Let me add a quick topological proof that a product of two commutators is not a commutator in a free group.

Solutions to the equation $[a,b]=[c,d][e,f]$ in a free group $F$ correspond naturally to homomorphisms $\Gamma\to F$ where $\Gamma=\langle a,b,c,d,e,f\mid [a,b]=[c,d][e,f]\rangle$. Notice that $\Gamma$ is the fundamental group of $\Sigma$, the orientable surface of genus three!

We will realise $F$ as the fundamental group of a graph $X$. So we can interpret the question as being about homotopy classes of continuous maps $\Sigma\to X$. Next we use a Folklore Theorem, which seems to date back to a circle of ideas explored by Stallings and Zieschang.

**Folklore Theorem.** Let $U$ be a handlebody with $\partial U=\Sigma$ and let $\iota:\Sigma\to U$ be the inclusion map. Up to free homotopy, any continuous map $f:\Sigma\to X$ factors as $\Sigma\stackrel{\alpha}{\to}\Sigma\stackrel{\iota}{\to}U\to X$ where $\alpha$ is a self-homeomorphism of $\Sigma$.

The theorem is not very hard to prove; the idea is to pull back midpoints of edges of $X$ and observe that this gives you an embedded multicurve in $\Sigma$ that dies.

In our case, $\pi_1U$ is a free group of rank three, and so any elements $a,b,c,d,e,f$ of $F$ which satisfy $[a,b]=[c,d][e,f]$ must generate a subgroup of rank at most three. For a counterexample, therefore, take $F$ to be a free group of rank four generated by $c,d,e,f$.

generatedby elements of the form [x,y]. $\endgroup$simplegroup actually is a commutator. $\endgroup$finiteand if k(G) < 2|G:G'|, then G' is ''made'' of only commutators. For those ''in characters'': whenever G haslessnonlinear irreducible (complex) characters than linear ones, G' consists of only commutators. Just think of Q_8, or of the abelian ones. I plan to publish this, but, who on earth will accept it? Of course, it's just asufficientcondition for G' to consist of only commutators - see the above comments on the ''simples''. Marian $\endgroup$1more comment