# Ends of finitely generated torsion groups

It is known that the number of ends of a finitely generated group is 0,1, 2 or $$\infty$$.

Problem 1. What is known about the number of ends of infinite finitely generated torsion groups?

In particular,

Problem 2. Is there any description of finitely generated torsion groups with 1 end?

• The theorem you attribute to Stallings is due to Hopf and Freudenthal in 1943/44; more precisely they proved that if a f.g. group has $\ge 3$ ends then its space of ends is a Cantor space. It's also a theorem of Freudenthal (1944) that f.g. torsion groups are $\le 1$-ended (basically he proved that $\infty$-ended groups act as convergence groups on their boundary). Of course this also follows from Stallings' (genuine) later theorem ($\sim 1969$). – YCor Apr 6 at 14:34
• More specifically, Freudenthal (1944) observed that if a f.g. group $G$ has a space of ends $E$ with $|E|\ge 3$ then there exists a clopen subset $X$ of $E$ and $g\in G$ such that $gX$ is a proper subset of $X$. This clearly implies that $g$ has infinite order. – YCor Apr 6 at 14:39
• @YCor Thank you very much for the comments. Could you write them as an answer (with precise coordinates of the Freudenthal paper) and I will accept it as an answer? – Taras Banakh Apr 6 at 14:44
• All these references are given in the introduction to arxiv.org/abs/1901.11073 (whose emphasis is on space of ends of $\infty$-ended infinitely generated groups, which can fail to be Cantor). – YCor Apr 6 at 16:16

Stallings' Theorem, given in his book Group theory and three-dimensinal manifolds, states that a f.g. group with 2 or infinitely many ends splits essentially as a graph of groups with finite edge groups and therefore, being infinite, acts without fixed points on a simplicial tree, and therefore has an infinite order element.

• It's quite hard to find a more inefficient proof! – YCor Apr 7 at 20:46
• @YCor It depends on your definition of efficiency :-) But I aggree that a proof using only the topolgy of the space of ends and the group action is more "efficient". – NWMT Apr 8 at 12:57
• Actually Stallings is enough (from being an amalgam/HNN you directly get a nonabelian free subgroup), so Dunwoody has the only effect of making the proof harder and the generality more narrow (possibly empty, since one doesn't know any infinite f.p. torsion group). Freudenthal's proof is actually proved in the same stream of proving the characterization of 2-ended groups as virtually $\mathbf{Z}$ ones. – YCor Apr 8 at 13:09
• I only cited Dunwoody because Stallings didn't prove his theorem for arbitrary f.g. groups. Stallings Theorem would not have applied to any known torsion groups. – NWMT Apr 8 at 13:19
• Stallings' initial paper was maybe for f.p. groups, but Stallings' 1971 book is for arbitrary f.g. groups and he doesn't not attribute it to Dunwoody, nor even quotes him. – YCor Apr 8 at 13:43

A direct proof. Following Yves' comments above, it is possible to give an easy proof of the fact that an infinitely-ended group must contain an infinite-order element, which implies that it cannot be a torsion group. (It is standard that two-ended groups are virtually infinite cyclic, so I focus on the infinitely-ended case.)

Let $$B$$ be a ball inside your group $$G$$ such that $$G \backslash B$$ has $$n \geq 3$$ unbounded connected components. Let $$C_1$$ be one of them. Now define $$a_1,a_2, \ldots \in G$$ and $$C_2,C_3,\ldots \subset G$$ by induction in the following way:

• Fix an element $$a_i \in G$$ such that $$a_iB$$ disconnects $$C_i$$ into $$n$$ unbounded components.
• Let $$C_{i+1}$$ be such a component which is disjoint from $$C_i$$.

If for every $$i$$ you number the components of $$G \backslash a_iB$$ from $$1$$ to $$n$$, then each $$a_i$$ naturally defines a bijection of $$\{1, \ldots, n\}$$ as it sends the components of $$G \backslash B$$ to the components of $$G \backslash a_iB$$. Up to extracting a subsequence, we may suppose that all the $$a_i$$'s define the same permutation. We get the following picture:

If $$a$$ sends $$3$$ to $$3$$ or $$2$$, then you find easily a subset $$X$$ of the space of ends such that $$aX \subsetneq X$$. In this case, $$a$$ must have infinite order. Similarly, if $$b$$ sends $$3$$ to $$2$$ or $$3$$, there exists some $$X$$ inside the space of ends such that $$b X \subsetneq X$$. In this case, $$b$$ must have infinite order. Now, if $$a$$ and $$b$$ both send $$3$$ to $$1$$, then $$ab^{-1}$$ sends $$1$$ to $$1$$. Next you find easily a subset $$X$$ of the space of ends such that $$ab^{-1}X \subsetneq X$$, so that $$ab^{-1}$$ must have infinite order.

More about infinitely-ended groups. As already mentioned, Stallings' theorem implies that an infinitely-ended group $$G$$ acts on a tree $$T$$ without inversions, fixed-point freely and with finite edge-stabilisers.

As a consequence, finitely-generated torsion subgroups of $$G$$ must be included into vertex-stabilisers. So you know that $$G$$ cannot be itself a torsion group, but it also follows some control on torsion subgroups.

An even stronger consequence of the action of $$G$$ on $$T$$ is that $$G$$ must be hyperbolic relatively to vertex-stabilisers. It implies that $$G$$ is very far from being a torsion group. For instance:

• $$G$$ is SQ-universal, ie., every countable group embeds into a quotient of $$G$$.
• The center $$Z$$ of $$G$$ is finite, and $$G/Z$$ satifies Property $$P_\text{naive}$$, ie., for every finite collection of elements $$g_1, \ldots, g_n \in G/Z$$, there exists some $$g \in G/Z$$ such that $$\langle g,g_i \rangle = \langle g \rangle \ast \langle g_i \rangle$$ for every $$i$$.
• A random element of $$G$$ (chosen thanks to a random walk) has infinite order almost surely. In fact, a random subgroup of $$G$$ (chosen thanks to a fixed number $$n \geq 2$$ of independent random walks) is almost surely free (of rank $$n$$).