If G is a finite group and $x \in G$, there is an integer n $\geq 1$ such that $x^n = e$

Question

Let G be a finite group. Show that, given $x \in G$, there is an integer n $\geq 1$ such that $x^n = e$.

I'm trying to use the info that is finite, but I can't find a way. For instance, if G is a multiplicative group and G = {-1,1} the affirmation is true.

Answer 1

Suppose that $x \in G$. If $x = e$ then set $n = 1$. Suppose that $x \neq e$. Since $G$ is finite the sequence $x^{n}$ for $n \in \mathbb{N}$ has duplicates. Suppose that $x^{m} = x^{n}$ for distinct $m$ and $n$ with $m < n$. Then $x^{n - m} = e$.

Answer 2

It's always nice to have multiple ways to prove something, so I will add an alternate method, although it does not necessarily use the fact that G is finite.

If $\mathbf G$ is a finite group and $x$ is an element of $\mathbf G$, then the cyclic group generated by $x$, or $<x>$, is a subgroup. A subgroup must contain an identity element $e$; otherwise, it would not be a subgroup. By the definition of $<x>$, there exists an integer n such that $x^n=y,\space \forall \space y\in <x>$. In particular, $e \in <x>$, so there must be an integer $n$ such that $x^n=e$ is true.

But to show that n is non zero then we need to go back to the fact that the subgroup is finite and use the pigeonhole principle