If $|a|=n $ show that $\langle a^k\rangle = \langle a^{\gcd(n,k)}\rangle$ and that $|a^{k}|=n/\gcd(n,k)$

Question

If $|a|=n $, show that $\langle a^k \rangle = \langle a^{\gcd(n,k)}\rangle$ and that $|a^k|=n/\gcd(n,k)$.

Let $\gcd(n,k)=t \Rightarrow \; \exists \; u,v \in \Bbb Z$ such that $n = tu$ and $k=tv$ also $\exists \; e_1, e_2 \in \Bbb Z$ such that $t = e_1 n + e_2 k$

Let $x \in \langle a^k\rangle \Rightarrow x=\{a^k\}^r = a^{kr} $for some $r \in\Bbb Z $

Therefore, $x=a^{tvr} \in\langle a^t\rangle \Rightarrow\langle a^k\rangle\subseteq\langle a^t\rangle$

Let $y \in \langle a^t\rangle \Rightarrow y=a^{ti}$ for some $i \in \Bbb Z$

Therefore $y = a^{ti}= a^{i(e_1 n + e_2 k)}= a^{e_1 ni}a^{e_2 ki}= \{a^{k}\}^{e_2i} \in \langle a^k \rangle$

$\langle a^{t} \rangle \subseteq \langle a^{k} \rangle \Rightarrow \langle a^k \rangle = \langle a^t \rangle$

so first part is done.

I don't have any idea how to approach 2nd part.

Answer 1

For the second part observe that $a^d$ has order $n/d$ for a divisor $d$ of $n$ since $(a^d)^{n/d}=a^n=1$ and if $(a^d)^m=1$ then $n\mid dm$ so $n/d \mid m$.

Answer 2

For the second part, note that, by definition, the order of $a^{k}$ is the least $m$ such that $km=nl$ for some integer $l$. Therefore, $m$ is of the form $\frac{n}{k/l}$ ($*$), and hence $|a^{k}|$ ("...the least...") is gotten when the denominator in ($*$) is the greatest divisor of $k$ which is also a divisor of $n$: $$|a^k|=\frac{n}{\gcd(k,n)}$$