Let $G$ be a f.g. virtually nilpotent group. Can an element $g\in G$ 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.
Best Answer
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 $t^{-1}gt = g^n$. Then the subgroup $H = \langle t,g \rangle$ of $G$ is isomorphic to a quotient of the solvable Baumslag-Solitar group $X = = \langle g,t \mid t^{-1}gt=g^n \rangle$.
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.