Finite cyclic group and generator

Let $G=\langle a \rangle$ be a finite cyclic group of order $n$ with generator $a$. Prove that $a^k$ generates $G$ if and only if $\gcd(n, k) =1$.

Solution

$( \Rightarrow)$ Suppose that $a^k$ generates $G$. Then $a$ is some power of $a^k$ so we have

$$a=(a^k)^{\ell}=a^{k\ell}$$

for some integer $\ell$. This means that $k \ell = 1 \mod n$ so that $[k]_n$ is a unit in $\mathbb{Z}_n$. This implies that $\gcd(n, k)=1$.

$( \Leftarrow)$ Suppose that $\gcd(n, k)=1$. Then there are integers $x, y$ such that $k x + n y =1$. We compute:

$$a=a^1= a^{kx + ny} = a^{kx} a^{ny}=(a^k)^x$$

since $a^n =1$. Let $g \in G$ be an arbitrary element , then there exists an integer $i$ such that $g=a^i$ and we have that

$$g=a^i = (a^k)^{xi} \in \langle a^k \rangle$$

Hence $a^k$ generates $G$.