For $\bigoplus_{n \leq 0}\mathbb{Z}_{p^n}$, the classes $\operatorname{Gen}(\bigoplus_{n \leq 0}\mathbb{Z}_{p^n})$ and $p$ torsion groups are equal. Let's consider a prime number $p \in \mathbb{Z}$ and the group $M=  \bigoplus_{n \leq 0}\mathbb{Z}_{p^n}$ (this means $M=\mathbb{Z} \oplus \mathbb{Z}_{p^2} \oplus \mathbb{Z}_{p^3} \oplus\cdots$) which is a module over $\mathbb{Z}$. I want to prove that $\operatorname{Gen}(M)=T_p$ where
$$T_p:= \lbrace G \mid G \text{ is a $p$-torsion group}\rbrace,$$
where a group $G$ is $p$-torsion means that for every $g \in G$ we have that $p^k g= e$ for some $k \in \mathbb{Z}$. Also,
$$\operatorname{Gen}(M):= \lbrace N \in \operatorname{Mod}(\mathbb{Z}) \mid M^{(X)} \twoheadrightarrow N \rbrace$$
where $\operatorname{Mod}(\mathbb{Z})$ is the category of modules over $\mathbb{Z}$ and $M^{(X)} \twoheadrightarrow N$ is a surjective map from arbitrary number of direct copies of $M$ into $N$. I need help proving both contentions, so far I have noticed that $M \in \operatorname{Gen}(M)$ since we have $1_{M}:M  \twoheadrightarrow M$ and that $M=  \bigoplus_{n \leq 0}\mathbb{Z}_{p^{n}} \in T_{p}$ my intuition says that $\operatorname{Gen}(M) \subseteq T_{p}$ is the easy contention.  For the $T_p \subseteq \operatorname{Gen}(M)$ contention my idea is to use that if $G \in T_p$ then we can think $G=\sum_{g \in G} \mathbb{Z}_{g}$ where  $\mathbb{Z}_g \cong \mathbb{Z}_{p^m}$ for some $0<m \in \mathbb{Z}$ so we have a surjection
$$ \bigoplus_{g \in G} \mathbb{Z}_{g} \twoheadrightarrow \sum_{g \in G} \mathbb{Z}_{g}=G$$
and each $\bigoplus_{g \in G} \mathbb{Z}_{g}$ is a direct summand of $M^{(G)}$. Thanks!
 A: Your intuition is right on track, I think, although I am confused by your approach to writing $G$ as a sum in the end of the proof, so my answer might be approaching that differently than you had intended.
$\operatorname{Gen}(M) \subseteq T_p$
Suppose that $N$ is a $\mathbb{Z}$-module such that there is an index set $X$ and a surjection $f: M^{(X)} \twoheadrightarrow N$.  Note that $\mathbb{Z}/p^n\mathbb{Z}$ is $p$-torsion, and that a direct sum of modules is $p$-torsion if (and only if) its summands are $p$-torsion, and thus conclude that $M$ is $p$-torsion and furthermore $M^{(X)}$ is $p$-torsion.  Next, note that homomorphic images of $p$-torsion modules are $p$-torsion, and thus conclude that $N$ is $p$-torsion.
$T_p \subseteq \operatorname{Gen}(M)$
Suppose that $G$ is a $p$-torsion group. Thinking of $G$ as a $\mathbb{Z}$-module, this just means that for all $g \in G$, there exists $n \in \mathbb{N}$ such that $p^n g = 0$.  We can view $G$ as a colimit of its finitely generated submodules.  Each of these f.g. submodules decomposes as $\bigoplus_{i=1}^k \mathbb{Z}/d_i\mathbb{Z}$ where $d_i \mid d_{i+1}$ (here we are appealing to, say, the structure theorem for f.g. modules over PIDs).  The fact that each element of this f.g. submodule is annihilated by $p^n$ for some $n$ implies that each $d_i$ is a power of $p$.  Thus $G$ is a colimit of finite direct sums of the form $\bigoplus_{i=1}^{k} \mathbb{Z}/p^{n_i}\mathbb{Z}$.  A colimit of modules is by definition a special quotient of their direct sum, so this gives us a representation of $G$ as the quotient of a module of the form $H = \bigoplus_{\alpha \in I} \mathbb{Z}/p^{n_\alpha}\mathbb{Z}$ where $I$ is some index set.  It is now straightforward to find an index set $X$ such that $M$ surjects onto $H$, and hence onto $G$ (because $G$ is a quotient of $H$).  For example, you could take $X = I$ and the map $f: M^{(I)} \twoheadrightarrow H$ defined componentwise as follows: the $f_\alpha$ component of $f$ maps the $\mathbb{Z}/p^{n_\alpha}\mathbb{Z}$ component of $M$ identically to the $\alpha$ component of $H$ (which is also $\mathbb{Z}/p^{n_\alpha}\mathbb{Z}$) and vanishes on the other components of $M$.
