Let G = $<a,b : a^2= b^n = 1 >$ be the (2,n,$\infty$)-triangle group. Define a map $\sigma:G \to Z\_2 \times Z\_n$ via $a \mapsto (-1,1), b \mapsto (1,[1]).$ The kernel H of $\sigma$ is then a torsion free subgroup. With a realization of $a$ and $b$ as elements of PSL(2,R) ($a$ as a rotation and $b$ also as a rotation but with a different center), Veech in this paper [Teichmuller curves, page 573](https://link.springer.com/article/10.1007%2FBF01388890) pointed out that, for this realization of G, H has only one cusp when n is odd, and two cusps when n is even.


I know there is a cusp given by the conjugacy class of $(ba^{-1})^{2n}$, but why H has no other cusps when n is odd and has one other cusp when n is even? And is H a free group?



a normal subgroup of a triangle group
The cusp in $G$ can be described as the conjugacy class of the infinite cyclic group generated by $b a^{-1}$. The intersection of this cyclic group with $H$ is described in two cases. When $n$ is odd then, as you say, the intersection is generated by $(ba^{-1})^{2n}$, and the conjugacy class of this cyclic group in $G$ is also a single conjugacy class in $H$. But when $n$ is even, the intersection is generated by $(ba^{-1})^n$, and the conjugacy class of this cyclic group in $G$ breaks into two conjugacy classes in $H$. 


