Number of Non-equivalent Generating Triples of Involutions in Finite Simple Groups Let $G$ be a group. We say $(x,y,z)\in G\times G\times G$ is a generating triple of involutions if $|x|=|y|=|z|=2$ and $\langle x,y,z\rangle = G$. A generating triple $(x,y,z)$ of $G$ is said to be non-degenerate if $\{x,y,z\}$ is a minimal generating set.
Let's say two generating triples $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ are equivalent  if there exists $g\in \mathrm{Aut}(G)$ such that $(x_1,y_1,z_1)^g = (x_2,y_2,z_2)^\sigma$, where $\sigma\in S_3$ is a permutation of three entries.
It's known that every non-abelian finite simple group has a generating triple of involutions.
My questions:


*

*How to find the number of non-equivalent generating triples of involutions of non-abelian finite simple groups? (In partucular, $A_n$.) (Added: or some bounds of this number?)

*If $xy = yx$, then what's the number will be?
For the first question, I tried $A_5$ and $A_6$. $A_5$ is easy but $A_6$ needs a lot of time to calculate by hand. For the second question, only the following non-abelian finite simple groups cannot have (see the first 5 references of this article):
\begin{equation*}
A_6,A_7,A_8,L_3(q),U_3(q),L_4(2^n),U_4(2^n),S_4(3),M_{11},M_{22},M_{23},McL.
\end{equation*}

Following up on Derek Holt's suggestion, here is how one could do it in GAP:
gap> f:=FreeGroup("x","y","z");;
gap> g:=f/ParseRelators(f,"x2,y2,z2");;
gap> h:=AlternatingGroup(5);; # or whatever group you like
gap> GQuotients(g,h);
[ [ x, y, z ] -> [ (2,4)(3,5), (2,3)(4,5), (1,2)(3,4) ], ... 19 in total

This returns homomorphisms up to equality of kernels, which is I think exactly what you want. This is clearly limited by group order, but will be much quicker than doing it by hand.
If you also require that the first two generators commute, replace the definition of g by
g:=f/ParseRelators(f,"x2,y2,z2,[x,y]");

