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 .
Another way is to count the finite order elements or the finite subgroups.
For example, let , where the generators of and are denoted and respectively.
acts faithfully on a -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 or , there are just two conjugacy classes of finite subgroups in .
2. Examples of groups with finitely many conjugacy classes of finite order elements
Let denote the set of groups that have finitely many conjugacy classes of finite order elements.
Some groups in are:
- groups — every finite order isometry of a space has a fixed point, and orbits of fixed points correspond to conjugacy classes of finite order elements. (Bridson-Haefliger, Corrollary II.2.8)
- Hyperbolic groups (Bogopolskii-Gerasimov, N. Brady)
- Arithmetic groups (Borel)
- Mapping class groups (Kerckhoff, Harer, written up by Bridson)
3. Groups with infinitely many conjugacy classes of finite order elements
How about groups not in ? One cheap example is
Whilst isn’t finitely generated, it sits inside as the kernel of the map in which and .
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 and . Let where acts by interchanging and . As we have previously discussed, the Bieri-Stallings group , in which and , is not finitely generated. The map from can be extended to a map by sending to . Thus
is not finitely generated.
Proposition (Key idea due to Leary-Nucinkis) is not in . More specifically, is not conjugate in to if .
The fixed point set of is in . For all , the fixed point set of is also in since .
We are now ready to construct a finitely generated group not in .
Let , where the first is generated by and , the second factor by and , is generated by , and acts by switching and and also switching and . Consider the Biery-Stallings group . As was shown previously, is finitely generated but not finitely presentable. Then is also finitely generated but not finitely presentable, and by the same argument as in the previous proposition.
In general, for , where the -th is generated by and , is generated by , and maps each to and vice versa, is in but not in and .
Note that can be embedded in some arithmetic groups, mapping class groups, and (Brady-Clay-Dani).
The general scheme in which all the above examples fit is as follows. Let be a locally finite cell complex. Fix a cellular map that induces a surjection . Let be a finite order isometry of with a fixed vertex such that acts on freely. Then is not in .
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 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 and a finitely generated group that fit into the short exact sequence
Take and fix a positive integer . Then we have a short exact sequence
The hope is that is hyperbolic, finitely generated, and .
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 complex (so that is hyperbolic), find an appropriate isometry , and examine .
Take a rose with 8 petals (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 . Consider a function mapping each edge in the rose to 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 .) Then lifts to a Morse function . The descending and ascending links of the vertex in are both homeomorphic to circles, with 8 vertices (corresponding to the 8 edges in ) 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 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 is piecewise hyperbolic, the space is , and so is hyperbolic. Let be an isometry of corresponding to a “ rotation”, edge 1 being mapped to 2, 2 to 3, etc. mod 8. Note that lifts to , preserves level sets of , and acts on the link of a fixed vertex in without fixed points. Thus, as in the previously discussed cases, is not in , and it sits inside a hyperbolic group (namely ) 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.