---
tag: B41GE4M134
kind: statement
role: theorem
name: Stallings–Swan theorem
msc: [20J05, "20E05"]
uses: [EF4ZWACHD8, 0KTW26C80J]
about: [EF4ZWACHD8, 0KTW26C80J]
sources:
  - url: https://en.wikipedia.org/wiki/Cohomological_dimension
---

## Statement

Let $G$ be a group.
Then $G$ has [[cohomological dimension|EF4ZWACHD8]] at most $1$ if and only if $G$ is a [[free group|0KTW26C80J]].

## 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|GWZN3EKEWD]], 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|W97X7WX5G8]].

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

## Explanations

### L1 human

Free groups are the groups with no relations among their generators.
The theorem says they are also exactly the groups whose cohomology dies above degree one.
So an algebraic condition on cohomology detects a combinatorial property of the group.

### L2 human

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|WA796ZCNRX]].