Stallings–Swan theorem

Remarks

Stallings proved the theorem for finitely generated groups, and Swan extended it to all groups.

One direction is direct: if $G$ is free on a set $X$, then the augmentation ideal of the group ring $\Z[G]$ is a free $\Z[G]$-module with basis $\set{x - 1 : x \in X}$. This gives a free resolution $0 \to \bigoplus_{x \in X} \Z[G] \to \Z[G] \to \Z \to 0$ of length $1$. For the converse with $G$ finitely generated and nontrivial, a group of dimension $1$ is torsion-free and has more than one end. By Stallings theorem about ends of groups, it splits as a free product or HNN extension over a finite subgroup, which must be trivial. Induction on the number of generators finishes the proof.

Example: every subgroup $H$ of a free group $F$ satisfies $\mathrm{cd}(H) \le \mathrm{cd}(F) \le 1$, so $H$ is free. This recovers the Nielsen–Schreier theorem.

Example: $\mathrm{cd}(\Z^2) = 2$, so $\Z^2$ is not free.

Going deeper

For a student

Serre proved that a torsion-free group has the same cohomological dimension as its subgroups of finite index. With this, the theorem shows that a torsion-free group with a free subgroup of finite index is itself free. In topology, it says that every group of cohomological dimension at most $1$ has a graph as a classifying space.