Let $M$ be a finitely generated noetherian module over a ring $R$.
Let $\phi$ be an automorphism of $M$. Let $P = \{ a \in M \mid \exists n \in \mathbb{N}, \text{such that } \phi^{n} (a) =a \}$ be the sets of periodic points of $\phi$ in $M$.
For $a \in P$, let $m_a$ be the smallest positive integers such that $\phi^{m_a} (a) =a$.
Question: Is it possible that $\{m_a \mid a \in P\}$ is unbounded?
Thoughts: If $M$ is a finitely generated abelian group, then by the fundamental theorem of finitely generated abelian groups the period must be bounded. However, finitely generated modules need not be finitely generated as an abelian group. I was wondering if this also holds for the finitely generated modules?
This holds if $M$ is a noetherian module over an arbitrary ring $R$.
For each $n$, let $M_n=\{a\in M:\phi^n(a)=a\}$; note that $M_n$ is a submodule of $M$, since $\phi^n$ is an automorphism of $M$. Moreover, we have $M_{mn}\supseteq M_m+M_n$ for all $m,n\in\mathbb{N}$; indeed if $a\in M_m$, then $\phi^{mn}(a)=(\phi^m)^n(a)=a$, so that $M_m\subseteq M_{mn}$, and similarly $M_n\subseteq M_{mn}$.
Since $M$ is noetherian, the family of submodules $\{M_n:n\in\mathbb{N}\}$ has a maximal element; say $M_{n}$. By the remark above, it follows that $M_m\subseteq M_n$ for all $m\in\mathbb{N}$; otherwise $M_{mn}\supseteq M_m+M_n$ would properly extend $M_n$, contradicting maximality. So $M_m\subseteq M_n$ for all $m\in\mathbb{N}$, and the result follows.