Let $R$ a principal ideal domain and $M$ a cyclic $R$-module of order $\alpha$. Show that for each divisor $\beta$ of $\alpha$ there is a unique submodule of order $\beta$.
Note: below $\langle\cdot\rangle$ stay for ideal generator and $\langle\!\langle\cdot\rangle\!\rangle$ stay for (sub)module generator.
My work so far:
By the uniqueness of the cyclic decomposition on primary submodules of a torsion module it is enough to see the case when $\alpha=p^n$ for some prime $p$.
I already know that $\tau: \langle\!\langle v\rangle\!\rangle\to R/\langle p^n\rangle,\, rv\mapsto r+\langle p^n\rangle$ is a $R$-module isomorphism and so it is enough to check that the submodules $\langle\!\langle p^{n-j}+\langle p^n\rangle\rangle\!\rangle$ are the unique submodules of order $p^j$ in $R/\langle p^n\rangle$.
If $x\in R$ induces a submodule of order $p^j$ in $R/\langle p^n\rangle$ then it is clear that $x=r p^{n-j}$ for some $r\in R$, and so $x\le p^{n-j}$ as induced submodules with equality when $r$ is a unit.
However I get stuck trying to show that $r$ must be necessarily an unit. Some help will be appreciated, thank you.
$r$ is aunit modulo $p^n$, or, what's equivalent $r$ is coprime to $p$. If it were not, we would have $r=p^k r'$, where $\gcd(r',p)=1$, therefore $\;x=r'p^{n-j+k}$ would have order $p^{j-k}\ne p^j$ if $\ne 0$.