8
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
1
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ 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
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.