Triangle group $(\beta,\beta,\gamma)$ is a subgroup of the triangle group $(2,\beta, 2\gamma)$.

Question

Let $1/\alpha+1/\beta+1/\gamma<1$, and let us consider the triangle group $(\alpha,\beta,\gamma)$, i.e. the subgroup of $\mathbb{P}\mathrm{SL}(2,\mathbb{R})$ induced by the hyperbolic triangle with angles $\pi/\alpha,\pi/\beta,\pi/\gamma$.

I have read that the triangle group $(\beta,\beta,\gamma)$ is a subgroup of the triangle group $(2,\beta, 2\gamma)$. How could we prove it? This is probably very standard, but I am not used to work with triangle groups. Any help would be appreciated.

Answer 1

It's the subgroup $\langle xyx,y \rangle$ of $\langle x,y \mid x^2,y^\beta,(xy)^{2\gamma} \rangle$.

The index of this subgroup $2$, so checking that the subgroup has the presentation $\langle z,y \mid z^\beta,y^\beta,(xy)^\gamma \rangle$ (with $z=xyx$) is routine.

Answer 2

The triangle with angles $\frac \pi \beta$, $\frac \pi \beta$ and $\frac \pi \gamma$ is an isosceles triangle. Tracing the height to the uneven side, two triangles with angles $\frac \pi 2$, $\frac \pi \beta$ and $\frac \pi {2 \gamma}$ are obtained. And the assertion follows.