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 $$\pi_1(SU(2)/H_8)=H_8?$$
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$SU(2)$ is a lie group, hence a manifold (or is a manifold, hence a Lie group, depends your perspective :) ).The action of quaternions produces an orbit space hence with induced topology by the natural projection you have a new topological space. So naturally you come up with the question of the fundamental group. If I recall correctly, in Hatcher's book you may find something useful about it (or something similar anyway). – Mar 31 '17 at 19:32
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1$H_8$ doesn't need to be normal to define $SU(2)/H_8$: it's simply the set of left cosets of $H_8$. Normality is only required to define an appropriate group structure. – Arnaud D. Mar 31 '17 at 20:04
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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 $$H_8\cong \pi_1(SU(2)/H_8).$$
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One thing I guess I should mention, thinking of $SU(2)$ as $S^3$, the group $H_8$ acts on $S^3$ by 90-degree rotations, so at least intuitively, this action should be properly discontinuous. – Mar 31 '17 at 20:28
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