Let $S^1$ be the group of all complex numbers $w$ such that $|w|=1.$ Show that $\Bbb R/\Bbb Z\cong S^1$

Question

Let $\mathbb R$, $\mathbb{Z}$ be the groups of real numbers and integers respectively under addition, and $S^1$ denote the group of complex with modulus $1$ under multiplication.

This question was posted a long time ago here Group isomorphism: $\mathbb{R}/\mathbb{Z}\cong S^1$, I am attaching an image of it as:

However, I don't understand how to prove this mapping as an injective function. Further, it seems the solution is a little complicated.

I have thought of a proof for it for this problem and hence, I am adding a solution verification tag. My proof goes as follows:

We have, $1=\cos 0+i \sin 0= \cos\theta +i \sin \theta$ and hence $\theta=0+2n\pi=2n\pi,n\in\Bbb Z$, thus if $e^{2\pi ri}=1$, then $r\in \Bbb Z.$ If we consider a mapping $f:\Bbb R\longrightarrow S^1$ such that $f(r)=e^{2\pi ri}$,$\forall r\in\Bbb R.$ Then, $f$ can be verified as a surjective homomorphism. Now, the identity element in $S^1$ is $1$, so, if $e^{2\pi ir}=1$, then $r\in \Bbb Z.$ Thus $\text{Ker} f=\Bbb Z$, and from 1st Isomorphism theorem, we have $\Bbb R/\Bbb Z\cong S^1.$

I posted this proof because if this is correct, then it becomes a more straight-forward approach than original solutions(atleast according to me.) So, is this proof, correct?

I dont know whether this is allowed in this site or not, but I didn't post this as a comment on that page because as it seems the user is inactive for ages and I am unlikely to get any response, so I went ahead. Also, I don't mind if moderators decide merging it with the original post.

Answer 1

You want to prove the map $f: [r] \mapsto e^{2 \pi i r}$ is injective. (I assume you've already satisfied yourself that the map is well defined.)

Assume $f([r])=f([s])$. Then $e^{2 \pi i r}=e^{2 \pi i s}$, so $e^{2 \pi i(r-s)}=1$. But $e^{2 \pi i (r-s)}= \cos 2 \pi (r-s) + i \sin 2 \pi (r-s)$, so $e^{2 \pi i (r-s)}=1$ means, in particular, that $\cos 2 \pi (r-s) =1$, and that happens only when $r-s \in \Bbb Z$. That means we're done, because $r-s \in \Bbb Z \Rightarrow [r]=[s]$ in $\Bbb R / \Bbb Z$, so $f([r])=f([s]) \Rightarrow [r]=[s]$, proving that $f$ is $1$-$1$.

You also can use the first Isomorphism Theorem to prove the result, as you did in your edit, by demonstrating a surjective homomorphism with $\Bbb Z$ as its kernel.

Answer 2

Variant 1: Proving that $f$ is surjective and injective.

Surjectivity is clear, since every complex number $z$ in $S^1$ is of the form $$ z=e^{i\theta},$$ where $\theta$ is some real number. If this isn't clear to you look up polar coordinates.

For injectivity, it is enough to show that if there is an $r \in \mathbb{R}$ such that $e^{2\pi i r}=1$, then $r \in \mathbb{Z}$. This can be seen using polar coordinates.

Variant 2: Constructing an inverse map.

Define $g:S^1\longrightarrow \mathbb{R}/\mathbb{Z}$ by setting $g(e^{2\pi r})=[r]$. Note that this is well defined because the argument of a complex number is defined up to an integer multiple of $2 \pi i$. Prove that $g$ is inverse to $f$.