TITLE: Additive quotient group $\mathbb{Q}/\mathbb{Z}$ is isomorphic to the multiplicative group of roots of unity
QUESTION [11 upvotes]: I would like to prove that the additive quotient group $\mathbb{Q}/\mathbb{Z}$ is isomorphic to the multiplicative group of roots of unity.
Now every $X \in \mathbb{Q}/\mathbb{Z}$ is of the form $\frac{p}{q} + \mathbb{Z}$ for $0 \leq \frac{p}{q} < 1$ for a unique $\frac{p}{q} \in \mathbb{Q}.$ This suggest taking the map $f:\mathbb{Q}/\mathbb{Z} \mapsto C^{\times}$ defined with the rule $$f(\frac{p}{q} + \mathbb{Z}) = e^{\frac{2\pi i p}{q}}$$  where $\frac{p}{q}$ is the mentioned representative. 
Somehow I have problems showing that this is a bijective function in a formal way. I suspect I do not know the properties of the complex roots of unity well enough.
Can someone point me out (perhaps with a hint) how to show that $f$ is injective and surjective?

REPLY [5 votes]: To prove it is a bijection, one can use rather "primitive" methods. suppose that:
$f\left(\frac{p}{q} + \Bbb Z\right) = f\left(\frac{p'}{q'} + \Bbb Z\right)$,
then: $e^{2\pi ip/q} = e^{2\pi ip'/q'}$, so $e^{2\pi i(p/q - p'/q')} = 1$.
This, in turn, means that $\frac{p}{q} - \frac{p'}{q'} \in \Bbb Z$, so the cosets are equal. Hence $f$ is injective.
On the other hand, if $e^{2\pi i p/q}$ is any $q$-th root of unity, it clearly has the pre-image $\frac{p}{q} + \Bbb Z$ in $\Bbb Q/\Bbb Z$ (so $f$ is surjective).
One caveat, however. You haven't actually demonstrated $f$ is a function (i.e., that it is well-defined, although if you stare hard at the preceding, I'm sure it will come to you).