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Let us consider $$ 1 \to N \to SU(2) \to Q \to 1 $$

(1). Can we find some examples of $N$ and $Q$ so that either $N$ or $Q$ contain finite groups?

We know $SU(2)/Z_2=SO(3)$, so we can choose $N=Z_2$ and $Q=SO(3)$ as an example. Here $Z_n$ is $Z/nZ$ as a finite group of order $n$.

Since $SU(2)$ contains the quaternion $H_8$ as a subgroup (correct?), then $SU(2) \supset H_8$. We know $H_8/Z_2=(Z_2)^2$ and $H_8/Z_4=Z_2$. Do we have something similar by replacing $H_8$ by $SU(2)$? Does it make sense to consider:

(2) $$SU(2)/N=(Z_2)^2$$ $$SU(2)/Z_2=SO(3) \text{ ( see the above )}$$ $$SU(2)/N=Z_2$$ $$SU(2)/Z_4=Q$$ What are these $N$ and $Q$ if they make senses?

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(3) If $H_8$ is a normal subgroup of $SU(2)$ (is this true?), then we can ask $$ 1 \to H_8 \to SU(2) \to Q \to 1 $$ What is $Q=SU(2)/H_8=?$

Partial answers are welcome!

miss-tery
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    Partial answer: $SU(2)$ has no nontrivial finite normal subgroups. Notably, if $N$ contains a non-zero element $h$, then it must contain the entirety of the orbit $$ {ghg^{-1} : g \in SU(2)} $$ which is necessarily infinite. – Ben Grossmann Dec 06 '16 at 23:06
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    And, as I mentioned in my other answer: if $Q$ is finite (and non-trivial), then $SU(2) \to Q$ cannot be a continuous homomorphism. – Ben Grossmann Dec 06 '16 at 23:07
  • Thanks, so you rule out (3), $SU(2)$ does not have $H_8$ as normal subgroups. I am glad to meet an expert here. :) – miss-tery Dec 06 '16 at 23:08
  • I've made a mistake! $SU(2)$ has a non-trivial center. So, its center ${I, -I}$ is a normal subgroup. I think that the argument does apply, however, to any subgroup besides this one. – Ben Grossmann Dec 06 '16 at 23:13
  • By "$SU(2) \to Q$ cannot be a continuous homomorphism, if Q is finite (and non-trivial)" --- do you mean that $Q$ cannot be a non-identity finite group as a quotient group of $SU(2)$? Why is that? Intuitively you seem right, but what is the more rigorous argument? (no need to prove, I am an engineer.) – miss-tery Dec 06 '16 at 23:14
  • again, continuous maps necessarily take connected sets to connected set. A finite (or for that matter, countable) set with more than one element is not connected. – Ben Grossmann Dec 06 '16 at 23:15
  • I am curious as to what an engineer would be doing with exact sequences, though – Ben Grossmann Dec 06 '16 at 23:16
  • (1) Does $SU(2)/Z_4=Q$ make sense? Should $Z_4$ be a normal subgroup of $SU(2)$??? (2) Then your $Z_2$ center example, I had $SU(2)/Z_2=SO(3)$ as I already mention. – miss-tery Dec 06 '16 at 23:24
  • If there's any chance for that to make sense, $Z_4$ would have to be a normal subgroup of $SU(2)$. For the quotient $G/H$ to make sense, $H$ needs to be normal in $G$. I don't think we can find a normal $Z_4$ in $SU(2)$ for reasons already stated. In particular, there are infinitely many matrices in $SU(2)$ whose eigenvalues are $\pm i$, and all of them are conjugate to any candidate generator of $Z_4$ within $SU(2)$. – Ben Grossmann Dec 06 '16 at 23:30
  • @Omnomnomnom, thank you very much, it seems to me that you are claiming that my questions (2) and (3) are impossible, thus negative attempt--- except the only example I gave: $SU(2)/Z_2=SO(3)$? Perhaps my question (1) is also negative? – miss-tery Dec 06 '16 at 23:33
  • @Omnomnomnom, and miss-tery, how about this: http://math.stackexchange.com/questions/2047269/? – wonderich Dec 07 '16 at 00:02
  • @miss-tery in fact, it seems that $SO(3)$ is a simple group, so there are no other possibilities for a normal $N \subset SU(2) \cong SO(3) \times Z_2$ – Ben Grossmann Dec 07 '16 at 00:42
  • Thanks. Why do you use isomorphic for $SU(2) \cong SO(3) \times Z_2$? is that true??? I know $SU(2)/Z_2 \cong SO(3)$ instead. – miss-tery Dec 07 '16 at 00:47
  • @Omnomnomnom, $SU(2)$ is isomorphic to unit quaternions but SU(2) is NOT isomorphic to quaternions H_8? See this: http://math.stackexchange.com/questions/185472/how-to-show-su2-mathbbz-2-cong-so3/185492#185492 – wonderich Dec 07 '16 at 00:52
  • @miss-tery ooof, you're right. I meant $SU(2)/Z_2 \cong SO(3)$. In any case, that's still enough to deduce that because $SO(3)$ is simple, there are no more normal subgroups. – Ben Grossmann Dec 07 '16 at 00:55

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