Show $\pi_1(SU(2)/H_8)$ Say $H_8$ is a quaternion group of order 8,
How should one think of the space $SU(2)/H_8$ even if $H_8$ is not a normal subgroup?
How do we show 
$$\pi_1(SU(2)/H_8)=H_8?$$
 A: If a discrete group $G$ acts properly discontinuously on a path-connected space $X$, then the projection 
$$p:X\to X/G$$ 
is a covering map. For a reference, check Switzer's algebric topology text page 62, or in Hatcher, page 72. From here, applying the long exact sequence of a fibration to the covering map $p$ gives:
$$\cdots\xrightarrow{\ \ \ }\pi_1(G)\xrightarrow{\ \ \ }\pi_1(X)\xrightarrow{\ \ \ }\pi_1(X/G)\xrightarrow{\ \ \ }\pi_0(G)\xrightarrow{\ \ \ }\pi_0(X)\xrightarrow{\ \ \ }\cdots$$
which gives us the short exact sequence:
$$0\xrightarrow{\ \ \ }\pi_1(X)\xrightarrow{\ p_* \ }\pi_1(X/G)\xrightarrow{\ \ \partial }\pi_0(G)\cong G\xrightarrow{\ \ \ }0,$$
since $\pi_1(G)=0$ and $\pi_0(X)=0$. One can show that $\partial$ is a homomorphism in this case, and therefore by the first isomorphism theorem,
$$G\cong \frac{\pi_1(X/G)}{p_*(\pi_1(X))}.$$
In your case, we have:
$$H_8\cong\frac{\pi_1(SU(2)/H_8)}{p_*(\pi_1(SU(2)))}.$$
As explained by Max in this answer, $SU(2)\cong S^3$, so $\pi_1(SU(2))\cong \pi_1(S^3)$ which is trivial, and therefore
$$H_8\cong \pi_1(SU(2)/H_8).$$
