I'm just wondering if I've come to the right conclusion here, because something seems off, and I don't know why. Any room for improvement otherwise is appreciated as well.
Chapter 4, Problem 68 in Gallian's Contemporary Abstract Algebra: Suppose that $|x|=n$. Find a necessary and sufficient condition on $r$ and $s$ such that $\langle x^r\rangle\subseteq\langle x^s\rangle$.
Fundamental Theorem of Cyclic Groups (FTCG): Every subgroup of a cyclic group is cyclic. Moreover, if $|\langle a\rangle|=n$, then the order of any subgroup of $\langle a\rangle$ is a divisor of $n$; and, for each positive divisor $k$ of $n$, $\langle a\rangle$ has exactly one subgroup of order $k$ - namely, $\langle a^\frac{n}{k}\rangle$.
Claim: $\langle x^r\rangle\subseteq\langle x^s\rangle$ if, and only if, $|x^r|\vert|x^s|$
Proof: First, if $|\langle x^r\rangle|=j$ and $|\langle x^s\rangle|=k$, then $j,k|n$, and these are the only subgroups contained in $\langle x\rangle$ of orders $j$ and $k$, respectively, by the FTCG. Suppose $\langle x^r\rangle\subseteq\langle x^s\rangle$; then $j|k$, by the FTCG. Suppose $j|k$; then by the FTCG, $\langle x^s\rangle$ has exactly one subgroup of order $j$, and since there is only one subgroup of order $j$ in $\langle x\rangle$, this subgroup is $\langle x^r\rangle$, so $\langle x^r\rangle\subseteq \langle x^s\rangle$. Thus, since $|\langle x^r\rangle|=|x^r|$ and $|\langle x^s\rangle|=|x^s|$, $\langle x^r\rangle\subseteq\langle x^s\rangle$ if, and only if, $|x^r|\vert|x^s|$. //
Now, another theorem points out that if $|x|=n$, then $|x^k|=\frac{n}{\gcd(n,k)}$. Thus, we have the following: $|x^r|\vert|x^s|\Leftrightarrow m\frac{n}{\gcd(n,r)}=\frac{n}{\gcd(n,s)}\Leftrightarrow m\gcd(n,s)=\gcd(n,r)\Leftrightarrow\gcd(n,s)|\gcd(n,r)$.
Therefore, the necessary and sufficient condition on $r$ and $s$ is $\gcd(n,s)|\gcd(n,r)$.