Cusps in fundamental domains for lattices in $SL_2(\mathbb{R})$ Let $\Gamma\leq SL_2({\mathbb{R}})$ be a Fuchsian group of the first kind (so a lattice). I understand that $\Gamma \backslash SL_2(\mathbb{R})$ is compact $\iff$ $\Gamma$ contains no parabolic elements (conjugates of upper triangular matrices). Let $D$ be a fundamental domain for $\Gamma$ and let $\alpha\in SL_2(\mathbb{R})$ be some element not in $\Gamma$.
I understand that the fundamntal domain for $\Gamma' = \langle \Gamma, \alpha\rangle$ (let's call it $D'$) should be a subset of $D$. Or rather, we should be able to tile $D$ with copies of $D'$.
However what if $\alpha$ is a parabolic element? Then clearly $D'$ needs a cusp, but this seems contradictory to the fact that we can tile the compact region $D$ with copies of $D'$. What's going on here?
 A: The trouble is that $\Gamma'$ might not be a discrete group, in which case the concept of a "fundamental domain" for $\Gamma'$ is not defined. 
If, on the other hand, $\Gamma'$ is a discrete group, then cocompactness of $\Gamma$ implies cocompactness of $\Gamma'$, in which case $\Gamma'$ has a compact fundamental domain, and therefore $\Gamma'$ does not have any parabolic elements; so it wasn't possible for $\alpha$ to be parabolic. 
Let me add that the concept of "cocompactness" itself can easily be generalized to indiscrete groups, like this: under the fractional linear action of $SL_2(\mathbb R)$ on $\mathbb H^2$, a subgroup of $SL_2(\mathbb R)$ acts cocompactly if and only if there exists a compact subset whose translates by the subgroup cover the whole of $\mathbb H^2$. Using this definition, if $\Gamma$ acts cocompactly then so does $\Gamma' = \langle \Gamma,\alpha\rangle$, even if $\alpha$ is parabolic. 
But there's no contradiction here, because the theorem which asserts "this more general definition of cocompactness implies existence of a compact fundamental domain" is only applicable to discrete groups. As said earlier, an indiscrete group does not even have a fundamental domain. 
