# 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 $\mathbb{Z} * \mathbb{Z}$?

• Is it just the commutator subgroup? What is the proof of this? I would like a simple description of the elements of that subgroup, as words on three generators. – J Richey Mar 10 '17 at 0:40
• Well, elements of the commutator subgroup are products of commutators. As for the proof that the intersection is the commutator subgroup, try to prove one of the inclusions. Which one did you manage to prove? – Moishe Kohan Mar 10 '17 at 0:59
• @MoisheCohen I don't think it is, $[a,b]$ is not in this group since it isn't in the kernel of $f_c$ – Paul Plummer Mar 10 '17 at 2:29
• @PaulPlummer Oh, you are right, I misread the definitions of these homomorphisms. – Moishe Kohan Mar 10 '17 at 2:57
• I don't think it answers the question, but these are called "Brunnian words" in the paper "Brunnian Links" by Paul Gartside and Sina Greenwood (Fundamenta Mathematicae 193(2007), 259-276), which contains a lot of information that might be of interest. impan.pl/en/publishing-house/journals-and-series/… – Jeremy Rickard Mar 12 '17 at 11:15

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.

• As awful as the name 'nice' is, your description of the elements is nice... Very cool! Any conjecture about $H_k^n$? – J Richey Mar 17 '17 at 9:52
• I could not find any better word than "nice", any suggestions ? Perhaps $H^n_k$ has a generating set described by commutators too, but I'm not sure. – Ewan Delanoy Mar 17 '17 at 9:56