From the introduction: If \(G\) is a group and \(g\) is an element of the commutator subgroup \([G,G]\), the ‘commutator length’ of \(g\), denoted \(\text{cl}(g)\), is the least number of commutators in \(G\) whose product is \(g\). The ‘stable commutator length’, denoted \(\text{scl}(g)\), is the limit \(\text{scl}(g):=\lim_{n\to\infty}\text{cl}(g^n)/n\).
A group \(G\) is said to obey a ‘law’ if there is a free group \(F\) (which may be assumed to have finite rank) and a nontrivial element \(w\in F\) so that for every homomorphism \(\rho\colon F\to G\), we have \(\rho(w)=\text{id}\). For example, Abelian (or, more generally, nilpotent or solvable) groups obey laws. The free Burnside groups \(B(m,n)\) with \(m\geq 2\) generators and odd exponents \(n\geq 665\) are perhaps the best known examples of non-amenable groups that obey laws.
The point of this note is to prove the following: Main Theorem. Let \(G\) be a group that obeys a law. Then \(\text{scl}(g)=0\) for every \(g\in [G,G]\).
The proof is very short, given some basic facts about stable commutator length, which we recall for the convenience of the reader.