Quotient of Cayley graph of the free group on two generators by a subgroup.

Question

If $F=F(\{a,b\})$ is the free group on two generators $a$ and $b$ and $G$ is the subgroup $$G=\:\langle b^n a b^{-n}|\: n\in \mathbb{N}\rangle \leq F$$ I am trying to work out what the quotient graph $\Delta / G$ looks like, where $\Delta = \Delta(F;\{a,b\})$, the Cayley graph of F with respect to it's generators.

I think that $\Delta$ is the 4-regular tree, and I think I understand that if you quotient out the Cayley graph by the group itself then all of the vertices become one and so you get a wedge of $|S|$ circles where $S$ is the generating set, but clearly when we quotient out by a smaller group, we see that not all vertices become one.

In fact, $G$ is not finitely generated (as far as I can tell), and so do we get a sort of infinite wedge of circles? I assume we would if we looked at the quotient $\Delta(G;S)/G$ for $S$ the set of generators of $G$, but we are taking a bigger initial group so I would think that it would still be an infinite wedge of circles, but with a 'bigger' infinite number... In that case its only vertex would have infinite valence.

Or do we just think of it as the Cayley graph of the finitely generated group with generators $a,b$ but an infinite set of relations (those given by the elements of $G$?) So is it just an infinite 4-regular graph?

Answer 1

If you think of $F$ as acting by left multiplication on the vertices of $\Delta$, then the natural interpretation of the quotient graph is as a graph whose vertices are the distinct cosets $Gg$ of $G$ in $F$, with an edge (labelled $x$) from $Gg \to Ggx$ for each $x \in \{a,a^{-1},b,b^{-1}\}$.

So, if you ignore the labels, it is still a an infinite 4-regular graph, but it is no longer a tree, and it is not a simple graph, because there are edges labelled $a$ and $a^{-1}$ from the vertex $Gb^n$ to itself for each $n \in {\mathbb N}$. (Does your ${\mathbb N}$ include $0$?)

Note that if you take $n \in {\mathbb Z}$ rather than $n \in {\mathbb N}$, then you get a normal subgroup, and the only vertices are $Hb^n$.

Answer 2

Here is an explicit visualization of $\Delta / G$. First visualize $\Delta$ as the infinite 4-regular tree with the following orientation/labelling pattern: each edge is oriented, each edge is labelled with either $a$ or $b$, and each vertex has one incoming $a$, one incoming $b$, one outgoing $a$, and one outgoing $b$. The quotient graph $\Delta / G$ will be a 4-regular graph inheriting the same kind of orientation/labelling pattern from $\Delta$.

The graph $\Delta / G$ will be a union of two subgraphs $\Gamma_1$ and $\Gamma_2$ meeting at a single point $x$. The subgraph $\Gamma_1$ is a piece of $\Delta$ itself, obtained by picking one $b$-edge, throwing away its interior and leaving two complementary components, and then throwing away the complementary component containing the terminal endpoint of the $b$-edge, leaving $\Gamma_1$ to be the other complementary component containing the initial endpoint $x_1$ of the $b$-edge. The subgraph $\Gamma_2$ is the ray $[0,+\infty)$ with base point $x_2 = 0$, subdivided at the natural numbers $1,2,3,...$, with all edges oriented to the right and labelled $b$, and with an oriented loop labelled $a$ attached to each natural number $1,2,3,...$. The graph $\Delta/G$ is obtained from the union of $\Gamma_1$ and $\Gamma_2$ by identifying $x_1$ and $x_2$ to a single point $x$.

One application of this description is that your generating set is a free basis for $G$: the unique maximal tree of $\Delta/G$ is the union of $\Gamma_1$ and the ray $[0,+\infty)$; and the complementary edges of the maximal tree are precisely the $a$-loops attached to the natural numbers.