Two Questions about Unit Circle

2010-08-12, 03:26

These aren't actually homework questions, but rather specific examples that came to mind during my reading. Obviously, I'm interested in the proof; not a Yes/No answer.

Let C be the unit circle in the complex plane. Regard it as a topological group, and a measure space.

1) Does C have closed infinite subgroups other than C itself?

2) Does C have subgroups of positive measure other than C itself?

Thanks

Orgasmic Troll · 2010-08-12, 23:15

Are you considering the group C to just be rotations, or the set of all bicontinuous functions on the unit circle?

2010-08-13, 01:10

C = {e^{it}: 0\leq t<2\pi}

R.D. Silverman · 2010-08-13, 13:12

Quote:

Originally Posted by Unregistered

C = {e^{it}: 0\leq t<2\pi}

This does not answer the previous question.

I will offer a hint. If working in the group of rotations, what is the order
of each element in any purported infinite sub-group? Then ask : Can
this sub-group have finite index?

R.D. Silverman · 2010-08-13, 18:58

Quote:

Originally Posted by R.D. Silverman

This does not answer the previous question.

I will offer a hint. If working in the group of rotations, what is the order
of each element in any purported infinite sub-group? Then ask : Can
this sub-group have finite index?

Also, I hope that you are not looking to prove Cartan's Theorem??????

2010-08-15, 14:28

Quote:

Originally Posted by R.D. Silverman

This does not answer the previous question.

I will offer a hint. If working in the group of rotations, what is the order
of each element in any purported infinite sub-group? Then ask : Can
this sub-group have finite index?

Sorry, I should have said complex multiplication is the operation.

Come to think of it, I read a theorem in Stein and Shakarchi's "Fourier Analysis" a few years ago that essentially said: If \theta\in [0,2\pi) is not a rational multiple of 2\pi, then the subgroup generated by e^{i\theta} is dense.

So if $A\subset C$ is an closed proper subgroup, it can contain only points of the form $e^{i\frac{2\pi k}{n}}$ . If $e^{i\frac{2\pi k}{n}}\in A$ and we assume WLOG that $gcd(k,n)=1$ , then $A$ contains every $n^{th}$ root of unity. If $A$ is infinite, then it contains all $n^{th}$ roots of unity for infinitely many n. Since any point in $C$ is within a distance $\frac{2\pi}{n}$ of some $n^{th}$ root of unity, $A$ is dense and hence equal to all of $C$ .

Thank you

2010-08-15, 14:43

I got the second problem too by the way. It follows from an exercise I just completed today:

If G is a Polish locally compact group and equipped with its Haar measure and A\subset G has positive measure, than $A^{-1}A$ contains an open neighborhood of 1.

But when $A$ is a subgroup, $A^{-1}A=A$ . So $A$ contains an open neighborhood $U$ of 1. Since any element of C is of the form $u^n$ for some $u\in U,n\geq 0$ , $A=C$ .

2010-08-12, 03:26	#1
Unregistered 2²×3²×5×53 Posts	Two Questions about Unit Circle These aren't actually homework questions, but rather specific examples that came to mind during my reading. Obviously, I'm interested in the proof; not a Yes/No answer. Let C be the unit circle in the complex plane. Regard it as a topological group, and a measure space. 1) Does C have closed infinite subgroups other than C itself? 2) Does C have subgroups of positive measure other than C itself? Thanks

Similar Threads
Thread	Thread Starter	Forum	Replies	Last Post
the unit circle, tangens and the complex plane	bhelmes	Math	0	2018-03-04 18:50
A hexagon inscribed in a circle	Batalov	Puzzles	9	2010-06-02 23:46
How to divide a circle into 16 pieces with only 5 lines????	4 lil pigs mom	Homework Help	52	2006-12-05 08:41
Circle intersection points	Mini-Geek	Puzzles	1	2006-11-27 03:20
Circle Puzzles 1	mfgoode	Puzzles	18	2005-07-11 11:51

2010-08-12, 23:15	#2
Orgasmic Troll Cranksta Rap Ayatollah Jul 2003 1201₈ Posts	Are you considering the group C to just be rotations, or the set of all bicontinuous functions on the unit circle?

2010-08-13, 01:10	#3
Unregistered B13₁₆ Posts	C = {e^{it}: 0\leq t<2\pi}

2010-08-15, 14:43	#7
Unregistered 20340₈ Posts	I got the second problem too by the way. It follows from an exercise I just completed today: If G is a Polish locally compact group and equipped with its Haar measure and A\subset G has positive measure, than $A^{-1}A$ contains an open neighborhood of 1. But when $A$ is a subgroup, $A^{-1}A=A$ . So $A$ contains an open neighborhood $U$ of 1. Since any element of C is of the form $u^n$ for some $u\in U,n\geq 0$ , $A=C$ .