## Groups having finitely many conjugacy classes of finite order elements

This post is based on a guest talk by Matt Clay.

1. How finite is this group?

One way to measure how close a group is to being finite is by the finiteness property ${\mathcal{F}_n}$.

Another way is to count the finite order elements or the finite subgroups.

For example, let ${G = \mathbb{Z}_2 \ast \mathbb{Z}_3 = \mathrm{PSL}(2, \mathbb{Z})}$, where the generators of ${\mathbb{Z}_2}$ and ${\mathbb{Z}_3}$ are denoted ${a}$ and ${b}$ respectively.

${G}$ acts faithfully on a ${(2,3)}$-regular tree and freely on its edges.

Each finite order element fixes a single vertex — the center of the convex hull of an orbit — and so does every finite subgroup. Since each vertex is fixed by a conjugate of either ${a}$ or ${b}$, there are just two conjugacy classes of finite subgroups in ${G}$.

2. Examples of groups with finitely many conjugacy classes of finite order elements

Let ${S}$ denote the set of groups that have finitely many conjugacy classes of finite order elements.

Some groups in ${S}$ are:

1. ${\mathrm{CAT}(0)}$ groups — every finite order isometry of a ${\mathrm{CAT}(0)}$ space has a fixed point, and orbits of fixed points correspond to conjugacy classes of finite order elements. (Bridson-Haefliger, Corrollary II.2.8)
2. Hyperbolic groups (Bogopolskii-Gerasimov, N. Brady)
3. Arithmetic groups (Borel)
4. Mapping class groups (Kerckhoff, Harer, written up by Bridson)
5. ${\mathrm{Out}(F_n)}$ (Culler)

3. Groups with infinitely many conjugacy classes of finite order elements

How about groups not in ${S}$? One cheap example is

$\displaystyle H = \langle \{a_i : i \in \mathbb{Z} \} \mid a^2_i \rangle = \mathbb{Z}_2 \ast \mathbb{Z}_2 \ast \cdots$.

Whilst ${H}$ isn’t finitely generated, it sits inside ${G = \langle a_0, t \mid a^2_0 \rangle = \mathbb{Z}_2 \ast_{\{1\}}}$ as the kernel of the map ${G \rightarrow \mathbb{Z}}$ in which ${t \mapsto 1}$ and ${a_0 \mapsto 0 }$.

Here is another cheap example…. still not finitely generated, but a better example as it will lead to examples enjoying better a range of finiteness properties, as we will see. Let ${F_2 = \langle a, b \rangle}$ and ${\mathbb{Z}_2 = \langle x \mid x^2 \rangle}$. Let $\displaystyle G = F_2 \rtimes \mathbb{Z}_2$ where ${x}$ acts by interchanging ${a}$ and ${b}$. As we have previously discussed, the Bieri-Stallings group ${K = \ker(F_2 \rightarrow \mathbb{Z})}$, in which ${a \mapsto 1}$ and ${b \mapsto 1}$, is not finitely generated. The map from ${F_2}$ can be extended to a map ${f:G \rightarrow \mathbb{Z}}$ by sending ${x}$ to ${0}$. Thus

$\displaystyle H = \ker f = K \rtimes \langle x \rangle$

is not finitely generated.

Proposition (Key idea due to Leary-Nucinkis) ${H}$ is not in ${S}$. More specifically, ${a^ixa^{-i}}$ is not conjugate in ${H}$ to ${a^jxa^{-j}}$ if ${i \neq j}$.

Proof:

The fixed point set of ${a^ixa^{-i}}$ is in ${f^{-1}(i)}$. For all ${h \in H}$, the fixed point set of ${ha^ixa^{-i}h^{-1}}$ is also in ${f^{-1}(i)}$ since ${f(ha^i) = f(h) + f(a^i) = i}$. $\Box$

We are now ready to construct a finitely generated group not in ${S}$.

Let ${G = (F_2 \times F_2) \rtimes \mathbb{Z}_2}$, where the first ${F_2}$ is generated by ${a}$ and ${b}$, the second ${F_2}$ factor by ${c}$ and ${d}$, ${\mathbb{Z}_2}$ is generated by ${x}$, and ${x}$ acts by switching ${a}$ and ${b}$ and also switching ${c}$ and ${d}$. Consider the Biery-Stallings group ${K = \ker(F_2 \times F_2 \rightarrow \mathbb{Z})}$. As was shown previously, ${K}$ is finitely generated but not finitely presentable. Then ${H = \ker(G \rightarrow \mathbb{Z}) = K \rtimes \mathbb{Z}_2}$ is also finitely generated but not finitely presentable, and ${H \notin S}$ by the same argument as in the previous proposition.

In general, for ${G = (F_2)^n \rtimes \mathbb{Z}}$, where the ${i}$-th ${F_2}$ is generated by ${a_i}$ and ${b_i}$, ${\mathbb{Z}_2}$ is generated by ${x}$, and ${x}$ maps each ${a_i}$ to ${b_i}$ and vice versa, ${H = \ker(G \rightarrow \mathbb{Z})}$ is in ${F_{n-1}}$ but not in ${F_n}$ and ${H \notin S}$.

Note that ${G}$ can be embedded in some arithmetic groups, mapping class groups, and ${\mathrm{Out}(F_n)}$ (Brady-Clay-Dani).

The general scheme in which all the above examples fit is as follows. Let ${X}$ be a locally ${\mathrm{CAT}(0)}$ finite cell complex. Fix a cellular map ${f: X \rightarrow S^1}$ that induces a surjection ${\pi_1(X) \rightarrow \mathbb{Z}}$. Let ${\sigma}$ be a finite order isometry of ${X}$ with a fixed vertex ${v}$ such that ${\sigma}$ acts on ${\mathrm{lk}(v)}$ freely. Then ${H = \ker(\pi_1(X) \rtimes \langle \sigma \rangle \rightarrow \mathbb{Z})}$ is not in ${S}$.

4. A group that has infinitely many conjugacy classes of finite order elements and is a subgroup of a hyperbolic group

How can we construct a finitely generated group not in ${S}$ which sits inside a hyperbolic group? One approach, due to Dani Wise, is to use the Rips construction.

Recall that we can find a hyperbolic group ${G}$ and a finitely generated group ${K_0}$ that fit into the short exact sequence

$\displaystyle 1 \rightarrow K_0 \rightarrow G \rightarrow \mathbb{Z} \rightarrow 1$.

Take ${a \in K_0}$ and fix a positive integer ${r}$. Then we have a short exact sequence

$\displaystyle 1 \rightarrow K \rightarrow G / \langle \langle a^r \rangle \rangle \rightarrow \mathbb{Z} \rightarrow 1$.

The hope is that ${G / \langle \langle a^r \rangle \rangle}$ is hyperbolic, ${K}$ finitely generated, and ${K \notin S}$.

This approach does work, but some effort is required to show that. Another approach, similar in style to the examples analyzed above, is to construct a ${\mathrm{CAT}(-1)}$ complex ${X}$ (so that ${\pi_1(X)}$ is hyperbolic), find an appropriate isometry ${\sigma}$, and examine ${\ker(\pi_1(X) \rtimes \langle \sigma \rangle \rightarrow \mathbb{Z})}$.

Take a rose with 8 petals ${R_8}$ (i.e. one vertex, 8 edges) with the edges labeled 1 through 8. Glue on 8 right-angled hyperbolic hexagons as shown below (with numbering modulo 8).

Denote the resulting cell complex by ${X}$. Consider a function ${f: X \rightarrow S^1}$ mapping each edge in the rose to ${S^1}$ and extended linearly to the 2-cells. (Thus in the above hexagon, a route from the bottom to the top would map to a path winding three times around ${S^1}$.) Then ${f}$ lifts to a Morse function ${\tilde{f}: \tilde{X} \rightarrow \mathbb{R}}$. The descending and ascending links of the vertex in ${X}$ are both homeomorphic to circles, with 8 vertices (corresponding to the 8 edges in ${X}$) and edges between consecutive vertices (corresponding to the top and bottom “angles” in the glued hexagon illustrated above).

In the full link, there are edges between vertices in the ascending and descending links (corresponding to the remaining “angles” in the hexagons). Each vertex in the ascending link is connected to every other vertex in the descending link. In the illustration below, only the edges from the vertex ${1^+}$ in the ascending link to those in the descending link are drawn.

The smallest cycle in the link has length 4, so by Gromov’s link condition and the fact that ${X}$ is piecewise hyperbolic, the space ${\tilde{X}}$ is ${\mathrm{CAT}(-1)}$, and so ${\pi_1(X)}$ is hyperbolic. Let ${\sigma}$ be an isometry of ${X}$ corresponding to a “${2\pi / 8}$ rotation”, edge 1 being mapped to 2, 2 to 3, etc. mod 8. Note that ${\sigma}$ lifts to ${\tilde{X}}$, preserves level sets of ${\tilde{f}}$, and acts on the link of a fixed vertex in ${\tilde{X}}$ without fixed points. Thus, as in the previously discussed cases, ${H = \ker(\pi_1(X) \rtimes \langle \sigma \rangle \rightarrow \mathbb{Z})}$ is not in ${S}$, and it sits inside a hyperbolic group (namely ${\pi_1(X) \rtimes \langle \sigma \rangle}$) and is finitely generated. The last fact follows from an argument essentially identical to the one used in a previous post to show that the second Bieri-Stallings group is finitely generated.