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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.

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  • $\begingroup$ For your second question, elements of infinite order could have either finite or infinite conjugacy classes. $\endgroup$ – Derek Holt Jul 22 at 14:37
  • $\begingroup$ &DerekHolt Could you give an example? $\endgroup$ – QMath Jul 22 at 14:44
  • $\begingroup$ An example of what exactly? $\endgroup$ – Derek Holt Jul 22 at 15:42
  • $\begingroup$ &DerekHolt An example of a virtually abelian group and an element of infinite order with infinite conjugacy class. $\endgroup$ – QMath Jul 22 at 16:07
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    $\begingroup$ $y \in \langle x,y \mid y^{-1}xy=x^{-1}\rangle$. $\endgroup$ – Derek Holt Jul 22 at 17:29
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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.

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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) $G / N \mapsto \text{Isom}(\mathbb{E}^n)$ 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 $\text{Isom}(\mathbb E^n)$ is its group of isometries.

Let me denote the composed homomorphism $$f : G \to G/N \to \text{Isom}(\mathbb E^n) $$ 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 $\text{Isom}(\mathbb E^n)$ $\iff$ $g$ 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 $\Gamma$.

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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".

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  • $\begingroup$ The quaternion group is even nilpotent. $\endgroup$ – Dietrich Burde Jul 22 at 14:35
  • $\begingroup$ Sorry, I forgot to write this assumption, I am interested in elements of infinite order. $\endgroup$ – QMath Jul 22 at 14:43
  • $\begingroup$ You should edit your post to add that assumption. $\endgroup$ – Charlie Cunningham Jul 22 at 14:55

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