# Roots of Unity under Multiplication form Cyclic Group

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## Theorem

Let $n \in \Z$ be an integer such that $n > 0$.

The $n$th complex roots of unity under the operation of multiplication form the cyclic group which is isomorphic to $C_n$.

## Proof

$U_n = \set {e^{2 i k \pi / n}: k \in \N_n}$

where $U_n$ is the set of complex $n$th roots of unity.

Let $\omega = e^{2 i \pi / n}$.

Then we have:

$U_n = \set {\omega^k: k \in \N_n}$

that is:

$U_n = \set {\omega^0, \omega^1, \omega^2, \ldots, \omega^{n - 1} }$

Let $\omega^a, \omega^b \in U_n$.

Then $\omega^a \omega^b = \omega^{a + b} \in U_n$.

Either $a + b < n$, in which case $\omega^{a + b} \in U_n$, or $a + b \ge n$, in which case:

 $\ds \omega^a \omega^b$ $=$ $\ds \omega^{a + b}$ $\ds$ $=$ $\ds \omega^{n + t}$ for some $t < n$ $\ds$ $=$ $\ds \omega^n \omega^t$ $\ds$ $=$ $\ds \omega^t$ as $\omega^n = 1$

So $U_n$ is closed under multiplication.

We have that $\omega_0 = 1$ is the identity and that $\omega^{n - t}$ is the inverse of $\omega^t$.

Finally we note that $U_n$ is generated by $\omega$.

Hence the result, by definition of cyclic group, and from Cyclic Groups of Same Order are Isomorphic:

$U_n = \gen \omega \cong C_n$.

$\blacksquare$