Consider the ring of $p$-adic integers $\mathbb{Z}_p$.
Let $M$ be a $\mathbb{Z}_p$-module with cardinality powers of $p$ and consisting of $p^n$-torsion points for some $n \in \mathbb{N}$.
In other words, for any $m \in M$, we get $p^n \cdot m=0$.
I want to know when $M$ can be cyclic $\mathbb{Z}_p$-module ?
Since $M$ is finitely generated, it would be cyclic if we show $M$ has an unique maximal submodule.
Let us arrange the elements of $M$ with an indexing $\{m_1,m_2,\cdots, m_i, \cdots \}$ in the sense that $m_i$ has order $p^i$.
- If we assume $pm_i=m_{i+1}$, then I think $M=\left\langle m_1 \right\rangle$. In that case $M$ is cyclic.
Do we have other way to show that $M$ is cyclic or $M$ has unique submodule which is maximal ?
Thanks
As said in coments, it is not clear what exactly you are asking. Maybe the following reminder helps:
For a unital commutative ring $R$, every cyclic $R$-module is isomorphic to $R/I$ for some ideal $I$ of $R$.
The only ideals of $\mathbb Z_p$ are $(0)$ and all $(p^n)$ for $n \in \{0,1,2,...\}$.
Hence up to isomorphism, the only cyclic $\mathbb Z_p$-modules are $\mathbb Z_p$ itself and $\mathbb Z_p/(p^n)$ for $n \in \{0,1,2,...\}$. It is well-known that $\mathbb Z_p/(p^n) \simeq \mathbb Z/(p^n)$.
If you have a $\mathbb Z_p$-module $M$ which is torsion and has finite cardinality ($p^c$, say), then in particular it is finitely generated, and hence by the structure theory of f.g. torsion modules over PIDs (or an easier version thereof in this case of a DVR), it is a direct sum of cyclic modules. So here:
$$M \simeq \bigoplus_{n \ge 0} (\mathbb Z/p^n)^{i_n} \qquad \qquad (*)$$
where $(\mathbb Z/(p^n))^{i_n}$ is short notation for $\underbrace{\mathbb Z/(p^n) \oplus ... \oplus \mathbb Z/(p^n)}_{i_n \text{ times}}$, and only finitely many $i_n$ are $\neq 0$. Actually, for $M$ to have cardinality $p^c$, we need $\prod_{n \ge 0} n \cdot i_n = c$.
Obviously the cyclic ones among these are the ones where $i_n=1$ for $n=c$ and $i_n=0$ for all $n \neq c$.
Or in other words, if the module has cardinality $p^c$ and is cyclic, it must be $\simeq \mathbb Z/p^c$. So: If we know the module has cardinality $p^c$, then for it to be cyclic it suffices to show, for example, that it has an element of order $p^c$ (which is nearly trivially equivalent to being cyclic, if you think about it).
As regards torsion, if we know that $M$ has exact torsion $p^t$ (i.e. $p^t \cdot m=0$ for all $m$ but there is $m \in M$ with $p^{t-1} \cdot m \neq 0$), then in $(*)$ this only imposes the restriction that $i_t \neq 0$ and $i_{t+1}=i_{t+2}= ... =0$. As said above, such module is cyclic iff $i_t=1$ and all other $i_n=0$ i.e. iff $M =\simeq \mathbb Z/(p^t)$ i.e. iff it has cardinality $p^t$. (It would suffice to show its cardinality is $\le p^t$.)
As said in comments, I find it hard to imagine how you are given a module $M$ for which you can decide at all whether it is cyclic, but without that decision being kind of immediate via any of the criteria above.