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?
Best Answer
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 $Hb^n$.