The following is an example from Dummit and Foote's Abstract Algebra, 3rd edition, page 59.
Does anyone know if the last statement about inclusions is true for an arbitrary group $\mathbb{Z}/n\mathbb{Z}$? In other words, is the following statement true for any $n\in\mathbb{Z}^{+}$? $$\langle\bar{a}\rangle\leq\langle\bar{b}\rangle~~\text{if and only if}~~(b,n)~\text{divides}~(a,n), ~~1\leq a, b\leq n$$
On
Hint: There are several results that you can check:
Let $G=\langle x \rangle$ where $x$ is of order $n$:
(i) Let $k$ be a positive integer. Then $\langle x^k\rangle=\langle x^{(n,k)}\rangle$
(ii) Let $p$ and $q$ be divisor of $n$. Then $\langle x^p\rangle\le \langle x^q\rangle$ iff $q$ divides $p$.
Yes; this is related to the fact that every finite cyclic group $G$ has exactly one subgroup of order $d$ for every divisor $d$ of $\#G$. Indeed, the subgroup of $\Bbb Z/n\Bbb Z$ of order $d$ is generated by any integer $a$ such that $(a,n) = n/d$. In particular, the subgroup of order $d$ is generated by $a=n/d$ itself, from which your question easily follows.