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:

Then

$$ $$ $$

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.