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?

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.
