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Old 2010-08-12, 03:26   #1
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Default 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?

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Old 2010-08-12, 23:15   #2
Orgasmic Troll
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Are you considering the group C to just be rotations, or the set of all bicontinuous functions on the unit circle?
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Old 2010-08-13, 01:10   #3
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C = {e^{it}: 0\leq t<2\pi}
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Old 2010-08-13, 13:12   #4
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Quote:
Originally Posted by Unregistered View Post
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?
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Old 2010-08-13, 18:58   #5
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Quote:
Originally Posted by R.D. Silverman View Post
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??????
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Old 2010-08-15, 14:28   #6
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Quote:
Originally Posted by R.D. Silverman View Post
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
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Old 2010-08-15, 14:43   #7
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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.
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