Condition for an abelian group to be equipped with $\mathbb{Z}/n\mathbb{Z}$ -module structure

Question

I have this algebra question : Let $n$ be any integer strictly greater than $1$, and $(G,+)$ an abelian group such that for all $x$ in G, $nx=0$. Show that $(G,+)$ can be equipped with $\mathbb{Z}/n\mathbb{Z}$-module structure. Now I know that any abelian group can be equipped with a $\mathbb{Z}$-module structure, but it doesn't seem like I need to use this argument as the given is related to this n integer. I am not sure how to define the external multiplicative law and how to implement n in it. Thank you.

Answer 1

In general, if $R$ is a unital ring (commutative for simplicity), $I$ is an ideal of $R$ and $M$ is an $R$-module, how do we know if $M$ is also an $R/I$-module? Assuming it is, choose any $r\in R$ and any $m\in M$. We want $(r+I)m = rm$, from which we may deduce that $Im = 0$ and hence $IM = 0$. In other words, the condition that $xm=0$ for all $x\in I, m\in M$ seems to be necessary.

We may now define try defining an $R/I$-structure on $M$ by setting $(r+I)m=rm$ for all $r\in R, m\in M$. To see that this is well-defined, assume $s+I=r+I$, so that $r-s=x\in I$. Then we see that $rm = (s+x)m=sm+xm=sm$, and hence $(r+I)m=(s+I)m$ - what we wanted.

Now that we know the assignment is well-defined, I'll let you check if it gives you a module structure (spoiler: it does!), and then you can apply it to the special case where $R=\mathbb{Z}, I=n\mathbb{Z}, n>1$. :)