Does there exist a finitely generated group $G$ and an automorphism $\Phi\colon G \to G$ such that there are $\Phi$-periodic elements with unbounded period?


If $G$ is merely countable and not finitely generated there are easy examples: consider a free or free abelian group of countably infinite rank, let $x\_1,x\_2,\ldots$ be a standard generating set, and consider the automorphism that cyclically permutes the first $2$ generators, the next $3$ generators, the next $4$ generators and so on.



Does there exist an automorphism of a finitely generated group with periodic points of unbounded period?
1. Yes, and even with an inner automorphism.


Namely, start from your example of a countable group $G$ with elements $x\_n$ ($n$ ranging over some arbitrary subset) and automorphism $f$ such that $x\_n$ lies in an $n$-cycle for $f$. Let $G'$ be the semidirect product $G\rtimes\langle f\rangle$. Finally, embed $G'$ into a finitely generated group $H$ (every countable group embeds into a f.g. group, an old result of Higman-Neumann-Neumann). Then in $H$, for the inner automorphism defined by $f$, the element $x\_n$ lies in an $n$-cycle. [NB: as $G$ can be chosen to be abelian and hence $G'$ metabelian, $H$ can be chosen to be 4-step solvable, by the Neumann-Neumann proof of the previous embedding theorem.]
2. Note: there are no examples among linear groups (= subgroup of $\mathrm{GL}\_d$ over a field) with an inner automorphism. Indeed, for the inner automorphism $i\_g$ defined by an element $g$, the centralizer of $g^{n!}$ is an increasing sequence of subgroups. But in a matrix group the centralizers in the algebra of matrices are subspaces. So there is no strictly infinite sequence of centralizers. Hence the finite $i\_g$-cycles have bounded length.
3. Here's another example, due to B.H. Neumann 1937, with the additional feature of being residually finite. Namely, consider a set $X$ consisting of the infinite disjoint union of sets $X\_n$ ($n$ ranging over an infinite set $I$ of integers $\ge 5$), each $X\_n$ being in bijection with $\{1,\dots,n\}$. Let $c$ be the element acting as the $n$ cycle $n\mapsto 1\mapsto 2\mapsto\dots$ on each $X\_n$, and $t$ the element acting as the transposition $1\mapsto 2\mapsto 1$ on each $X\_n$. Let $G\_I$ be the subgroup of permutations of $X$ generated by $\{t,c\}$. Then $G\_I$ is residually finite (since it acts faithfully with finite orbits). Also $G\_I$ contains, for each $n\in I$, as normal subgroup, the alternating group $\mathrm{Alt}\_n$ of $X\_n$ (acting trivially elsewhere). Let $s\_n$ be the 3-cycle $1\mapsto 2\mapsto 3\mapsto 1$ acting on $X\_n$, identity elsewhere. Then we see that $s\_n\in G\_I$, and the inner automorphism defined by $c$ admits $s\_n$ as element of period exactly $n$.


