Let $G=\operatorname{PGL}_2(p)$, where $p\ge 5$ is a prime. Is there a generating triple of involutions $(x,y,z)$ of $G$ such that $|xy|=p$, $|xz|=p+1$ and $|yz|=p-1$? That means, $\langle x,y,z\rangle=G$ with

1. $x^2=y^2=z^2=1$;

2. $\langle x,y\rangle \cong D_{2p}$;

3. $\langle x,z\rangle\cong D_{2(p+1)}$;

4. $\langle y,z\rangle\cong D_{2(p-1)}$.

We know that when $p\equiv 1\pmod 4$ we have $x,y\in\operatorname{PSL}_2(p)$, while when $p\equiv 3\pmod 4$ there is exactly one of them in $\operatorname{PSL}_2(p)$ (but we do not know which one is).

Is there any result on this?

Let $G=\operatorname{PGL}_2(p)$, where $p\ge 5$ is a prime. Is there a generating triple of involutions $(x,y,z)$ of $G$ such that $|xy|=p$, $|xz|=p+1$ and $|yz|=p-1$? That means, $\langle x,y,z\rangle=G$ with

1. $x^2=y^2=z^2=1$;

2. $\langle x,y\rangle \cong D_{2p}$;

3. $\langle x,z\rangle\cong D_{2(p+1)}$;

4. $\langle y,z\rangle\cong D_{2(p-1)}$.

If there exists, then what is the number of such triples?

We know that when $p\equiv 1\pmod 4$ we have $x,y\in\operatorname{PSL}_2(p)$, while when $p\equiv 3\pmod 4$ there is exactly one of them in $\operatorname{PSL}_2(p)$ (but we do not know which one is).

Is there any result on this?