What is the degree of an n-fold branched cover over a trefoil? The order-2 cyclic branched cover over a trefoil has degree 6, meaning the preimage of any point off the trefoil has cardinality six. (You can find a wonderful video of this here, made by Moritz Sümmermann.) The order-3 cyclic branched cover over a trefoil has degree 24.
What is the formula for the degree of an order-$n$ cyclic branched cover over a trefoil?
(I'm unsure of the proper terminology for this.)
I'm pretty sure this is $|G_n|$, where
$$G_n=\langle x,y\mid xyx=yxy,~x^n=y^n=1\rangle;$$
however, I have no idea how to find the size of this group.


The group
$$
G_n = \langle x,y \mid xyx=yxy,\, x^n=1,\, y^n=1\rangle
$$
is infinite for all $n\geq 6$.
To see this, observe first that the third relation $y^n=1$ follows from the first two, since by the first relation $y=(xy)x(xy)^{-1}$ and hence $y^n=(xy)x^n(xy)^{-1}=1$.  Thus
$$
G_n = \langle x,y\mid xyx=yxy,\, x^n=1\rangle.
$$
Now, it is well-known that the braid group $B_3=\langle x,y\mid xyx=yxy\rangle$ can also be presented as $\langle a,b\mid a^2=b^3\rangle$, where $a=xyx$ and $b=xy$.  Since $x=b^{-1}a$, it follows that
$$
G_n = \big\langle a,b \;\bigl|\; a^2=b^3,\,(b^{-1}a)^n = 1\big\rangle
$$
Now consider the following quotient of $G_n$:
$$
Q_n = \big\langle a,b \;\bigl|\; a^2=b^3=1,\,(b^{-1}a)^n = 1\big\rangle
$$
(It doesn't affect the argument, but $\langle a,b \mid a^2=b^3=1\rangle$ is a presentation for the modular group $\mathrm{PSL}(2,\mathbb{Z})$, which is a quotient of $B_3$.)
What does the Cayley graph of $Q_n$ look like?  If we treat $a$ edges as undirected, then the Cayley graph is the 1‑skeleton of a regular tiling of a simply connected surface by $2n$-gons corresponding to $(b^{-1}a)^n$ and triangles corresponding to $b^3$, with two $2n$-gons and one triangle meeting at every vertex.
If we try to make the polygons regular and Euclidean, then each $2n$-gon has angles of $\pi\bigl(1-\frac{1}{n}\bigr)$ and each triangle has angles of $\pi/3$, so the total angle at each vertex is
$$
2\pi\biggl(1-\frac{1}{n}\biggr) + \frac{\pi}{3} = 2\pi\biggl(\frac{7}{6}-\frac{1}{n}\biggr).
$$
For $n<6$ this sum is less than $2\pi$, so the Cayley graph of $Q_n$ is the 1‑skeleton of a tiling of the sphere, and hence $Q_n$ is finite.  Indeed, the Cayley of $Q_n$ for $n=2$, $n=3$, $n=4$, and $n=5$, are respectively the 1‑skeleta of a triangular prism, a truncated tetrahedron, a truncated cube, and a truncated dodechedron.  Zeno Rogue's computer code shows that $G_n$ is finite in these cases as well.
For $n=6$, the sum is equal to $2\pi$, so the Cayley graph of $Q_n$ is the 1‑skeleton of the truncated hexagonal tiling of the Euclidean plane by equilateral triangles and regular dodecagons.  For $n>6$, the sum is greater than $2\pi$, which means that the Cayley graph of $Q_n$ is the 1‑skeleton of a tiling of the hyperbolic plane.  For example, when $n=7$ this is the truncated heptagonal tiling of the hyperbolic plane. In particular, $Q_n$ is infinite for all $n\geq 6$, and hence $G_n$ is as well.
