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I am trying to understand the proof 2.6.7 in "Joseph A. Wolf - Spaces of Constant Curvature" page 88.

This shows that every finite subgroup of $\mathrm{SU}(2)$ is either Cyclic or Binary where the binary groups are defined to be the preimage under the double cover $\pi: \mathrm{SU}(2) \rightarrow \mathrm{SO}(3)$ of the finite subgroups of $\mathrm{SO}(3): G^*= \pi^{-1} (H)$ (using random finite subgroups letters).

Define $U \leq \mathrm{SU}(2)$ finite and $O=\pi(U)$. He splits the two cases $U\neq \pi^{-1}(O)$ and $U=\pi^{-1}(O)$.

In the first case he deduce without proof that $\pi: U \simeq O$ and $$\pi^{-1}(O) = U \times \{ \pm 1 \} $$ Where clearly $\{ \pm 1 \} $ is the center of $\mathrm{SU}(2)$ and is what we use to define $\mathrm{SO}(3) = \mathrm{SU}(2)/ \{ \pm 1 \} $. Can someone help me? I tried to see it, but I definetly don't.

on1921379
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Badillo
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1 Answers1

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If $-1 \in U$ then $\ker \pi = \{ \pm 1 \} \le U$ and so $U = \pi^{-1}(\pi(U)) = \pi^{-1}(O)$. So if $U \ne \pi^{-1}(O)$ then $-1 \not\in U$ and so $ U \cap \{\pm 1\} = \{1\}$, so $U \times \{\pm 1\} \le {\rm SU}(2)$.

Clearly $\pi(U \times \{\pm 1\}) = \pi(U) = O$, so $ \pi^{-1}(O) = U \times \{\pm 1\}$. To see this, note that clearly $U \times \{\pm 1\} \le \pi^{-1}(O)$ and, if $g \in \pi^{-1}(O)$, then there exists $h \in U$ with $\pi(g) = \pi(h)$, so $gh^{-1} \in \ker \pi = \{ \pm 1\}$ and $g \in U \times \{\pm 1\}$.

Also, but the First Isomorphism Theorem, $U \cong (U \times \{\pm 1\})/\{\pm 1\} \cong \pi(U)=O$.

Derek Holt
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  • I don't get something.
    1. How do you prove that if $U \neq \pi^{-1} (O)$ implies $-1 \notin U$?

    I can use the First isomorphism theorem to show $U\simeq O$ and I see ${ \pm 1 } \subset \pi^{-1}(O)$. Then we see that $\pi: SU(2) \rightarrow SU(2)/ {\pm 1 }$ so it quotient out ${ \pm 1 }$ from $SU(2)$ and if $O\simeq U$, $-1 \notin U$ since $1$ always is? I'm not sure about this argument. 2) Why $\pi (U \times { \pm 1 }) = \pi (U) = O$ implies $\pi^{-1}(O) = U times { \pm 1 }$? If I can just apply $\pi^{-1}$ I would have $\pi^{-1}(O) = U$. I don't understand. I really thank you.

     – Badillo Aug 27 '22 at 09:04
  • I've added a bit more detail, but all of these things follow easily from $\ker \pi = { \pm1 }$. – Derek Holt Aug 27 '22 at 11:22