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Let $G = PSL(2,q)$. I'm interested in the Cayley graphs of $G$ generated by triples $(A,BAB^{-1},B^{-1}AB)$, where $A, B \in G$ are elements of order $2, 3$ respectively: such a triple generates all of $G$ (so that the Cayley graph is connected) iff the corresponding pair $(A,B)$ does.

Let $n$ denote the order of $AB$, and call $(A, B, (AB)^{-1})$ a $(2,3,n)$-triple: the idea is that it's the image of the generators of the triangle group $T(2,3,n) = \langle a,b,c|a^2=b^3=c^n=abc=1 \rangle$ under some map $T(2,3,n) \rightarrow G$. Glover and Sjerve show that the map is surjective, i.e. $(A,B)$ generates $G$, iff there are elements of order $n$ in $G$, but not in any of its proper subgroups of the form $PSL(2,q')$ or $PGL(2,q')$.

My question is: given that $n$ satisfies this condition, what effects does its value have on the properties of the Cayley graph generated by $(A,BAB^{-1},B^{-1}AB)$? I'm particularly interested in girth, diameter and the eigenvalues $\lambda_2, \lambda_{|G|}$.

If it helps, I'm willing to restrict to the case when $q = 2^m$ for some $m$, since then the only subgroups we need worry about are $PSL(2,2^{m'})$ = $PGL(2,2^{m'})$ where $m'|m$.

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    $\begingroup$ what do you mean "there are elements of order $n$ in $G$, but not in any of its proper subgroups"? This will only be true when $G$ is cyclic of order $n$, but I guess you're still assuming $G=PSL(2,q)$... Perhaps you mean subspace subgroups?? $\endgroup$
    – Nick Gill
    Sep 4, 2014 at 18:22
  • $\begingroup$ Yes, sorry, what I wrote was nonsense. I'll correct it now. $\endgroup$ Sep 4, 2014 at 18:36
  • $\begingroup$ It might help to give a little motivation also - why is this particular triple of elements of interest?? $\endgroup$
    – Nick Gill
    Sep 4, 2014 at 18:43
  • $\begingroup$ The motivation is merely that such a generating set is the smallest possible which is both closed under inverses and "symmetric", in the sense that it's closed under conjugation by B. $\endgroup$ Sep 4, 2014 at 19:01

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