First of all, I am very new to group theory.
The order of an element $g$ of a group $G$ is the smallest positive integer $n: g^n=e$, the identity element. I understand how to find the order of an element in a group when the group has something to with modulo, for example, in the group $$U(15)=\text{the set of all
positive integers less than } n \text{ and relatively prime to } n.$$
$$\text{ which is a group under multiplication by modulo }n=\{1,2,4,7,8,11,13,14\},$$ then $|2|=4$, because
\begin{align*}
&2^1=2\\
&2^2=4\\
&2^3=8\\
&2^4=16\mod15=1\\
&\text{So } |2|=4.
\end{align*}
However, I don't understand how this works for groups that don't have any relation to modulo. Take $(\mathbb{Z},+)$ for instance. If I wanted to find the order of $3$, then I need to find $n:3^n$ is equal to the identity, which in this case is $0$.
I suppose my question can be summarized as follows:
Does the order of an element only make sense if we are dealing with groups dealing with modulo?
Best Answer
Yes, it makes sense. The order of an element $g$ in some group is the least positive integer $n$ such that $g^n = 1$ (the identity of the group), if any such $n$ exists. If there is no such $n$, then the order of $g$ is defined to be $\infty$.
As noted in the comment by @Travis, you can take a small permutation group to get an example. For instance, the permutation $(1,2,3,4)$ in the symmetric group $S_4$ of degree $4$ (all permutations of the set $\{1,2,3,4\}$) has order $4$. This is because $$(1,2,3,4)^1 = (1,2,3,4)\neq 1,$$ $$(1,2,3,4)^2 = (1,3)(2,4)\neq 1,$$ $$(1,2,3,4)^3 = (1,4,3,2)\neq 1$$ and $$(1,2,3,4)^4 = 1,$$ so $4$ is the smallest power of $(1,2,3,4)$ that yields the identity.
For the additive group $\mathbb{Z}$ of integers, every non-zero element has infinite order. (Of course, here, we use additive notation, so to calculate the order of $g\in\mathbb{Z}$, we are looking for the least positive integer $n$ such that $ng = 0$, if any. But, unless $g = 0$, there is no such $n$, so the order of $g$ is $\infty$.)