Using multiplicative notation, for a finite group $G$, the order of an element $g\in G$ is $\text{ord}(g) = n$ iff $g^n = e, n \in \mathbb{Z}, n>1$ and such that for all $m\neq n$, if $g^m = e \rightarrow m\ge n$.
For a finite additive group $G$, $\text{ord}(g\in G) = n$, where $n$ is the least positive integer such that $ng = 0$.
In general, given a group $G$, if there is no positive integer $n$ such that $g^n = e$ for a given $g\in G$ (additively, such that $ng = e$), then $g$ is of infinite order, and the converse is also true.
The order of an element in a group $G$ can also be thought of as equal to the order of the subgroup it generates: for $g\in G, \text{ord}(g) = |\langle g \rangle|$.