Prove that $g^m=1$ for any element $g$ in a finite group of order $m$

Question

Let $\mathbb{G}$ be a finite group with $m = |\mathbb{G}|$, the order of the group. Prove that $g^m = 1$ for any element $g \in \mathbb{G}$.

I can prove it if $\mathbb{G}$ is abelian. If $\mathbb{G}$ is abelian, for an arbitrary $g \in \mathbb{G}$ and $g_1, \cdots g_m \in \mathbb{G}$, $$g_1 \circ g_2 \circ \cdots \circ g_m = (g \circ g_1)\circ(g \circ g_2) \cdots (g \circ g_m) = g^m \circ (g_1 \circ g_2 \cdots g_m)$$ Since the right-hand side is equal to the left-hand side, $g^m = 1$.

However, I really don't know how to prove it when $\mathbb{G}$ is not abelian.

Answer 1

By Lagrange's Theorem, the order of any element $g \in G$ divides the order of the group. Suppose that $|G|=m$ and that $|g|=n$. Since $n$ must divide $m$, we must have $m= rn$ for some integer $r$. But then we have $$ g^m=g^{rn}=(g^n)^r=1_G^r=1_G, $$ as desired.

Answer 2

Here is the simplest proof I know for the general case. It uses the coset proof of Lagrange's theorem in mild disguise.

First, note that $a^m=1$ iff $m$ is a multiple of $o(a)$.

Take a fixed $a \in G$ and consider the map $f\colon G \to G$ given by $f(x)=ax$.

Consider the orbits of $f$, that is, the sets $\mathcal{O}(x)=\{ x, f(x), f(f(x)), \dots \}$ for $x \in G$.

The map $ax \mapsto ay$ is a bijection between $\mathcal{O}(x)$ and $\mathcal{O}(y)$.

Therefore, all orbits have the same number of elements.

In particular, $|\mathcal{O}(x)|=|\mathcal{O}(1)|$. By definition, $|\mathcal{O}(1)| = o(a)$.

Therefore, $G = \bigcup_{x\in G} \mathcal{O}(x)$ implies that $|G|$ is a multiple of $o(a)$ because different orbits are disjoint.