PSL(2,p) as quotient of triangle groups As a by-product of some Magma computations, I've observed that, for each prime $p$
such that ${\rm PSL}(2,\mathbb{F}_p)$ can be a quotient of the $(3,3,4)$-triangle group
(i.e. $p \equiv \pm 1 (\!\!\!\mod 8)$), there seem to be exactly two normal subgroups
of the $(3,3,4)$-triangle group such that the quotient is isomorphic to
${\rm PSL}(2,\mathbb{F}_p)$.  Surely this is a known fact.  Can anyone please provide a proof or a reference?
 A: I think I can show that there are at most $2$ for all primes $p$ using rigid triples / rigid local systems.
We are looking at triples $x,y,z$ in $SL_2(\mathbb F_p)$ that generate the group and multiply up to $1$, with $x$ and $y$ of order $3$ and $z$ of order $4$, up to global conjugacy
Any such triple can be lifted to $SL_2(\mathbb F_p)$ in four different ways, and there is a unique way of doing it so that $x$ and $y$ remain order $3$. Then $z$ will be order $8$. It is sufficient to show that there are only two of these up to conjugacy.
There is one conjugacy class of order $3$ in $SL_2(\mathbb F_p)$ but there are two conjugacy classes of order $8$, because there are four primitive $8$th roots of unity in two pairs that each multiply to $1$.
I claim for each conjugacy class, there is a unique triple of two order $3$ elements, and an order $8$ element in that conjugacy class, that multiply to one up to global conjugacy. That is, it is a "rigid triple".
Given two such triples, view them as representations of the free group $F_2$ and hence as sheaves on $\mathbb P^1$ minus three points. Take the tensor product $V \otimes W^{\vee}$, and take the parabolic cohomology / cohomology of its middle extension to $\mathbb P^1$. This is a rank $4$ sheaf on a surface of Euler characteristic $2$, for an Euler characteristic of $8$, but it drops to a rank $2$ sheaf at the $3$ singular points, for a contribution of $-6$, so the total is $2$. 
Because the Euler characteristic is positive, $H^0$ or $H^2$ must be nonvanishing. But that can only happen if the sheaf has a global section, which would be an isomorphism between the two representations, showing that the two triples are conjugate.
Hence there are at most two pairs, one for each conjugacy class. By Ian Agol's argument we can show that they are both inhabited when $p \equiv \pm 1\mod 8$, so there are exactly two in that case. 
A: I'm guessing that this can be explained by some number theory/algebraic geometry for all but finitely many primes $p$. This triangle group sits inside a quaternion algebra defined over $\mathbb{Q}(\sqrt{2})$, which in turn sits inside $M_2(\mathbb{C})$. From computations of Takeuchi, it looks like this quaternion algebra splits at all odd primes. Now, if an odd prime splits over $\mathbb{Z}[\sqrt{2}]$, then one may localize the quaternion algebra at this prime, and take the quotient algebra which is $M_2(\mathbb{Z}_p)$, then map the units to $PSL_2(\mathbb{F}_p)$. But there will always be two such primes, so one obtains two such representations. The primes split precisely when $p\equiv \pm 1 (\mod 8)$, so I think this accounts for all your representations. I also believe that there's a principal of Grothendieck which ought to imply (together with the uniqueness of the $PSL_2(\mathbb{C})$ rep. of the $(3,3,4)$ triangle group) that all but finitely many $PSL_2(\mathbb{F}_p)$ quotients occur this way, but I am not aware of a specific reference.  
