Finite simple groups (non-abelian) can generated by two elements.

Let $G=\langle x,y|x^l=y^m=(xy)^n=1,...\rangle$ be a finite simple group (non-abelian), and $\langle x,y|x^p=y^q=(xy)^r=1,...\rangle$ be another presentation of $G$. *(Here, "..." means possibly more relations).*

1) If $(1/l)+(1/m)+(1/n)<1$, then does it imply that $(1/p)+(1/q)+(1/r)<1$?

2) If $(1/l)+(1/m)+(1/n)=1$, then does it imply that $(1/p)+(1/q)+(1/r)=1$?

3) If $(1/l)+(1/m)+(1/n)>1$, then does it imply that $(1/p)+(1/q)+(1/r)>1$?

(For a group $G=\langle x,y|x^l=y^m=(xy)^n=1,...\rangle$ to be a group of symmetries of a compact orientable surface, there are some restrictions on the integers $l,m,n$ and genus of surface, due to Riemann-Hurwitz relation. If the integers (l,m,n) satisfy some conditions as above, then it will allow us to consider less number of presentations of finite simple groups to check for possibility of action of the group on the surface).