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This may not be a research-level question, which is why I submitted it to math.stackexchange first, but so far the question there has barely been viewed, let alone answered. Apologies if submitting it here as well is poor etiquette.

Let $G = PSL(2,q)$. I am interested in the different ways of writing $G$ as an image of the modular group. If $A, B \in G$ are elements of order $2, 3$ respectively, then $\langle A,B \rangle$ is an image of the modular group (and is quite often the whole group $G$). Call $(A,B)$ a $(2,3)$-pair, and say that the pairs $(A,B)$, $(A',B')$ are equivalent if there is an automorphism of $G$ taking one pair to the other.

For example, I looked at the case $q = 5$ and there was exactly one equivalence class of $(2,3)$-pairs for each possible order $2, 3, 5$ of the element $AB$ (unless I made a mistake).

Is this true in general? If not, is there a nice way to list the equivalence classes in $G$?

My motivation for asking about this is here.

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    $\begingroup$ Regarding the reposting of a question previously asked on MSE: usually you should wait a bit more than just 17 hours -- say at least a week or so. It may take a bit until people find time to answer. Also, it may help people find your question on MSE if you tag it in a better way -- e.g. with a general group theory tag as you have done here. $\endgroup$
    – Stefan Kohl ♦
    Commented Oct 1, 2014 at 16:19
  • $\begingroup$ Ok, thanks. I will be sure to follow this advice in future. $\endgroup$
    – Robin Saunders
    Commented Oct 1, 2014 at 17:16

3 Answers 3

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Concerning the discussion for previous answer (I'm not allowed to comment yet), the discrepancy comes between counting (2,3,5)-subgroup generator tuples, or subgroups of type A5:

A5 has two non-conjugate (2,3,5)-triples (that are conjugate under S5), so it is plausible that there are two orbits of generator triples and one orbit of subgroups.

Concerning the original question, this gets quickly more complicated for larger $q$: The GAP code:

f:=FreeGroup("x","y");
f:=f/ParseRelators(f,"x^2,y^3");
GQuotients(f,PSL(2,7)); # replace 7 with the q you like

determines 2,3 generating pairs of $PSL_2(q)$ up to automorphisms of $PSL$. For $q=71$ I get 30 such pairs and the orders of the product of the generators can be 7,9,12,18,35,36,71.

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  • $\begingroup$ Welcome on MathOverflow! $\endgroup$
    – Stefan Kohl ♦
    Commented Oct 10, 2014 at 13:12
  • $\begingroup$ As I can comment now :-) Thank you! $\endgroup$
    – ahulpke
    Commented Oct 10, 2014 at 20:57
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While your observation is true also for ${\rm PSL}(2,7)$, ${\rm PSL}(2,8)$ and ${\rm PSL}(2,9)$, it is false for ${\rm PSL}(2,11)$. -- In ${\rm PSL}(2,11)$ there are two orbits of $(2,3)$-pairs whose product has order $5$, as you can check with GAP as follows:

gap> G := PSL(2,11);;
gap> A := AutomorphismGroup(G);;
gap> pairs := Cartesian(Filtered(AsList(G),g->Order(g)=2),
>                       Filtered(AsList(G),g->Order(g)=3));;
gap> orbits := Orbits(A,pairs,OnTuples);;
gap> List(orbits,orbit->Order(Product(orbit[1])));
[ 5, 5, 11, 3, 2, 6 ]
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    $\begingroup$ Thanks for this. I confess to not really having encountered GAP before, so I'll try playing around with it before asking any more questions of this sort. $\endgroup$
    – Robin Saunders
    Commented Oct 1, 2014 at 17:23
  • $\begingroup$ I'm slightly confused - I think this question neglects the OP's consideration of the automorphism group of $G$. In the case where you have a $(2,3)$-pair whose product is $5$ they will generate a group isomorphic to $A_5$. Subgroups of this form exist in $PSL_2(q)$ exactly when $q=-1,0,1\pmod 5$. What is more if one considers the automorphism group of $G$, then all subgroups isomorphic to $A_5$ are conjugate. Now Robin's calculations for $q=5$ suggest that all such pairs should be conjugate in the automorphism group of $PSL_2(q)$. $\endgroup$
    – Nick Gill
    Commented Oct 2, 2014 at 17:30
  • $\begingroup$ .. Thus I think the two orbits found by Stefan will fuse in $PGL_2(11)$. And, in general, there will be one orbit of $(2,3)$-pairs with product equal to $5$ in the automorphism group of $G$. For other product orders things will be different I guess. $\endgroup$
    – Nick Gill
    Commented Oct 2, 2014 at 17:31
  • $\begingroup$ One further thing: For products of order equal to $2$, $3$ and $4$, I guess the problem reduces to studying the number of orbits in $Aut(G)$ of subgroups isomorphic to $S_3$, $A_4$ and $S_4$. I'm guessing there's only one such subgroup in the latter two cases, not sure about more generally. (This should be a tractable problem though.) For product $6$ the pair commutes so this is easy. For product $7$ one is studying Hurwitz groups and so the pair will generate $PSL_2(p)$ where $p$ is the prime. Now one can check how many non-conjugate triples generate any given $PSL_2(p)$. $\endgroup$
    – Nick Gill
    Commented Oct 2, 2014 at 17:47
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    $\begingroup$ My mistake! Alexander's answer below shows my error - the triples fuse in $S_5$ (explaining Robin's calculation in the question) but, of course, there is no $S_5$ in $PGL_2(q)$... $\endgroup$
    – Nick Gill
    Commented Oct 10, 2014 at 13:25
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The paper "Arc-transitive cubic cayley graphs on PSL(2,p)" by Du and Wang seems to give a definitive answer to my question for the case where q is prime.

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