This theorem is from Contemporary Abstract Algebra by J. Gallian page $78$.
Let $a$ be an element of order $n$ in a group and let $k$ be a positive integer. Then $\langle a^k\rangle=\langle a^{(n,k)}\rangle$.
Proof: Let $d=\gcd(n,k)$ and let $k=dr$. Since $a^k=(a^d)^r$, $\langle a^k\rangle \subseteq \langle a^d\rangle$. Moreover, by Bezout's lemma, $d=ns+kt$, so $a^d=a^{ns+kt} \rightarrow a^d=e(a^k)^t$, so $\langle a^d\rangle \subseteq \langle a^k\rangle$. Then, $\langle a^k\rangle=\langle a^d\rangle=\langle a^{(n,k)}\rangle$.
My question: I agreed with the proof, but there is something stuck in my mind s.t If it holds for $d=(n,k)$, then could we say that $d=2(n,k)$ or $d=3(n,k)$ etc. because, this identity holds for every $d$ values if we use the definition of cyclic subgroup. Is my claim correct? If so, why don't we write $d=t(n,k)$ where $t \in \mathbb Z^{\geq1}$
My second question is that should $a$ be a generator? Because some sources say that $a$ has to be generator,for example here page 64, but this book doesn't give such statement, so I confused at this point, as well.
To the first question, you can not take $d=t(n,k)$ because of the first containment proved, as you wouldn’t have $k=dr$ with $r$ an interger. The other side would be true as bezouts gives a way to write all multiples of d.
Also $a$ doesn’t need to be a generator, this is true for every element in the group. But one cool consequence of this property if $a$ is a generator is that for every $j$ coprime with $n$, $a^j$ is also a generator.