Roots of Unity under Multiplication form Cyclic Group

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

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

Proof
From Roots of Unity, we have:
 * $$U_n = \left\{{e^{2 i k \pi / n}: k \in \N_n}\right\}$$

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

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

Then we have:
 * $$U_n = \left\{{\omega^k: k \in \N_n}\right\}$$

that is:
 * $$U_n = \left\{{\omega^0, \omega^1, \omega^2, \ldots, \omega^{n-1}}\right\}$$

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:

$$ $$ $$ $$

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 Same Order Isomorphic:
 * $$U_n = \left \langle \omega \right \rangle \cong C_n$$.