# The triangle group $(\alpha,\alpha,\alpha)$ is a subgroup of the triangle group $(3,3,\alpha)$

This answer made me wonder if there is a geometrical way to prove that the triangle group $(\alpha,\alpha,\alpha)$ is a subgroup of the triangle group $(3,3,\alpha)$. In other words, how can we visualize that the fundamental domain of the triangle group $(\alpha,\alpha,\alpha)$ can be obtained joining copies of the fundamental domain of the triangle group $(3,3,\alpha)$?

I am sorry, my comments suggesting that there was no such subgroup were wrong, and I have deleted them.

If $G = \langle a,b \mid a^3, b^3, (ab)^\alpha \rangle$, then the subgroup $H = \langle x,y \rangle$ with $x=ab$ and $y=ba$ has index $3$ in $G$, and has the presentation $\langle x,y \mid x^\alpha,y^\alpha,(xy)^\alpha \rangle$.

This is is easy to prove algebraically using the Reidemieister-Schreier method, but I know you are looking for a geometrical proof, and I expect Lee Mosher's argument will do that.

Here's a partial answer, which is all I have time for at the moment: if it is a subgroup, its index must equal 3. Probably that can be used to either narrow down the possibilities for $\alpha$, and perhaps to rule it out altogether.

Here's the proof that the index equals $3$. By the Gauss-Bonnet theorem, the area of a hyperbolic triangle with angles $a,b,c$ is $$A(a,b,c) = \pi - (a+b+c)$$

So the area of an $(\alpha,\alpha,\alpha)$ triangle is $$A(\pi/\alpha,\pi/\alpha,\pi/\alpha) = \frac{(\alpha-3)\pi}{\alpha}$$

And the area of a $(3,3,\alpha)$ triangle is $$A(\pi/3,\pi/3,\pi/\alpha) = \frac{(\alpha-3)\pi}{3\alpha}$$

Now, if the $(\alpha,\alpha,\alpha)$ triangle group were a subgroup of index $k$ in the $(3,3,\alpha)$ triangle group, then in the hyperbolic plane we would obtain two fundamental domains of the $(\alpha,\alpha,\alpha)$ triangle group: one of area $A(\pi/\alpha,\pi/\alpha,\pi/\alpha)$; and the other of area $k \cdot A(\pi/3,\pi/3,\pi/\alpha)$. Since those fundamental domains must have the same area, we would conclude that $k=3$.

• This works for all $\alpha$. You can make a branch cut between the two points of order 3. In the other direction, the quotient of the doubled equilateral triangle by a 3-fold rotation merges the vertices and turns the top and bottom midpoints into cone points of order 3. Aug 26, 2023 at 1:15