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 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
 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}
2010-08-13, 13:12   #4
R.D. Silverman

Nov 2003

22·5·373 Posts

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?

2010-08-13, 18:58   #5
R.D. Silverman

Nov 2003

22·5·373 Posts

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   #6
Unregistered

2×359 Posts

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 #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 $u\in U,n\geq 0$, $A=C$.

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