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Prove that every finite subgroup of the multiplicative group $T=\{z \in \Bbb C||z|=1\}$ is cyclic.

I was thinking to prove that the order of every subgroup of $T$ is prime, then they are all cyclic. But I can't. Could somebody give me some hints. Many thanks.

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Don't the $n$th roots of unity form a multiplicative subgroup of order $n$? $n$ need not be prime... – MPW Jan 19 '14 at 23:25
Perhaps there's some way to show that any group of order $n$ contains the element $$e^{(2\pi i)/n} = \cos(2\pi/n) + i\sin(2 \pi/n)$$ – Omnomnomnom Jan 19 '14 at 23:25
So if $G$ has order $n$ then it is easy to see that every element of $G$ is an $n$th root of unity. If you can show the reverse containment, all $n$th roots of unity in $G$, then you would have a primitive $n$th root of unity in your group and you would be done. I think that would work...although I haven't suggested a way to show the other containment. – monroej Jan 19 '14 at 23:27
(I'm essentially repeating what @Omnomnomnom is saying) – monroej Jan 19 '14 at 23:28
Possible duplicate of…. – lhf Jan 20 '14 at 2:30

Assume the order of $G$ is $n$. Every element of $G$ is an $n$th root of unity. Since there are $n$ of them, you have all $n$th roots of unity. Therefore, you have a primitive $n$ th root of unity, say $x$. The order of $x$ is $n$. Therefore, $x$ generates your group $G$. I think this does it.

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Check out this link. It's even better than you might think. He states

Theorem: Any finite subgroup of the multiplicative group of any field is cyclic.

It applies here even though $\mathbb{T}$ is not a field, because $\mathbb{C}$ is.

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Multiplication of numbers in $\mathbb{T}$ is a geometric rotation. A finite group of geometric rotations must be cyclic.

Diferently said, $\mathbb{T}$ is isomorphic to the special orthogonal group SO2. But a finite subgroup of SO2 is cyclic (inside the proof ofThm 8.1).

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