivial outer automorphism group, though I never worked it out. $\endgroup$
    – HJRW
    Mar 19 '17 at 21:06
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    $\begingroup$ @HJRW What about inverse transpose? $\endgroup$
    – Igor Rivin
    Mar 19 '17 at 21:11
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    $\begingroup$ @HJRW one actually has to be careful, because that automorphism is inner in $GL(2, \mathbb{Z}),$ but the conjugation is by a matrix of determinant $-1.$ $\endgroup$
    – Igor Rivin
    Mar 19 '17 at 21:27
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    $\begingroup$ @YiftachBarnea -- yes, sorry, I was careless. But I expect there to be lots of virtually free examples with trivial outer automorphism group. For instance, I expect $S_7*S_8$ to work. $\endgroup$
    – HJRW
    Mar 19 '17 at 21:31
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    $\begingroup$ If $A,B$ are nontrivial, nonisomorphic, finite or one-ended groups then $A*B$ has outer automorphism group of the form $\text{Aut}(A) \times \text{Aut}(B)$: the obvious homomorphism $\text{Aut}(A) \times \text{Aut}(B) \to \text{Aut}(A*B) \to \text{Out}(A*B)$ is an isomorphism (and so it's never trivial). Levitt wrote down those kinds of formulas in one of his papers. So for example the outer automorphism group of $\text{PSL}(2,\mathbb{Z}) \approx \mathbb{Z}/2 * \mathbb{Z}/3$ is cyclic of order 2. $\endgroup$
    – Lee Mosher
    Mar 20 '17 at 0:11
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There are many, many examples of large, residually finite groups with trivial outer automorphism group. Indeed, any nonabelian virtually special group is large and residually finite. Furthermore, if they're hyperbolic, Paulin showed that they have finite outer automorphism group unless they split over a virtually cyclic subgroup; in particular, most such examples have trivial outer automorphism group. Instances of this include:

  • fundamental groups of closed hyperbolic 3-manifolds;

  • random groups at density less than 1/6;

  • most small-cancellation groups;

  • groups constructed using the Rips construction;

etc etc.

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    $\begingroup$ Henry, just to make sure do you mean large in the sense that a subgroup of finite index maps on a non-abelian free group? $\endgroup$ Mar 19 '17 at 21:22
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    $\begingroup$ @YiftachBarnea, yes, I mean large in the sense of Pride. $\endgroup$
    – HJRW
    Mar 19 '17 at 21:29
  • $\begingroup$ I think the result you mention about outer automorphism groups of hyperbolic groups should be attributed to Bestvina and Feighn: it is proved in their paper untitled Stable actions of groups on real trees (in fact, the first step of the proof is based on a construction due to Paulin). $\endgroup$
    – Seirios
    Mar 20 '17 at 20:06
  • $\begingroup$ @Seirios, you're quite right. It's usually called Paulin's theorem, although it's true that the version of the Rips machine that the argument uses was worked out by Bestvina and Feighn. $\endgroup$
    – HJRW
    Mar 21 '17 at 21:50
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    $\begingroup$ It is worth pointing out that Bumagin-Wise (Every group is the outer automorphism group of a finitely generated group, J. Pure App. Alg. 2007) used a variation on Rips construction to prove that given any finitely presented group $Q$ there exists a finitely generated, residually finite group $Q$ such that $\operatorname{Out}(G_Q)\cong Q$. Here, $G_Q$ is the kernel of Rips construction, so if we take $Q$ to be trivial then $G_Q$ is $C{\prime}(1/6)$ and we get an explicit presentation. If $Q$ is finite then $G_Q$ is still hyperbolic, large, virtually compact special, etc. $\endgroup$
    – ADL
    Apr 7 '17 at 12:54
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A countably-infinite family of groups is given at the end of Section 6 of a paper of A. D. Logan1: for each natural number $n>1$, the two-generator-, one-relator group

$$G_n := \langle a, b; (a^{−2}ba^4ba^{−3}ba^5b)^n\rangle$$

has trivial outer automorphism group.

EDIT This has nothing to do with the above, but I don't want to add a third answer :) My discussion with Henry in the comments to the OP, might suggest that $Out(PGL(2, \mathbb{Z}))$ might be trivial. Apparently, this was proved by Hua and Reiner in 1951-1952, and then disproved(!) by Joan Dyer in 1978, where she constructed an outer automorphism (known ever since as the Dyer automorphism, so $Out(PGL(2, \mathbb{Z})) = C_2;$ the "so" is not trivial but true.

For completeness, her automorphism $\mathcal{D}$ is described as follows. The generators of $PGL(2, \mathbb{Z})$ are:

$$S = \pm \begin{pmatrix}0 &1\\ -1 & 0\end{pmatrix}, T=\pm \begin{pmatrix}1 &1\\ 0 & 1\end{pmatrix},B=\pm \begin{pmatrix}0 &1\\ 1 & 0\end{pmatrix}.$$ Then the automorphism sends $S, T, B$ to $SB, TB, B.$

1A. D. Logan: The outer automorphism groups of two-generator, one-relator groups with torsion. Proc. Amer. Math. Soc. 144 (2016) 4135-4150

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  • $\begingroup$ is this group virtually (non-abelian free)? $\endgroup$ Mar 19 '17 at 23:42
  • $\begingroup$ @YiftachBarnea I am suspecting that yes, but can't do better than that at the moment. $\endgroup$
    – Igor Rivin
    Mar 20 '17 at 0:58
  • $\begingroup$ It's not. If it were virtually free, it would be a free product of cyclic groups, and have a non-trivial outer automorphism. $\endgroup$
    – HJRW
    Mar 20 '17 at 6:24
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    $\begingroup$ to suspect is not conjecture, and this "suspicion" was already discarded. And indeed a non-abelian free product of cyclic groups has non-inner automorphisms (if at least one has $\ge 3$ elements, take inversion on it and identity on others; if all have 2 elements, take a non-trivial permutation of factors). $\endgroup$
    – YCor
    Mar 20 '17 at 20:12
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    $\begingroup$ My reasoning was quite simple: the only torsion in a one-relator group comes from powers of the relator; in particular, there is only one conjugacy class of (maximal) finite subgroup, which is cyclic. Therefore, if it is virtually free, it must be a free product of an infinite cyclic group and a finite cyclic group. $\endgroup$
    – HJRW
    Mar 21 '17 at 21:54
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All fundamental groups of finite volume hyperbolic manifolds (of dimension $\geq 3$) have finite outer automorphism groups, and most have trivial outer automorphism groups.

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