0

We are told to show that the quotient group $\mathbb{R}/\mathbb{Z}$ is isomorphic to the group $U$ = $\{z \in \mathbb{C}\mid \vert z \vert = 1\}$ endowed with multiplication as operation.

My answer:

The group $U$ symbolizes the unit circle. Euler's formula gives an isomorphism from an previous assignment as $z = e^{i2\pi r}$. We also showed that $\phi:\mathbb{R} \longmapsto \mathbb{C}$ is a homomorphism. The equation for that was $\phi(r)=\cos (2\pi r) + i \sin (2\pi r)$. Plotting Euler's formula on an imaginary-real axis gives us a circle. Here $r = 1$ (as seen on the kernel since $\epsilon = 1$). This indicates that it is isomorphic to $U$. $\mathbb{Z}$ is just any set of integers and will only make the values smaller and thereby into the unit circle.

What do you think of my answer?

user26857
  • 53,190
Aboriginal_Hero
  • 23

1 Answers1

4

So what you want to say is something like that $\psi:\Bbb R\to U$ by $\psi(r)=e^{2\pi i r}$ induces an isomorphism between $\Bbb R/\Bbb Z$ and $U$. Here $U$ is the circle group. This is true by the first isomorphism theorem, since $\operatorname{ker}\psi=\Bbb Z$.

This is true.

  • Thank you. I am aware of this. But how do I prove that ker ø = Z? – Aboriginal_Hero Apr 10 '20 at 21:52
  • 1
    @HomerSimpson $e^{2 \pi i r} = \cos(2\pi r ) + i \sin(2 \pi r) = 1$ exactly when $r$ is an integer. You probably don't need to say more than that - basic facts about trig should be free to use. – Jair Taylor Apr 10 '20 at 21:55