About generators of a cyclic group of finite order

Question

I am looking at a proposition from Goodman's book: Let $a$ be an element of finite order $n$ in a group. Then $\langle a^k \rangle=\langle a \rangle$ if and only if $k$ is relatively prime to $n$. The number of generators of $\langle a \rangle$ is $\varphi(n)$.

So my question is whether all generators of $\langle a \rangle$ are of the form $a^k$?

Intuitively that sounds obvious but how can I say that formally? Also, for example if $n=24$, then would all generators of $\langle a\rangle$ be $\{a,a^5,a^7,a^{11},a^{13},a^{17},a^{19},a^{23}\}$?

Answer 1

Yes. By the way, a proposition you mentioned answers your question. And your example is completely correct.

Answer 2

For the first question, $\left<a\right>$ is the group generated by $a$, by definition. Then the generators of the group must have the form $a^k$ for some integer $k$. Moreover, by the proposition you stated, the number $k$ must be relatively prime to $n$: suppose not. Then by the proposition, $\left<a^k\right>\neq \left<a\right>$ and since we know $\left<a^k\right>\subseteq \left<a\right>$, then there exists some element in the group $\left<a\right>$ that cannot be generated by $a^k$, contradiction. For the second question, yes, and you can check this by looking at the assignment of elements (or just number theory).