Given an hyperbolic triangle group, presented as $$ \langle a,b,c \mid a^2=b^2=c^2=(ab)^p=(bc)^q=(ca)^r=1\rangle \text{, with } \frac1p+\frac1q+\frac1r<1. \tag1 $$
As pointed out in comments, these groups are infinite. But what if I'm able to construct finite groups that respect the same relations as given in $(1)$. An example, based on graphs, is given here. Does this mean that
the finite groups are subgroups of the infinite one?
This problem seems to be the common "are quotient groups the same as subgroups?" problem, as adding relators in corresponds to quotienting by the normal closure of those relators. If so, then the easiest counter-example which comes to mind is $\mathbb{Z}$ under addition. Here, every proper quotient group is finite cyclic, while every non-trivial subgroup is infinite cyclic. Therefore, the only group occuring as both a quotient group and a subgroup is the trivial group.
Okay, lets now take the question seriously. There are groups for which every quotient group occurs as a subgroup (e.g. finite cyclic groups), so the question asked here is not unreasonable, and possibly even interesting! The answer for hyperbolic triangle groups is:
No. Every hyperbolic triangle group $\Delta(p, q, r)$ has finite quotient groups which are not isomorphic to subgroup of $\Delta(p, q, r)$.
A group $G$ is residually finite if for every non-trivial element $g\in G$ there exists a homomorphism $\phi_g:G\rightarrow H_g$ such that $g\not\in\ker(\phi_g)$ and $H_g$ is finite (equivalently, there exists a normal subgroup $N_g$ of finite index in $G$ such that $g\not\in N_g$). Triangle groups are residually finite, and the above result is in fact true for every residually finite hyperbolic group*.
Proof. As hyperbolic triangle groups are infinite and residually finite, they have quotient groups of unbounded order (for all $n\in\mathbb{N}$ there exists a quotient group of order greater than $n$). In particular, they have infinitely many isomorphism classes of finite quotient groups. On the other hand, hyperbolic groups have only finitely many conjugacy classes of finite subgroups; hence they have only finitely many isomorphism classes of finite subgroups. Therefore, there are more isomorphism classes of quotient groups than quotient groups, and the result follows. QED
We can actually say something concrete about the finite subgroups of these triangle groups $\Delta(p,q,r)$ when $p, q, r>6$**. Here, the presentation $\Delta_o(p,q,r)=\langle x, y, z\mid x^p, y^q, z^r, xyz\rangle$ defines an index-two subgroup (this is the group of orientation-preserving symmetries, and is often itself referred to as a "triangle group"). We can then cancel the $z$-generator to get the presentation $\langle x, y\mid x^p, y^q, (xy)^r\rangle$, which satisfies the $C'(1/6)$ small cancellation condition. Therefore, by results from small cancellation theory (see previous link), every finite subgroup is cyclic, and indeed conjugate to a subgroup of one of the subgroups $\langle x\rangle$, $\langle y\rangle$ or $\langle z\rangle$. Hence, every finite subgroup of $\Delta(p,q,r)$ contains one such subgroup of finite index. So, for example, the largest possible order of a finite subgroup of $\Delta(p,q,r)$ is $\max(2p, 2q, 2r)$.
*The point about residual finiteness underlies JCAA's now-deleted partial answer. The earliest citation I can find for residual finiteness of these triangle groups actually proves a strictly stronger property called LERF: Scott, Peter. "Subgroups of surface groups are almost geometric." Journal of the London Mathematical Society 2.3 (1978): 555-565.
**The same result holds without the restriction of $p,q,r>6$, using the theory of Fuschsian groups, but properly working out how the bits of this theory fit together is not a rabbit hole I wish to crawl down today.
