Mfcttrf

Let $G$ be a f.g. virtually nilpotent group. Can an element $gin G$ of infinite order be conjugate to its power $g^n$ for $n>1$?

Let $G$ be a f.g. virtually abelian group. Is it true that elements of infinite order have finite conjugacy classes?

Note that for the infinite dihedral group non-trivial elements of finite order have infinite conjugacy classes.

It is not possible for an infinite order element $g$ of a virtually nilpotent group to be conjugate to $g^n$ for $n>1$.

Suppose $t in G$ with $t^-1gt = g^n$. Then the subgroup $H = langle t,g rangle$ of $G$ is isomorphic to a quotient of the solvable Baumslag-Solitar group $X = = langle g,t mid t^-1gt=g^n rangle$.

Now $X$ has the abelian normal subgroup $Y := langle g^G rangle$, which is isomorphic to the additive group of rational numbers that can be written as $a/n^k$ for some $a,k in mathbb Z$. It is not hard to see that any nontrivial quotient of $Y$ is a torsion group, and also that no nontrivial normal subgroup of $X$ can intersect $Y$ trivially. So, since we are assuming that $g$ has infinite order, we have $H cong X$.

But $X$ is not virtually nilpotent. To see that, show that no subgroup of $X$ of finite index can have trivial centre.

All finite groups are virtually nilpotent, so consider the quaternion group $Q_8$. Then

$j i j^-1 = j i (-j) = - i j (-j) = -i = i^3$.

So an element is conjugate to its cube, so the answer to your first question is "Yes".

For a virtually abelian group $G$, the issue can be settled with some Euclidean geometry. The general structural theorem for virtually abelian groups is as follows (this is a summary of the Bieberbach Theorems):

• 
  $G$ has a unique maximal finite normal subgroup $N < G$,


• The quotient $G / N$ is a Euclidean crystallographic group, meaning that is has a faithful, discrete, cocompact action (or representation) $G / N mapsto textIsom(mathbbE^n)$ for some $n ge 0$.

Here I am using $mathbb E^n$ to denote Euclidean $n$-space, i.e. $mathbb R^n$ equipped with the standard metric, and $textIsom(mathbb E^n)$ is its group of isometries.

Let me denote the composed homomorphism
$$f : G to G/N to textIsom(mathbb E^n)
$$
Its image is a discrete group, and it follows that an element of $f(G)$ has finite order if and only if it fixes a point.

The key facts regarding order of elements are as follows:

• For each $g in G$, the translation length $L_g$ of the isometry $f(g) : mathbb E^n to mathbb E^n$ is a conjugacy invariant of $g$.


• 
  $L_g = 0$ $iff$ $f(g)$ has finite order in $textIsom(mathbb E^n)$ $iff$ $g$ has finite order in $G$.

So if $g$ has infinite order then it is a translation with translation length $L_g > 0$, and it follows that $g^n$ is a translation with translation length $n cdot L_g ne L_g$. Since translation length is a conjugacy invariant, $g$ and $g^n$ are not conjugate.

I suspect there is also a geometric solution to the virtually nilpotent case, as an alternative to the answer of @DerekHolt. In outline, $G$ still has a unique maximal normal subgroup $N$, and $G/N$ embeds as a lattice in a nilpotent Lie group $Gamma$, and I believe that one can then proceed using the geometry of the left invariant metric on $Gamma$.