Understanding a Theorem regarding Order of elements in a cyclic group This is part of practice midterm that I have been given (our prof doesn't post any solutions to it) I'd like to know whats right before I write the midterm on Monday this was actually a 4 part question, I'm posting just 1 piece as a question in its own cause it was much too long.
Let $G$ be an abelian group.
Suppose that $a$ is in $G$ and has order $m$ (such that $m$ is finite) and that the positive integer $k$ divides $m$.
(ii) Let $a$ be in $G$ and $l$ be a positive integer. State the theorem which says $\langle a^l\rangle=\langle a^k\rangle$ for some $k$ which divides $m$.
I have a theorem in my textbook that says as follows.
Let G be a finite cyclic group of order n with $a \in G$ as a generator for any integer m, the subgroup generated by $a^{m}$ is the same as the subgroup generated by $ a^{d}$ where $d=(m,n)$.
if this is the correct theorem can someone please show me why? thanks

For the theorem in your textbook, note that $(a^d)^{m/d}=a^m$ and $(a^m)^{b}=a^d$ where $b$ is the integer that $bm+cn=d$, which you can find using Euclidean Algorithm. 
This means they can generate each other. Therefore, the groups they generate are the same.
You can use this to prove the question.
The question says $\forall l>0, \exists k|m,\langle a^l\rangle=\langle a^k\rangle$. Using the theorem, we know that $\langle a^l\rangle=\langle a^d\rangle$ where $d=gcd(l, n)$. Since $a$ is of order $m$, $d$ must divides $m$. Therefore, pick $d$ as your $k$.
