20100812, 03:26  #1 
875_{10} 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 
20100812, 23:15  #2 
Cranksta Rap Ayatollah
Jul 2003
641_{10} Posts 
Are you considering the group C to just be rotations, or the set of all bicontinuous functions on the unit circle?

20100813, 01:10  #3 
2×709 Posts 
C = {e^{it}: 0\leq t<2\pi}

20100813, 13:12  #4 
Nov 2003
1D24_{16} Posts 

20100813, 18:58  #5 
Nov 2003
2^{2}×5×373 Posts 
Also, I hope that you are not looking to prove Cartan's Theorem??????

20100815, 14:28  #6  
4,139 Posts 
Quote:
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 is an closed proper subgroup, it can contain only points of the form . If and we assume WLOG that , then contains every root of unity. If is infinite, then it contains all roots of unity for infinitely many n. Since any point in is within a distance of some root of unity, is dense and hence equal to all of . Thank you 

20100815, 14:43  #7 
2,083 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 contains an open neighborhood of 1. But when is a subgroup, . So contains an open neighborhood of 1. Since any element of C is of the form for some , . 
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