Given integers $a, b, c$ all at least 2, I would like to identify the smallest group with an element $x$ of order $a$, element $y$ of order $b$ such that the product $yx$ has order $c$. For example, if $(a,b,c)=(2,3,7)$ the smallest group with elements $x, y, yx$ of orders $2,3,7$, respectively, is the simple group of order 168.


This problem is related to [triangle groups](http://en.wikipedia.org/wiki/Triangle_group) but here I want to concentrate on finite groups, especially the smallest such. 


This question arose in a discussion on permutation groups in a graduate algebra class. One can always create finite permutations $x, y$ with the desired properties but looking for small groups with these properties seemed to quickly become complicated. (This question may overlap with [this MathOverflow question](https://mathoverflow.net/questions/130697/how-to-find-quotients-of-infinite-triangle-groups-or-von-dyck-groups).) 



What are smallest finite images of triangle groups?
Here's an elaboration of my comment above in the pairwise coprime case, in an even more special case (the fact that in a finite solvable group we can't have three elements of pairwise coprime orders whose product is the identity is, I believe, due to P. Hall). If one insists, as in the question, on the minimality of the order, then if $a,b,c$ are distinct primes, any finite group $G$ of minimal order subject to containing elements $x,y,z$ of respective orders $a,b,c$ such that $xyz = 1\_{G}$ (equivalently, $xy = z^{-1}$), is necessarily a finite simple group. 


For let $N$ be a proper non-identity normal subgroup of $G.$ Then if none of $x,y,z$ is in $N,$ the minimality of $G$ is contradicted, as $xN yN zN = N$ in $G/N.$ If two of $x,y,z$ are in $N,$ then so is the third, and the minimality of $G$ is contradicted again.If (say) $x \in N,$ but neither $y$ nor $z$ lies in $N,$ we have a contradiction since $yNzN = N$ in $G/N$, but $yN$ and $zN$ have coprime orders. Hence there is no such normal subgroup $N$ and $G$ is simple.


Later edit: Here is a general argument which may be relevant, and admits various generalizations: Let $a>5$ be an odd prime, $b>1$ be an odd divisor of $a-1$ and $c>1$ be an odd divisor of $a+1$ (note that $a$ is then neither a Fermat nor a Mersenne prime). Then $G = {\rm PSL}(2,a)$ (which is isomorphic to a subgroup of $S\_{a+1}) $ contains three elements $x,y,z$ of respective orders $a,b,c$ with $xyz = 1.$ 
This is because if $u$ is an element of order $a,v$ an element of order $b$ and $w$ an element of order $c$ in $G,$ then each non-trivial irreducible character $\chi$ of $G$ vanishes at one of $u,v,w.$ Hence $\sum\_{\chi \in {\rm Irr}(G)} \frac{\chi(u) \chi(v) \chi(w)}{\chi(1)}> 0.$ By a standard character-theoretic formula,this means that we may choose conjugates $x,y,z$ of $u,v,w$ respectively such that $xy = w^{-1},$ or $xyw = 1.$
It is also easy to check that $G = \langle x,y \rangle$ for any such $x,y$, since $\langle x,y \rangle$ must have more that one Sylow $a$-subgroup, so has $a+1$ Sylow $a$-subgroups and hence is all of $G.$


