I believe that my notes are a little off because it seems that in one instance per$(M)$ is a set and in another case it's an element. Maybe a computation from the notes would help clarify the situation.
For starters, some definitions:
Let $M$ be a module over a pid $D$ and let $x\in M$. Then $\gamma_x : D\to M$ be given by $d\mapsto d.x$.
Then a generating element of $\ker(\gamma_x)=\{d|d.x=0\}$ is called the period of $x$, denoted per$(x)$.
We define the period of $M$ as per$(M):=\{ d| d.x=0 \text{ for all } x\in M\} $
(is this definition correct?)
Here's an example I'm stuck on, "Now let $M$ be a finitely generated torsion module (i.e. Tor$(M)=M$) over the pid $D$. If $M=\langle x_1,...,x_n \rangle $then it is an easy exercise to show that per$(M)=lcm (per(x_i))=d\neq 0$. Then if one has $d=p_1^{\nu_1}\cdots p_n^{\nu_n}$, one has that $$M=M(p_1^{\nu_1})\oplus \cdots \oplus M(p_n^{\nu_n})$$
Where $M(d):=\ker(\gamma_d)$. "
Note: the $=$ to the left of $d$ in $=d\neq 0$ might more appropriately be $=:$ but I'm not sure, $d$ just appeared suddenly from nowhere.
Say $D$ is a PID and $M$ is a $D$-module. Suppose further that $dM=0$ for $d\in D$, and that we have a factorization $d=\ell_1\ell_2\cdots\ell_n$ where the $\ell_i$s are pairwise coprime (i.e. $(\ell_i,\ell_j)=D$). Then
$$M=M(\ell_1)\oplus M(\ell_2)\oplus\cdots\oplus M(\ell_n)$$
where we define $M(\ell)=\{m\in M:\ell m=0\}$.
There is an inclusion in one direction. (Try to prove that no nonzero element of $M$ can be in $M(\ell_i)$ for two distinct $i$.) For the other direction, we need to construct a map $M\to\bigoplus_{i=1}^n M(\ell_i)$ out of maps $M\to M(\ell_i)$. The maps $M\to M(\ell_i)$ need to be the identity on the submodule $M(\ell_i)$ within $M$ but in a sense annihilate "everything else." Thus, make them out to be the multiplication maps $m\mapsto t_i m$ where $t_i$ is $1$ mod $\ell_i$ and $0$ mod $\ell_j$ for $j\ne i$. I leave it to you to prove that this works out.