Subgroup of the free group on 3 generators I have a question about a subgroup of the free group on three generators, inspired by the following riddle:
Can you hang a painting using a string and two nails so that if either of the nails is removed, the painting falls? [This has been mentioned on the stack before: see this post for a solution and discussion.]
In short, this question is equivalent to asking if there is an element of the free group on $a$ and $b$ whose image under either quotient map $a \mapsto 1$ or $b \mapsto 1$ yields the identity element. (I'm thinking of the free group on two generators as the fundamental group of the plane minus two points; each generator corresponds to wrapping the string around one of the nails.) The commutator $[a,b] = aba^{-1}b^{-1}$ is the simplest solution: the set of elements that work is (I think) exactly the commutator subgroup of $\mathbb{Z} * \mathbb{Z}$. 
I want to know about the analogous question for the free group on $a,b$ and $c$: if $f_a, f_b$ and $f_c$ denote the quotients by the generators $a,b$ and $c$, respectively, then what is the intersection $H$ of the kernels of $f_a, f_b$ and $f_c$? 
$H$ is a normal subgroup of $\mathbb{Z} * \mathbb{Z} * \mathbb{Z}$, since the intersection of normal subgroups is normal, and it's non-trivial: one element that works is $[a,b]c[a,b]^{-1}c^{-1}$. Is there a nice characterization of this subgroup as in the case with two generators? 
This paper has a lot of cool results about this type of question, though their work is mostly about finding the shortest length word that satisfies the condition I'm talking about. As far as I can see, they don't discuss a characterization of all solutions to the painting puzzles. 
If we can find a solution for the free group on three generators, maybe we can generalize: let $H_k^n$ be the intersection of the kernels of all the quotient maps of the free group on $n$ generators by any distinct $k$ generators. (So $H_1^3$ is what I asked about above.) Is there a simple characterization of the elements in $H_k^n$, similar to $H_1^2$ being the commutator subgroup of $\mathbb{Z} * \mathbb{Z}$?
 A: Let us call an expression of the form $[[x,y],z]$, a double commutator,an expression of the form $[[[x,y],z],t]$ a triple  commutator etc.
In the free group on $a_1,a_2,\ldots, a_n$, a $n-1$-commutator of the form $[\ldots[[[x_1,x_2],x_3],x_4] \ldots,x_n]$ where each $x_k$ is a conjugate of a nontrivial power of $a_k$, is clearly in $H_1^n$, so let us call those nice $n-1$-commutators.
Theorem. The subgroup $H_1^n$ is exactly the subgroup generated by the nice $n-1$-commutators.
The argument is by induction on $n\geq 2$, and I find it more convenient to start
with the induction step rather than the base case. So suppose that  the theorem is true
for rank $n-1$. Using  exponential notation $b^a$ to denote the conjugate $aba^{-1}$, we will need two identities :
$$
\begin{array}{lcll}
[x^{g},y^{g}] &=& [x,y]^{g} & (1) \\
[xy,z] &=& [y^x,z^x][x,z] & (2) \\
\end{array}
$$
Note that (2) generalizes to
$$
\Bigg[\bigg(\prod_{k=1}^n x_k\bigg),z\Bigg]=\prod_{k=1}^n [x_k^{x_1x_2\ldots x_{k-1}},z^{x_1x_2\ldots x_{k-1}}] \tag{3} 
$$
For convenience, let us put $c=a_n$.  Define an intermediate commutator to be a commutator of the form $[h,z]$ where $h\in H_1^{n-1}$ and $z$ is a conjugate of a nontrivial power of $c$. Then (3) combined with the induction hypothesis shows that any intermediate commutator is a product of nice $n-1$-commutators. So it will suffice to show that $H_1^n$ is generated by the intermediate commutators.
Denote by $F_k$ the subgroup generated by $a_1,a_2,\ldots,a_k$. Any $w\in F_n$ can be written
$$
w=u c^{i_1} d_1 c^{i_2} d_2 c^{i_3} d_3 \ldots  d_{r-1}c^{i_r}v \tag{4}
$$
where $i_1,i_2,\ldots,i_r$ are nonzero integers, $u,d_1,d_2,\ldots,d_{r-1},v\in F_{n-1}$, and none of the $d_i$ is the identity. Note that $w\not\in F_{n-1}$ iff $r>1$, and that case the decomposition (1) is unique for $w$. In any case, the $r$ in the decomposition is unique, and we call it the $c$-length of $w$. From this unicity, it follows that $w\in H_1^n$ iff $\sum_{k}i_k=0$ (in particular $r\geq 2$), $uv$ and all the $d_k$ are in $H_1^{n-1}$, and $ud_1d_2\ldots d_{r-1}v=e$. 
Let us now take an arbitrary $w\in H_1^{n-1}$ with $c$-length $r\geq 2$, and let us show that by induction on $r$ that $w$ is a product of intermediate commutators. Again, I prefer to treat the induction step first. Decompose $w$ as in (4). Then $u=v^{-1}d_{r-1}^{-1}\ldots d_1^{-1}$, so $w=v^{-1}w'v$ where $w'=d_{r-1}^{-1}\ldots d_1^{-1}c^{i_1} d_1 c^{i_2} d_2 c^{i_3} d_3 \ldots  d_{r-1}c^{i_r}$, and it will suffice to show that $w'$ is a product of intermediate commutators. In other words, replacing $w$ with $w'$ we may assume that $v=e$. But then $w=w''[d_{r-1}^{-1},c^{-i_r}]$ where $w''$ is a word of $c$-length $<r$, so we are done. This analysis also shows that in the base case $r=2$, $w$  can be rewritten as a single intermediate commutator $[v^{-1}d_1^{-1}v,v^{-1}c^{i_1}v]$. This finishes the induction step on $n$.
The base case $n=2$ is similar and simpler : the same argument shows that $w$ is a product
of intermediate commutators, and for $n=2$ intermediate commutators coincide with nice
$1$- commutators. This finishes the proof.
