Consider the $1$-out-of-$n$ painting hanging problem: Given $n$ nails in a wall, how can we hang a painting such that upon removal of any nail, it falls. This has a nice interpretation as a problem in free groups: Let $F_n$ be the free group on $n$ generators. Then for each $1 \leq i \leq n$ there is a homomorphism $\phi_i : F_n \to F_{n-1}$ that sends the $i$-th generator to the identity. Solutions to the problems are non-zero elements of $H_n = \bigcap_i \ker \phi_i$. Constructions of examples using commutators is pretty easy, better examples of only polynomial length are possible (link).
Yesterday however, I though of a couple somewhat interesting questions, given that $H_n$ is a normal subgroup, as the intersection of kernels:
- Can we describe the quotient group $G_n := F_n/H_n$?
- If we replace the "plane minus $n$ points" with the homotopy equivalent rose with $n$ petals (that is, a graph with a single node and $n$ edges), then can we geometrically describe the covering space associate with $H_n$ under the Galois correspondence?
What I know so far:
Elements of $H_2$ are words on two generators such that the total powers of each generator present sum to zero. Equivalently, $H_2$ consists of precisely elements that vanish if the two generators commute. Thus $G_2 = \mathbb{Z}^2$, and the covering graph s a lattice grid.
Letting $a,b,c$ be generators of $F_3$, $abca^{-1}b^{-1}c^{-1}$ is an element of the commutator subgroup that is not in $H_3$, so $G_3$ is not commutative.