Let $G = \langle a\rangle$ be a group of order $n$. If $H$ is a subgroup of $G$ and $|H|=m$ divides $n$, then $H = \langle a^\frac{n}{m}\rangle$.

Question

I have to show the following statement:

Let $G = \langle a\rangle$ be a cyclic group of order $n$. If $H$ is a subgroup of $G$ and $|H|=m$ divides $n$, then $H = \langle a^d\rangle$ where $d = \frac{n}{m}$.

My attempt:

Since $G = \langle a\rangle$, we know that $H = \langle a^k\rangle$ where $k$ is the least positive integer such that $a^k\in H$. Then, it is enough to prove that $k = d$. Note that $(a^k)^m = e$, because the order of $H$ is $m$. Since the order of $G = \langle a\rangle$ is $n$, it follows that $n\leq km$. Hence, $d = \frac{n}{m}\leq k$. However, I do not get why $k\leq d$, or equivalently, $km\leq n$. It is clear that $a^{n} = e$, but I don't know how to use this fact in order to show that $km\leq n$.

Edit: In this post it is said that $km\leq n$ since $a^{n} = e$ and $m$ is the least positive integer such that $(a^k)^m = e$. Nevertheless, I think that this argument is wrong because we should have showed firstly that $k\mid n$. Am I misunderstanding something?

Answer 1

From $a^{km}=e$, it follows that $km$ is a multiple of $n$, say $km=ln$ with $l\in\mathbb{N}$. Since $m|n$ you can deduce $l|k$. It follows that $$\left(a^{\frac{k}{l}}\right)^m=e.$$

Now, the subgroup $\langle a^{k/l}\rangle$ contains at most $m$ elements by the equation above, but also $a^k\in\langle a^{k/l}\rangle$, so that $\langle a^{k/l}\rangle=H$ and in particular $a^{k/l}\in H$. However, $k$ is the smallest positive integer with that property, thus $l=1$ and $km=n$.