*[Cross-posted from MSE]*

Consider the Von Dyck group

$$ G = \langle x,y\mid x^a=y^b=(xy)^c=1\rangle $$ where $a,b,c\ge3$. Because $G$ is infinite and residually finite, it has an infinite family of finite quotients $\{\overline{G}_n\}$ such that $|\overline{G}_n|\rightarrow\infty$. I'm wondering if there is an expicit such family that can be constructed, giving a direct proof that $G$ is infinite.

Specifically, I'm trying to use Derek Holt's argument (that I copied in this MO answer), which can be used to show $G$ has quotients in $PSL(2,q)$ for infinitely many primes $q$. To summarize: Let $q$ be a prime such that $q-1$ is divisible by $2a$, $2b$, and $2c$. Let $\zeta$ be a primitive element of $\mathbb{F}_q^\ast$, and define $\zeta_t=\zeta^{(p-1)/t}$, so that $|\zeta_t|=t$. Then define the matrixes $X,Y\in SL(2,q)$ as \begin{equation*} \begin{aligned} X &= \begin{pmatrix}\zeta_{2a} & 1\\ 0 & \zeta_{2a}^{-1}\end{pmatrix} \end{aligned}\qquad \begin{aligned} Y = \begin{pmatrix}\zeta_{2b} & 0\\\lambda & \zeta_{2b}^{-1}\end{pmatrix} \end{aligned} \end{equation*} where $\lambda = (\zeta_{2c}-\zeta_{2a}\zeta_{2b})+(\zeta_{2c}^{-1}-\zeta_{2a}^{-1}\zeta_{2b}^{-1})$. Then the images $\overline{X},\overline{Y}\in PSL(2,q)$ generate a subgroup $\langle \overline{X},\overline{Y}\rangle$ that is a quotient of $G$.

It seems to me in this case we should actually have $\langle \overline{X},\overline{Y}\rangle=PSL(2,q)$. Is this true, and how hard is the proof? Equivalently (but perhaps easier to show), is it true that $\langle \overline{X},\overline{Y}\rangle$ contains a Sylow $q$-subgroup of $PSL(2,q)$?

**Note**: While I am mostly interested in the specific question of whether $\langle \overline{X},\overline{Y}\rangle=PSL(2,q)$, there are other ways to exhibit an infinite family $\{\overline{G}_n\}$. For example, by reducing to the (perhaps more tractable) case where $a,b,c$ are odd primes or $4$. Or if there are other explicit infinite families -- especially linear -- I'd like to hear about them!