@Derek Ah a cycle of length $|S|$. I just thought they meant any cycle! That makes the question a lot more understandable. The idea about dihedral groups sounds promising, can you elaborate this, maybe as an answer?

Still I think the more general question of what sets of involutions generate the symmetry group is very interesting, I'd like to learn more about that and hope some answer to this question can shed some light on that

Maybe I am missing something but if $a_1, a_2$ and $a_3$ all leave some element $s \in S$ fixed, then so does every element of the group they generate. E.g. if we interpret the example in the comment above as acting on $S = \{1, \ldots, 7\}$ instead of $\{1, \ldots, 6\}$ then it is clear that they do not generate the whole symmetry group since they won't generate any element that sends 7 to 1 (or any other element)

+1, but what is EGNO?

Ehm after I typed this I am starting to think that maybe it is even simpler and $\phi(x, y)$ is simply $e^{[x, y]}$. But I guess you already thought of that and dismissed it for some reason. Can you elaborate about what your thoughts were?

I did not think to deeply about it so maybe this is non-sense but at first glance it seems that the explicit form of $\phi$ is what the Baker–Campbell–Hausdorff formula tries to achieve. (Wikipedia: en.wikipedia.org/wiki/…)

It seems to me that the original formulation was the right one and the edit introduced an error. But then again I know virtually nothing about this subject, so it seems likely that I am wrong. Either way I would be happy if someone could clarify this

I am confused now about the $x$ nilpotent vs $n$ nilpotent business. The way it is written now in the post it looks like the homomorphism is very picky and only accepts nilpotent elements as input. But that is not how homormorphisms are supposed to work, right? The domain should be the whole group

@AlexM. Links are mysterious creatures. For me it doesn't work when clicking it, but it does work when right-clicking it and selecting 'open link in new tab'. No idea why, but this might also work for you...

This TeXromancers project is amazing! Any ideas what books they (you?) will do next?

I like the idea of using two initials (first and last name)! It makes the abbreviation look more like a person and less like a mystery. Now that we know this is common in physics, I think we can all start doing that in mathematics also without feeling ashamed.

For my understanding: $1/2d^2$ is $1/(2d^2)$, right, not $(1/2)d^2$? Or is there some simple-in-retrospect miracle that I am overlooking?

aaah right, I was mixing up multiplication and addition. It is indeed obvious in retrospect. Thanks

If you have time, can you elaborate a bit on why this sentence is true? "If the group $X$ is commutative, then each polynomial is of the form $f(x)=ax^n$ for some $a \in X$ and $n \in N$