Especially the first paragraph of this answer reads like ChatGPT wrote it. It’s always better to rewrite it into your own words (the tone is quite strange).

The relations given in the other question indeed seem wrong, but if you just follow the paper it links to (by Robertson & Williams) then you can easily find a presentation for $\operatorname{GL}(2,p)$ in there. Properly looking into all the information you've already been given is surely the bare minimum before you ask another question!

You say the relations between the generators for $\operatorname{GL}(2,p)$ were not given in your previous question, but this is not true. You were given a link to the book by Coxeter & Moser, which contains presentations for $\operatorname{GL}(2, p)$ for all primes $p$ on pp. 95-96.

It would be nice to know whether $\pi_1(X)$ is free or not.

@PaulPlummer It is worth mentioning that this fact about sums of minimum relators in presentations of free products had been proved by Rapaport already 20 years before Hog, Lustig & Metzler.

@E.Rauzy I wouldn't bet money on that conjecture being true. It's decidable, for example, whether a finitely presented group with decidable word problem is free or not.

What if the cannonballs are reflective? Is there a ray starting in one face of the pyramid and ending in another?

@HJRW For the “more thought”, it think (?) it should be sufficient to take each generator to the $n!$th power, where $n$ is the bound you mention, and check if those matrices commute (up to precision). If they do, then your subgroup has an abelian subgroup of finite index. Then it just suffices to check that all your matrices have finite order — surely easy, up to precision, too?

In addition to @AndyPutman ‘s comments about the subtlety of posing the problem correctly, please note that the spectre of undecidability of most decision problems in group theory hangs over your question. Deciding if an element has finite order, for example, or deciding if a group is finite, is precisely the kind of thing that is undecidable in general.

Here’s a heuristic for why heuristics work: because they’ve worked in the past.

A former example: every subgroup of a surface group is a surface group (if finite index) or free (if infinite index). This was well-known by topological means for a long time (certainly Klein and Fricke knew it) but only given an algebraic proof in the 1970s.

I did some digging. I can write a more complete answer soon (and I have some pending hooks out). Here are some things in the meantime: he (Goeff Croes) did his PhD at the University of Groningen in 1945, and worked at Shell between 1955 and 1978. He then moved to working as a software engineer in astronomy, eventually working at the DRAO in Canada. There he was part of designing the AIPS++ programming language. He retired in 1993. I don't yet know why he couldn't be contacted.

Sure. But right now your definitions are not saying that. Saying that a set is isomorphic to another is just saying that they have the same cardinality, which is much less than you want. Your new formulation has the same issue. It seems much easier to just say that you have a graph which is the Cayley graph of $\mathbb{Z}_3^n$ with respect to some symmetric generating set $C$.

“Its vertex set is isomorphic to $\mathbb{Z}_3^n$” just means it has $3^n$ vertices. You mean that it’s equal to that set (this also matches how you use the vertices later).

@FrancoisZiegler This is a tiny remark, but "quoth" and "quote" are etymologically completely unrelated, so using "quoting" is better for a quote.

@non-euclideangeometry I don’t think your indirect, slightly cryptic comments are very helpful to OP at all. Could you rephrase them in a more direct way?