Can the infinite von Dyck groups be subgroups of $SU(n)$? I know by constructing some particular cases that I can find unitary matrices $X$, $Y$ and $Z$ such that
$X^m = Y^n = Z^p = XYZ = 1$
with 
$$
\frac{1}{m} + \frac{1}{n}+\frac{1}{p} < 1
$$
indicating an infinite von Dyck group unless the fact that the matrices are unitary implies some additional, non-trivial relations between $X$, $Y$ and $Z$. Is it possible for infinite von Dyck or triangle groups to be subgroups of $SU(n)$?
 A: Yes, such subgroups exist. The von Dyck subgroups in the first set of these examples already embed in $SU(2)$. Here is the construction. 
Takeuchi in this paper 
"Arithmetic triangle groups" J. Math. Soc. Japan
Volume 29, Number 1 (1977), 91-106. 
gives a complete (finite) list of triples $(p,q,r)$, $2\le p\le q\le r\le \infty$ such that the triangle groups defined by the triple are arithmetic (see section 5 of his paper).
For instance, the first member of his list (in the cocompact case) is $(p,q,r)=(2,3,7)$. Now, take any cocompact arithmetic subgroup $\Gamma< SL(2, {\mathbb R})$ from the list and consider the field $K$ (a nontrivial algebraic extension of ${\mathbb Q}$) generated by the matrix entries of the generators (this is sensitive to conjugation, of course, there is a more invariant construction using traces). Now, $\Gamma< SL(2, K)$. Consider elements $\sigma$ of the Galois group $Gal(K/{\mathbb Q})$.
Of course, each $\sigma$ induces an automorphism of ${\mathbb C}$ as well. 
 Each $\sigma$ determines a nontrivial automorphism of $SL(2, K)< SL(2, {\mathbb C})$ and, hence, a faithful representation 
$$
\rho_\sigma: \Gamma\to SL(2, K)< SL(2, {\mathbb C}). 
$$
Let $G_{\sigma}$ denote the closure of the image of $\rho_\sigma$ in $SL(2, {\mathbb C})$. The point of $\Gamma$ being arithmetic then is that for each $\sigma\ne id$, the group $G_\sigma$ is compact and hence conjugate to $SU(2)< SL(2, {\mathbb C})$. Now, you got your representation. 
It may help to think of the construction as follows, working with subgroups of $O(2,1)$ rather than $SL(2, {\mathbb R})$. Each arithmetic subgroup $\Gamma< SO(2,1)$ is defined as the group of integer automorphisms of a quadratic form $q$ in three variables of signature $(2,1)$ with coefficients in some number field $F$. Automorphisms $\sigma\in Gal(F/{\mathbb Q})$ define a new quadratic form $q^\sigma$. Arithmeticity of $\Gamma$ precisely means that for each nontrivial $\sigma$, the form $q^\sigma$ is definite. Hence, the associated faithful  representation
$$
\rho^\sigma: \Gamma \to GL(3, {\mathbb R})
$$
sends $\Gamma$ into the group of automorphisms of definite quadratic form which, therefore, is isomorphic to $O(3)$. 
One can prove more, in fact:
Theorem. For each von Dyck group $\Gamma=\Gamma(p,q,r)$
there exists a faithful representation $\Gamma\to SU(n)$ for some $n$ (depending on $\Gamma$). 
Proof. Take first one of the arithmetic examples I just described, say, $\Gamma(2,3,7)$ and embed it in $SU(2)$. Since  $\Gamma(2,3,7)$ is cocompact, it contains a surface subgroup of some genus $g$ (actually, $g=3$ in the $(2,3,7)$ case). Therefore, every surface group of genus $\ge g$ embeds in $SU(2)$. Now, turn to $\Gamma(2,3,7)$. This group contains a torsion-free subgroup $\Gamma_1$ of finite index. Thus, $\Gamma_1$ is a surface group, i.e. is isomorphic to the fundamental group of a compact (and without boundary) and genus $h\ge 2$. By passing to a further finite index subgroup if necessary we can assume that $h\ge g=3$. 
Consider (by restriction from $\Gamma(2,3,7)$) a faithful representation $\rho: \Gamma_1\to SU(2)$. Now, let 
$$
\rho'=Ind_{\Gamma_1}^{\Gamma(p,q,r)} \rho
$$ 
denote the induced representation of $\Gamma(p,q,r)$ (induced from $\rho$). One then checks that $\rho'$ is still faithful and its image lies in a compact subgroup $K$ of $GL(n, {\mathbb C})$ for some $n$ (since this was true for $\rho$). Both are quite easy once you know what the induced representation is. 
Lastly, being a compact subgroup,  $K$ is conjugate to $SU(n)$ and you get a faithful representation $\Gamma(p,q,r)\to SU(n)$. qed
Here is what I do not know how to prove or disprove (I think it is true though):
Conjecture. There exists $n$ such that every $\Gamma(p,q,r)$ admits a monomorphism to $SU(n)$. 
