Stallings theorem about ends of groups

Remarks

The number of ends of a finitely generated group is always $0$, $1$, $2$ or $\infty$. It is $0$ exactly for finite groups and $2$ exactly for infinite groups with an infinite cyclic subgroup of finite index.

The easy direction: if $G$ splits over a finite subgroup, then $G$ acts on the Bass–Serre tree of the splitting. Removing a single edge of this tree separates the Cayley graph into at least two infinite parts. The hard direction, due to Stallings, builds from a set of vertices with finite boundary in $\Gamma$ a tree on which $G$ acts with finite edge stabilisers.

Example: $\Z$ has two ends, since removing one vertex of the line leaves two rays; and $\Z$ is the HNN extension of the trivial group over the trivial subgroup. Example: the free group $F_2 = \Z * \Z$ is an amalgam over the trivial group, and it has infinitely many ends. Example: $\Z^2$ has one end, since the complement of a finite set in the grid has one infinite component; so $\Z^2$ does not split over a finite subgroup.

Going deeper

For a student

The theorem is a cornerstone of Bass–Serre theory and of geometric group theory. Since the number of ends is a quasi-isometry invariant, splitting over a finite subgroup is a quasi-isometry invariant property. It implies that a finitely generated torsion-free group with a free subgroup of finite index is itself free. Iterating splittings over finite groups leads to the accessibility theorem of Dunwoody for finitely presented groups.