Let $p$ be a prime number. Consider a direct sum of $n$ copies of the cyclic group of order $p$, written $C_p$: $G=C_p^{\oplus n}$. By a comment in this question, given an $R$-module $M$ and an ideal $I\subset R$, there is a structure of an $R/I$-module on $M$ given by $\overline r x=rx$ ($r\in R,\ x\in M$) provided $IM=0$. In our case, $G$ is a $\mathbb Z$-module; after setting $I=(p)$, we see that $IM=0$ is equivalent to $pG=0$, which holds. So there is a $\mathbb Z/p\mathbb Z$ -module structure on $G$. In other words, there is an $\mathbb F_p$-vector space structure on $G$.
As I'm writing this, I came to realize that I don't really understand why is the dimension of this space $n$. (I believe this is so.) How to justify this?
Further, any group endomorphism of $G$ is by definition a $\mathbb Z$-module endomorphism of $G$, and it is obvious that any such endomorphism is also an endomorphism as an endo of $\mathbb F_p$-vector space. Why is the converse true? I.e., why is any endomorphism of $G$ as an $\mathbb F_p$-vector space must be an endomorphism of $G$ as a group ($\mathbb Z$-module)? First of all, we need to define the structure of a $\mathbb Z$-module on the $\mathbb F_p$-vector space $G$; how can we do it?