Torsion elements in integer-modules In a worryingly short amount of time I've managed to forget almost everything I knew about modules, groups, rings e.t.c. I'm using the definition that an element $m$ of a module $M$ over an integral domain $R$ is torsion if
$\text{Ann}(m)\neq\{0\}$ where the annihilator is defined by
$\text{Ann}(m)=\{r\in R|rm=0\}$
So an element is torsion if a ring element other than zero kills it. I'm trying to show that in a $\mathbf{Z}$ module, the torsion elements are precisely those which have finite order. So suppose $rm=0$ where $r\in \mathbf{Z}$. I'm not really sure what to do from here. Can we cancel the $r$? Is this because $\mathbf{Z}$ is an integral domain? Then that would give that the order of $m$ is 1, which is finite. Conversely, if $m$ has finite order, then $m^n=0$ for some $n\in \mathbf{N}$. From here I need to find an integer $r$ such that $rm$=0. I'm a bit stumped to be honest.
I'm not even at all happy with the forwards direction, I get that the cancellation law holds in integral domains, but that cancellation is between the elements of the integral domain, i.e. just because we can cancel $r$ in the expression $rx=ry$ where $r,x,y\in \mathbf{Z}$, why should we be able to cancel $r$ in $rm=0$, where $m\in M$, some module.
I know that $\mathbf{Z}$-modules are just abelian groups, so I feel like I should have proved the property $nx=0 \implies x=0$ for $n\in N$, $x\in G$ for abelian groups $G$ in the past at some point (having studied groups before). That's if it is true, I'm actually not convinced it is, because if it were then there would be no such thing as a torsion element, as if an element were torsion, then it would have to be zero.
Thanks for any help.
 A: I would use additive notation rather than multiplicative notation for the abelian group operation, just to make things clearer. In particular let's denote the identity element of the group by $0$ and the inverse of an element $a$ by $-a$ (so $a + (-a) = 0$).
In the identification of the $\mathbb{Z}$-module structure with the abelian group structure, $rm$ (where $r \in \mathbb{N}$ and $m \in M$) really means $m + \dots + m$ ($r$ times) (if one is using multiplicative notation for the group operation, this would be $m^r$). $(-r)m$ really means $(-m) + \dots (-m)$ ($r$ times), or, equivalently, $-(m + \dots + m)$.
So an element being torsion by your definition means precisely that a finite sum $m + \dots + m = 0$, i.e., that $m$ has finite order.
Once you understand the correspondence between the $\mathbb{Z}$-module and abelian group structures, there's really nothing to prove here.
Edit: your wariness of "cancelling" the $n$ is completely correct! As you said, if it were true that $nx = 0$ always implies $x=0$, then there would be no such thing as torsion elements.
A: You are mostly there. Let $x\in M$ be a torsion element, then there exists $r\in \mathbb{Z}-\{0\}$ such that $r\cdot x=0$. Parsing this (and noting that there is no loss of generality assuming that $r>0$ since $-r\cdot x=0$ if and only if $r\cdot x=0$) we see that this says $$\underbrace{x+\ldots+x}_{r\ \text{times}}=0.$$ That is, $x$ has finite order (since the group operation in question is addition).
Conversely, if $x\in M$ has finite additive order, then there exists $r>0$ such that $$\underbrace{x+\ldots+x}_{r\ \text{times}}=0.$$ That is, $r\cdot x=0$ and hence, $x$ is torsion.
