Vtyjkyuk

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 $mathbbZ/nmathbbZ$? In other words, is the following statement true for any $ninmathbbZ^+$?
$$langlebararangleleqlanglebarbrangle~~textif and only if~~(b,n)~textdivides~(a,n), ~~1leq a, bleq n$$

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 $Bbb Z/nBbb Z$ 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.

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 $langle x^krangle=langle x^(n,k)rangle$

(ii) Let $p$ and $q$ be divisor of $n$. Then $langle x^pranglele langle x^qrangle$ iff $q$ divides $p$.