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How can I prove that the quotient group $S^3/\{+I,-I\}$ is isomorphic to $SO_3$ and that the group $S^3$ is not isomophic to $SO_3$?

Here $S^3$ is the subgroup of the quaternion group: $S^3=\{a+bi+cj+dk | a^2+b^2+c^2+d^2=1\}$

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What do you mean by $S^3/\{\pm I\}$? Are you looking at the orbit space of $S^3$ under the action of $\Bbb{Z}/2\Bbb{Z}$ given by the antipodal map? –  fpqc Nov 26 '12 at 0:22
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To prove that $S^3/\pm 1$ is isomorphic to $SO(3)$, use the hypersphere of rotations. This gives a way of naturally thinking of $SO(3)$ as a quotient of $S^3$. Embedding everything inside the quaternions makes defining the mapping much cleaner.

To show that $SO(3)$ and $S^3$ are not isomorphic as lie groups, note that this gives a homeomorphism of the underlying topological spaces. But from the hypersphere of rotations, one can also see that $SO(3)$ and $\mathbb{RP}^3$ are homeomorphic, and $\pi_{1}(\mathbb{RP}^3) = \mathbb{Z}/2\mathbb{Z}$, whereas $\pi_{1}(S^3) = \{1\}$ since the $3$-sphere is simply connected.

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They are also not isomorphic as groups. $S^3$ has nontrivial center ($-1$ is in it), but the center of $SO(3)$ is trivial. –  Jason DeVito Nov 25 '12 at 23:36
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