Let C(p,q) be the Coxeter group:


$C(p,q):= \langle a,b,c\hspace{1mm}|\hspace{1mm} a^2,b^2,c^2,(ac)^2,(ab)^p, (bc)^q \rangle$


for integers $p,q$ s.t. $\frac{1}{p}+\frac{1}{q}<\frac{1}{2}$. This group is infinite and one ended. 


Let $G$ an infinite group obtained from $C(p,q)$ by adding some relations to the presentation above. Can $G$ have 2 or infinitely many ends? 



Ends of quotients of Coxeter Groups
No. Your group has Serre's Property FA, meaning that any action on a tree has a global fixed point. (This can be deduced from the fact that it has a generating set such that every element is torsion, and some product of each pair of elements is also torsion.)


Suppose now that some quotient $G$ has more than one end. By Stallings' Theorem, $G$ acts on a tree without a global fixed point, so $C(p,q)$ does too. This is a contradiction.


