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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.

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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.

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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. –  Anurag Sharma Apr 12 at 4:53
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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 $p$.)

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