# Group of invertible elements of subring of C.

1. Feb 19, 2013

### Silversonic

1. The problem statement, all variables and given/known data

Let R be a subring of ℂ such that the group of invertible elements U(R) is finite, show that this group is a cyclic group. (Group operation being multiply).

2. Relevant equations

3. The attempt at a solution

I have the answer, and I got very close to getting there myself before I reached a roadblock and couldn't move on. Even though I've looked at the answer, I still cannot understand the part that was blocking me. I've genuinely spent a long time trying to understand this last part and it's probably something really simple, which has annoyed me a great deal.

The deal is this;

The group is finite, and therefore any element must have a modulus of one - because if an elements didn't have modulus one any power of said element would be a unique, and thus the group would not be finite. Hence every element of the group of invertile elements has the form

$e^{2\pi i \frac {p}{q}}$ where p and q are natural numbers, i.e. their arguments are rational multiples of two pi.

The step the answer takes is this. Assume that q (above) is the lowest common denominator of all the fractions $\frac {p}{q}$ that occur. Then there will be an element $x$ within the group such that

$x = e^{2\pi i \frac {p_0}{q}}$, where $\frac {p_0}{q}$ is in lowest terms, i.e. they are coprime. The proof carries on and I am able to understand the rest.

But the bolded bit, I've embarassingly spent ages on this and don't know how this can be true. Every rational number in the exponent was re-written in the form where the denominator was the LCD of all the exponents of the elements in the group, I understand that. But I don't see how it goes to show that one of the elements is coprime to this said LCD (and as such, was already in the form $\frac {p}{q}$ where q is the LCD, to begin with).

Would it involve writing $p = kq +r$ (r<q) so that every element becomes of the form;

$e^{2\pi i \frac {r}{q}}$ and showing that a combination of elements such as this can produce a $p_0$ which is coprime to $q$?

Last edited: Feb 19, 2013
2. Feb 19, 2013

### Dick

I think the main point is that there must be an element of the group that has the smallest value of |p/q|. If there's no smallest value then it must be infinite, right? If there is a smallest value then you can certainly express it in lowest terms.

3. Feb 20, 2013

### Silversonic

That's certainly true, every term of the form $\frac {p}{q}$ where q is the LCD of all the fractions has a term where $p$ is the lowest. But this lowest $p$ is coprime with $q$ you would say? Why would that be?

4. Feb 20, 2013

### Dick

For me, the lowest p being coprime with q would come out of the rest of the proof. I'll know that after I know p/q generates the group. I don't think you can just assume it up front.

5. Feb 20, 2013

### Silversonic

Is there any way to prove this bolded bit? I was along those lines myself before looking at the answer, in that the smallest element (by which I mean its arguments is smallest) must generate the whole group. I didn't happen to get anywhere though.

The rest of the proof uses the fact that there exist integers m, n such that

$mp_{0} + nq = 1$

Then

$x_0^m = e^{2\pi i \frac {mp_0}{q}} = e^{2\pi i \frac {1-nq}{q}} = e^{2\pi i \frac {1}{q}}$

Hence every element, is of the form

$e^{2\pi i \frac {a}{q}} = x_0^{ma}$

for some a, i.e. it is generated by x0.

Last edited: Feb 20, 2013
6. Feb 20, 2013

### Dick

I'm not really sure what the proof is up to there. Think about proving it a different way. If $e^{2\pi i \frac {p}{q}}$ doesn't generate the group then there is some other element $e^{2\pi i r}$ in the group. And for some integer k you must have kp/q<r<(k+1)p/q, right? Since it's a group that means $e^{2\pi i (r-\frac {kp}{q})}$ is in the group. But r-kp/q is less than p/q. That would contradict p/q being the smallest number in the group.