# Proof that $\Bbb R/\Bbb Z$ is isomorphic to $S^1$

I just learned the first isomorphism theorem for groups and now I need to prove that:

$$\Bbb R/\Bbb Z \cong S^1$$

In other words, the quotient group of $\Bbb Z$ in $\Bbb R$ is isomorphic to the $S^1$ (the group of all permutations of $1$ elements, I guess).

I'll write what I understand about the first isomorphism theorem for groups:

If we have $\phi: G\to G'$ as an homomorphsim, then the quotient group

$$G/\ker(G)\cong G'$$

and the isomorphism if given by $\phi'$ which I forgot exactly how it's given.

So, if I want to prove that

$$\Bbb R/\Bbb Z \cong S^1$$ using the first isomorphism theorem, I would need to somehow get $\Bbb Z$ to be the kernel of a homomorphism between $\Bbb R$ and $S_1$? Because the exercise does not give any information about which is the homomorphism between the two sets, so I'm thinking about creating one. Am I doing the rigth thing?

• $S^1$ is rather the unit circle. – Spenser Jul 9 '15 at 22:01
• $S_n$ is the group of permutations of $\{1,\cdots,n\}$, $S^n$ is the $n$-sphere. Can you figure out a map from $\Bbb R$ to the unit circle? – whacka Jul 9 '15 at 22:02

View $S^1$ as the unit circle in $\Bbb C$ with multiplication as the group operation. Let $$\varphi:\Bbb R\to S^1,\quad\varphi(x)=e^{2\pi i x}.$$ It is easy to show that $\varphi$ is a surjective group homomorphism, and that $\ker\varphi=\Bbb Z$. Now, you can use the first isomorphism theorem.