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?

  • $\begingroup$ This is just a note on notation and LaTeX. In the definition of G, it might be better to use \langle and \rangle in place of < and >, and drop the curly brackets altogether. With some spacing put in, this gives: $G=\langle\, b^n\,a\,b^{-n}\,\mid\,n\in\mathbb{N}\,\rangle$. $\endgroup$ – Bey May 27 '13 at 15:50
  • 1
    $\begingroup$ Thank you, I have changed the formatting of that part. $\endgroup$ – user79474 May 27 '13 at 15:52

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 $Hb^n$.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.