Quotient of Unit Quaternions by Subgoup (Lie Groups) Let $G \leq Sp(1)\cong S^3$ (unit quaternions) be a discrete subgroup of order 120 (the Binary Icosahedral Group, not the other one), with presentation $G=<s,t| s^2=t^3=(st)^5>$. $\hspace{2mm}G$ is a perfect group with one non-trivial normal subgroup of order $2$ (its center).
$a.)$ Show that $M=Sp(1)/G$ is a compact, oriented manifold without boundary, and that it has the same integral homology groups as $S^3$.
$b.)$ Let $X$ be the CW-complex formed by attaching two $2-$cells to $M$, using loops representing $s$ and $t$ as the attaching maps. Compute $H_*(X,\mathbb{Z})$.
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For $a.)$, since $S^3$ is compact and $M$ is its image under the projection map $M$ must also be compact. Since this is a covering projection, $M$ is locally just like $S^3$ and therefore a manifold w/o boundary. 
For orientability, I'm thinking since $\pi_1(M)=G$, if $M$ is not orientable then it has a double cover, which implies a subgroup of $\pi_1(M)=G$ of index two (which is normal), but this does not exist, so $M$ must be orientable. Is $M$ a Lie group? (if $G$ is normal I know that it is)
$b.$ I'm not sure.

Since the group G acts on the sphere by orientation preserving diffeo without fixed points, the quotient is an orientable manifold of dimension three, obviously compact. In particular, H_3 is Z The map from the sphere to M is clearly the universal covering space of M, so the fundamental group of M is G. One can check That G is perfecto, so that H_1  is zero. Poincaré duality then implies that H_2 is also zero.
