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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?

user557
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  • Do you believe/understand why $\mathbb{R}^n$ is $n$-diml over the reals? – Randall Aug 06 '18 at 01:33
  • Yes I understand. – user557 Aug 06 '18 at 01:35
  • It's the same idea. You have a basis consisting of vectors like $(0, \ldots, 0, 1, 0, \ldots, 0)$ where the single $1$ in the $i$th spot takes care of the $i$th factor of $C_p$ in $G$. This is your dimension question. The endo question is different. – Randall Aug 06 '18 at 01:37
  • The endo question: a vector space map is also a group homomorphism (just forget the scaling for a moment). – Randall Aug 06 '18 at 01:39
  • Oh, I didn't realize that the basis for the $\mathbb Z_p$-module can be chosen the same as the basis for the $\mathbb Z$-module. I did realize that the $e_i$'s form a basis for $G$ as a $\mathbb Z$-module. – user557 Aug 06 '18 at 01:39
  • The $0$s and $1$s mean something different (different group) but not really enough to matter. – Randall Aug 06 '18 at 01:50
  • Oh, and be careful: it doesn't have a "basis" as a $\mathbb{Z}$-module because it's not free. – Randall Aug 06 '18 at 01:51
  • Right, in this case $0$ is the identity element of $C_p$ and $1$ is a generator of $C_p$. Indeed, the free $\mathbb Z$-modules are the modules $\mathbb Z^m$. Then I can only say that the $e_i$ form a basis as an $\mathbb F_p$-vector space. – user557 Aug 06 '18 at 02:03
  • Right on. Carry on. – Randall Aug 06 '18 at 02:10

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