What is $[[H, H], [H, H]]$, where $H$ is the free group on two elements $a,b$?

Question

I think that when $H$ is the free group on two elements, $[[H, H], [H, H]]$ is the trivial group. I feel that this shouldn't be too hard to prove/disprove, but I've got nothin'. Ideas?

Answer 1

I don't think that's true. Let $x$ and $y$ be the generators. Then $a = xyx^{-1}y^{-1}$ is a commutator, and (using $xy$ and $yx$) so is $b = xyyx(xy)^{-1}(yx)^{-1} = xy^2xy^{-1}x^{-2}y^{-1}$. Your assertion implies that $aba^{-1}b^{-1} = 1$. However, we compute: \begin{align} aba^{-1}b^{-1} &= xyx^{-1}y^{-1}xy^2xy^{-1}x^{-2}y^{-1}yxy^{-1}x^{-1}yx^2yx^{-1}y^{-2}x^{-1}\\ &= xyx^{-1}y^{-1}xy^2xy^{-1}x^{-1}y^{-1}x^{-1}yx^2yx^{-1}y^{-2}x^{-1} \end{align} which is not the identity.

That was a terrible proof. Here is a better one. Take $S_n$. The commutator subgroup of $S_n$ is $A_n$, and the commutator subgroup of $A_n$ is again $A_n$ for $n > 4$ or else $A_n$ wouldn't be simple. Since any $S_n$ is generated by two elements and hence receives a surjection from your free group, the twice- iterated commutator subgroup of your free group can't be trivial.