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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?

rjkaplan
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  • $[F_2, F_2]$ is the free group on countably many generators (e.g. by a topological argument looking at fundamental groups of graphs), so you want to figure out $[F_{\infty}, F_{\infty}]$ which I expect is also a free group on countably many generators by a similar topological argument. – Qiaochu Yuan Apr 25 '12 at 03:07
  • @Qiaochu Yuan. I think the topological approach might be messy. To see $F_2^{(n)} = F_\omega.$ Just fix your favorite surjection $f:F_\omega \rightarrow F_2.$ Then $f$ induces a surjection $f':F_\omega' \rightarrow F_2' = F_\omega.$ Abelianizing and recalling that any subgroup of a free group is free, we observe that $F_\omega'$ must equal $F_\omega.$ The result follows by induction. – jspecter Apr 25 '12 at 04:21

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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.