Order of element in finite group is finite

Dependencies:

  1. Group

Let $G$ be a group and $g \in G$. If $G$ is finite then $\operatorname{order}(g)$ is finite.

Proof

Consider the infinite sequence $[g^0, g^1, g^2, g^3, \ldots]$. By closure of $G$, $g^i \in G$ for all $i$. Since $G$ is finite but this sequence is infinite, there must be repetition in this sequence.

Let $i$ be the first index with a repeated element, i.e. the smallest integer for which $\exists j < i$ such that $g^j = g^i$. $ g^i = g^j \implies g^{i-j} = e = g^0 $. $j < i \Rightarrow i - j > 0$, so $i - j$ is an index with a repeated element, the element being $e$. For $i$ to be the smallest index with a repeated element, $i \le i - j \Rightarrow j = 0$. Therefore, the first element to be repeated is $e$.

Therefore, there exists a positive integer $i$ such that $g^i = e$. Therefore, $\operatorname{order}(g) = i$, which is finite.

Dependency for:

  1. Order of cyclic subgroup is order of generator

Info:

Transitive dependencies:

  1. Group