# Conjugacy classes in virtually nilpotent groups

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.

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

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.

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 $$\Gamma$$.

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".

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