Existence (or the number) of generating triple of involutions of $\operatorname{PGL}_2(p)$ with some conditions

Question

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?

Answer 1

Such triples exist, I think.

First, embed $PGL_2(p)$ in $S_{p+1}$ through its action on $1$-spaces from ${\mathbf F}_p^2$. This maps elements of order $p+1$ in $PGL_2(p)$ to $(p+1)$-cycles, and elements of order $p-1$ to $(p-1)$-cycles. In particular, every element of order $p+1$ or $p-1$ gets mapped to an odd permutation.

There are two conjugacy classes of elements of order two in $PGL_2(p)$. Elements from one class fix two $1$-spaces and elements from the other class fix none. So, elements of one class map to even permutations and elements of the other map to odd permutations under the given embedding.

It follows now that if $a,b \in PGL_2(p)$ have order two with $|ab| \in \{p-1,p+1\}$ then one of $a,b$ fixes two points and the other fixes none.

The stabilizer $B$ of a $1$-space in $PGL_2(p)$ is an extension of a cyclic group of order $p$ by a cyclic group of order $p-1$ that acts faithfully. It follows that if $a,b \in B$ are distinct elements of order two, then $|ab|=p$.

Now we are in good shape. Pick $g \in PGL_2(p)$ with $|g|=p+1$. Let $z=g^{(p+1)/2}$. As $g$ acts as a $(p+1)$-cycle, $z$ is a fixed-point-free involution. Now find involutions $a,b$, each with two fixed points, such that $|az|=p+1$ and $|bz|=p-1$. (This is certainly possible by the arguments above.) Now pick any $1$-space $V$ from ${\mathbf F}_p^2$. As $\langle g \rangle$ acts transitively on $1$-spaces, there exist a $\langle g \rangle$-conjugate $x$ of $a$ and a $\langle g \rangle$-conjugate $y$ of $b$ such that both $x$ and $y$ fix $V$.

As $g$ centralizes $z$, we get that $|xz|=p+1$ and $|yz|=p-1$. So, $x \neq y$. Now, as $x$ and $y$ fix a common $1$-space, $|xy|=p$.

Answer 2

You might find helpful the paper of Liebeck and Shalev "Classical Groups, Probabilistic Methods, and the (2,3)-Generation Problem". There, they show that there are three involutions that generate all but finitely many simple classical groups. Indeed, the same is true if you require the involutions to be conjugate.

Their technique is to choose elements at random, and show that with high probability they avoid being all contained in a maximal subgroup. Perhaps this technique can be adjusted for $PGL$, or perhaps their result for $PSL$ will suffice for your application.

Results of the type that you're looking for often go along with work on the (2,3)-generation problem, which asks whether a group is generated by an involution together with an element of order 3. (This question is interesting partly because such groups are exactly the quotients of the modular group $PSL_2(\mathbb{Z})$.)