$\frac{SU(2)}{N}= U(1) \times Z_2$. Find $N$?

Question

Consider the short exact sequence

$$ 1 \to N \to SU(2) \to U(1) \times Z_2 \to 1 $$

What is the normal subgroup $N$ here so that $U(1) \times Z_2$ is a quotient group and $SU(2)/N= U(1) \times Z_2$? Is it an allowed short exact sequence?

The $U(1)= R/Z$ as an Abelian compact complex phase, and $Z_2$ is $Z/2Z$ as a finite group of order 2.

Any comment welcome, please let me know.

score 4 · Accepted Answer · answered Dec 06 '16 at 22:30

4

This is not an allowed short exact sequence of topological groups. $SU(2) \to U(1) \times \Bbb Z_2$ must be onto, which means that it cannot also be continuous since $SU(2)$ is connected whereas $U(1) \times \Bbb Z_2$ is not.

answered Dec 06 '16 at 22:30

Ben Grossmann

225,327

Thanks for the comment +1, let me wait for other answers. – miss-tery Dec 06 '16 at 22:31
No problem. Are you interested in a potentially discontinuous group homomorphism? – Ben Grossmann Dec 06 '16 at 22:32
Yes, could you illuminate this: http://math.stackexchange.com/questions/2047197? – miss-tery Dec 06 '16 at 22:58

$\frac{SU(2)}{N}= U(1) \times Z_2$. Find $N$?

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