Power of Generator of Cyclic Group is Generator iff Power is Coprime with Order

Theorem
Let $C_n$ be the cyclic group of order $n$.

Let $C_n = \left \langle {a} \right \rangle$, that is, that $C_n$ is generated by $a$.

Then:
 * $C_n = \left \langle {a^k} \right \rangle \iff k \perp n$

That is, $C_n$ is also generated by $a^k$ iff $k$ is coprime to $n$.

Proof

 * First we suppose that $k \perp n$.

Then by Integer Combination of Coprime Integers:
 * $\exists u, v \in \Z: 1 = u k + v n$

So $\forall m \in \Z$, we have:

Thus $a^k$ generates $C_n$.


 * Then we suppose that $C_n = \left \langle {a^k} \right \rangle$, that is, $a^k$ generates $C_n$.

Thus $\exists u: a = \left({a^k}\right)^u$ as $a$ is an element of the group generated by $a^k$.

Thus $u k \equiv 1 \left({\bmod\, n}\right) \implies 1 = u k + v n$ for some $u, v \in \Z$, and thus $k \perp n$ as we wanted to show.