Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let G denote the group of orientation-preserving isometries of the plane; equivalently, the group of affine transformations of the complex field C of the form

$z \rightarrow \alpha z + \beta$ $(\alpha ,\beta \in C$ $ |\alpha| = 1)$

Therefore every element of G is either a translation or a nonidentity rotation about some point of the plane.

Now for any prime number p, $G_p$ will denote the subgroup of $G$ consisting of motions whose rotational component is by an angle of the form $\frac{2\pi m}{p^n}$; in other words, the group of transformations (1) such that $\alpha$ is a $p^{nth}$ root of unity for some n.

$\textbf {Question}$ : How to prove that $G_p$ is dense in $G$ under the topology induced by that of the complex numbers.

share|cite|improve this question
up vote 1 down vote accepted

Rational numbers of the form $\frac a{2^n}$ for fixed $n$ are uniformly spaced out in the real line at a distance of $\frac1{2^n}$; as $n$ goes up this distances get reduced arbitrarily, giving a dense subset of the reals; Replace 2 by any other prime, same holds. Now use the function $f(t) = e^{2\pi i t}$ which sends this to a dense subset of the unit circle.

share|cite|improve this answer
Okay I see somewhat what you are saying. I am bothered by the fact the addition of $\beta$ Group G is not only rotation. It includes translation also. – Alvis Apr 12 '14 at 4:53

Any real number can be arbitrarily closely approximated by a number of the form $\frac m{p^n}$. (E.g. think of it written in base $p$.)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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