# Inner? Automorphisms

It is a fairly easy result of Edna Grossman's that any automorphism of a (finitely generated) free group which acts by conjugation on every primitive element ( primitive element in$F_n$ is one which is a member of a generating set of order $n$) is in fact inner. The question is: what is the right generalization of this fact? For surface groups, one can replace primitive elements by simple curves, but I a, not sure if that the"canonical" thing to do...

UPDATE Looking up citations of Henry's (@HW's) great reference, I found that the paper he cites is not quite "the last work". A more recent (and very relevant) word seems to be Bogopolski-Ventura (On endomorphisms of torsion-free hyperbolic groups) from which I quote the abstract:

Let $H$ be a torsion-free $\delta$-hyperbolic group with respect to a finite generating set $S$. Let $a_1,..., a_n$ and $a_{1*},..., a_{n*}$ be elements of $H$ such that $a_{i*}$ is conjugate to $a_i$ for each $i=1,..., n$. Then, there is a uniform conjugator if and only if $W(a_{1*},..., a_{n*})$ is conjugate to $W(a_1,..., a_n)$ for every word $W$ in $n$ variables and length up to a computable constant depending only on $\delta$, $\sharp{S}$ and $\sum_{i=1}^n |a_i|$. As a corollary, we deduce that there exists a computable constant $\mathcal{C}=\mathcal{C}(\delta, \sharp S)$ such that, for any endomorphism $\phi$ of $H$, if $\phi(h)$ is conjugate to $h$ for every element $h\in H$ of length up to $\mathcal {C}$, then $\phi$ is an inner automorphism. Another corollary is the following: if $H$ is a torsion-free conjugacy separable hyperbolic group, then $\mbox{Out}(H)$ is residually finite. When particularizing the main result to the case of free groups, we obtain a solution for a mixed version of the classical Whitehead's algorithm.

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You could ask the question, given an infinite group $G$ and an outer automorphism $\phi$, whether there always exists an element $g$ in $G$ such that $g$ and $\phi(g)$ are not conjugate to each other. –  André Henriques Jun 10 '12 at 19:29
@andre, yes, this is certainly a good question, but a positive answer for $G$ free would be a little weaker than the cited result -- that said, I have no idea whether there are any counter examples to your statement. Do you know of any for FINITE $G?$ –  Igor Rivin Jun 10 '12 at 19:41
An infinite counterexample to André's question is the group of finitely supported permutations of an infinite set. Conjugation by an infinite permutation acts as an automorphism, but doesn't change the cycle types. A counterexample to a refinement: if the infinite permutation doesn't have any infinite cycles, then all the orbits of this outer automorphism are finite and even orbits of some inner automorphism. –  Ben Wieland Jun 10 '12 at 20:04
There are also finite counterexamples to André Henriques's question. The smallest is a group of order 32 constructed by G. E. Wall (see MR0025461). This question was first considered by Burnside: see Note B in his book. –  Mark Wildon Jun 10 '12 at 21:06
Andre - Grossman called this property A: a group is said to have property A if every conjugating automorphism, ie every automorphism that takes every element to a conjugate, is inner. As Ben and Mark point out, there are lots of groups that fail to have property A. As is the case with many properties of free groups, this property doesn't generalise to all groups, but does generalise to groups with negative curvature; see the paper that I cited in my answer. –  HJRW Jun 11 '12 at 0:59