Roots of unity isomorphic to cyclic group?

**charleschafsky** · November 23rd 2008, 11:45 AM

Hello there!

I've been tasked with showing that A_n, where A_n = {z, exists in the complex numbers | z^n = 1 } is isomorphic to the cyclic group Cn, where Cn is of order n.

My understanding (or memory, rather) of complex numbers is quite dim, so this problem is becoming quite of a hurdle.

**ThePerfectHacker** · November 23rd 2008, 08:33 PM

Originally Posted by charleschafsky

Hello there!

I've been tasked with showing that A_n, where A_n = {z, exists in the complex numbers | z^n = 1 } is isomorphic to the cyclic group Cn, where Cn is of order n.

My understanding (or memory, rather) of complex numbers is quite dim, so this problem is becoming quite of a hurdle.

Hint: The solutions to $z^n = 1$ are $\zeta, \zeta^2, ... , \zeta^n$ where $\zeta = e^{2\pi i/n}$ .

**whipflip15** · November 23rd 2008, 11:02 PM

First note that all the nth roots of unity are powers of $e^{\frac{2 \pi i}{n}}$
Now define a function $f:C_n \rightarrow \mathbb{C}$
where
$f(a)=e^{a\frac{2\pi i}{n}}$

Can you show f is an isomorphism?

**xianghu21** · November 23rd 2008, 11:19 PM

define an isomorphism from all roots of unity to An. I did this problem before, I forget how the isomorphism went. But it is not hard to show.

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