Statement: Let $p$ be a prime and suppose that $G$ is a cyclic group of order $p^n$, where $n$ is any given non-negative integer. Then, given any two subgroups of $G$, say $H$ and $K$, either $H\subseteq K$ or $K\subseteq H$ holds.
I tried to prove the above as follows:
If $n=0$ or $1$, then the statement is true. (For $n=0$, the statement holds vacuously and for $n=1, G$ doesn't have any non-trivial proper subgroups so the only distinct subgroups being $G$ and the trivial subgroup containing only identity and hence the statement is true.)
So suppose that $n\gt 1$. Let $H$ and $K$ be any two given non-trivial proper subgroups of $G$. (The statement holds if either of the given subgroups is either $G$ or trivial subgroup $\{e\}$, where $e\in G$ is the identity of $G$). Let $H\ne K$.
$H$ and $K$ (subgroups of $G$) are cyclic so by Lagrange's theorem the following holds:
$|H|=\langle h\rangle =p^r \text{ and } |K|=p^s$, where $1\lt r,s\lt p^n$.
It follows that $r\ne s$. WLOG, let $r\lt s$. Converse of Lagrange's theorem holds for cyclic groups so $h\in K$ ($p^r|p^s\implies K$ has $\phi(p^r)$ elements of order $p^r$ but $G$ being cyclic has only $\phi(p^r)$ elements of order $p^r$ in it). It follows that $H\subset K$. This proves the statement.
Is my proof correct? Thanks.
On a short Note:
Let $G$ be a cyclic group of order $\mathbb{P}^{n}$, then as we know converse of Lagrange theorem is true for the cyclic group.
So there is exactly one subgroup of order $1,\mathbb{P},\mathbb{P}^2...\mathbb{P}^{n-1},\mathbb{P}^n$ in $G$, and clearly see that for any two subgroups one is contained in the other. For example order of $G$ is $32=2^5.$ So the subgroups of $G$ are of order $1,2, 2^2,2^3,2^4,2^5$ also "Any finite cyclic group is isomorphic to $\mathbb{Z}_{n}"$. Hence in this case subgroups are $32\mathbb{Z}_{32}$ $\subset $ $16\mathbb{Z}_{32}$ $\subset $ $8\mathbb{Z}_{32}$ $\subset $ $4\mathbb{Z}_{32}$ $\subset $ $2\mathbb{Z}_{32}$ $\subset $ $\mathbb{Z}_{32}$.
I hope you understand.