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?

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    $\begingroup$ You could use first isomorphism theorem. Then you don't need to prove that the map is bijective - you get this from that theorem. $\endgroup$ – Martin Sleziak Aug 4 '12 at 10:45
  • $\begingroup$ I will say every $\mathbb{Q/Z}$ is isomorphic to $X = \left\{x : x \in [0,1] \text{ and }x \in \mathbb{Q} \right\}$ $\endgroup$ – user14082 Aug 4 '12 at 10:46
  • $\begingroup$ From the geometrical form ($e^{i\varphi}$ corresponds to angle $\varphi$) you have: $e^{(2\pi p)/q}=1$ implies $2\pi p/q=2\pi k$ for some $k\in\mathbb Z$; i.e. $\frac pq\in\mathbb Z$. Was this what you had problem with? $\endgroup$ – Martin Sleziak Aug 4 '12 at 10:49
  • $\begingroup$ You probably mean i.e $\frac{p}{q} \in Q$ right? I don't quite follow your argument here. Could you please elaborate a bit more? Thanks. $\endgroup$ – Jernej Aug 4 '12 at 11:59
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    $\begingroup$ BTW you can read here how to reply in comments. (It was only accident that I came back to this question and I saw your comment directed to me. If you want to get attention of other users who previously left comment, you can use @username.) $\endgroup$ – Martin Sleziak Aug 4 '12 at 13:23

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).

  • $\begingroup$ This is what I thought. But somehow I wasn't sure that 1. Every q-th root of unity is of the form $e^{\frac{2 \pi i p}{q}}$ and consequently that for every n-th root of unity z, $e^k=1$ if and only if $n|k$. As for the well defined remark isn't that implied by the uniqueness of the representative? $\endgroup$ – Jernej Aug 4 '12 at 18:40

Be canonical!

You have a morphism of groups $ex:\mathbb R \to S^1: r\mapsto e^{2i\pi r} $, where $S^1$ is the multiplicative group of complex numbers with $\mid z\mid=1$. This morphism is surjective and has kernel $\mathbb Z$.
[The wish to have kernel $\mathbb Z$ instead of $ 2\pi \mathbb Z$ dictated the choice of $ex(r)=e^{2i\pi r}$ instead of $e^{ir}$].

Restricting the morphism to $\mathbb Q$ induces a morphism $res(ex):\mathbb Q\to S^1$ with kernel $\mathbb Q\cap \mathbb Z=\mathbb Z$ and image $\mu_\infty\stackrel {def}{=} e^{2i\pi \mathbb Q} \subset S^1$.
The crucial observation is that this image is $\mu_\infty=\bigcup_n \mu_n$, where $\mu_n$ is the set of $n$-roots of unity $e^{\frac {2i\pi k}{n}}\quad (k=1,2,...,n)$.
Hence $\mu_\infty$ is the set of all roots of unity i.e. the set of complex numbers $z \in \mathbb C$ with $z^n=1$ for some $n\in \mathbb N^*.$

Applying Noether's isomorphism you finally get the required group isomorphism (be attentive to the successive presence and absence of a bar over the $q$ in the formula) $$Ex: \mathbb Q/\mathbb Z \xrightarrow {\cong} \mu_\infty:\overline {q}\mapsto e^{2i\pi q} $$

A cultural note
This elementary isomorphism is actually useful in quite advanced mathematics.You will find it, for example, in Grothendieck's Classes de Chern et représentations linéaires des groupes discrets.

  • $\begingroup$ Hi. How do you know that every element in $S^1$ can be written as $e^{2\pi q}$ for some $q \in \mathbb{Q}$? $\endgroup$ – user661541 Jul 16 at 13:57
  • $\begingroup$ @user661541: I never wrote the nonsensical statement you make. $\endgroup$ – Georges Elencwajg Jul 17 at 8:16
  • $\begingroup$ thanks... I misunderstood. $\endgroup$ – user661541 Jul 17 at 11:11

Define $\,f: \Bbb Q\to S^1:=\{z\in \Bbb C\;:\;|z|=1\}\,\,,\,f(q):=e^{iq}\,$ , show this is a homomorphism of groups, find its kernel and use the fist isomorphism theorem.


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