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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.

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    $\begingroup$ Use Lagrange's theorem. $\endgroup$ – Angina Seng May 10 '18 at 1:18
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    $\begingroup$ As Lord Shark the Unknown and mathematics2x2life observed, this follows from Lagrange's theorem; but it is a shame to lose the conceptual idea of your proof, which I like. Notice that it works unchanged if $g$ is central in $\mathbb G$, even if $\mathbb G$ is not Abelian, and so proves that $g^{\lvert C_{\mathbb G}(g)\rvert} = 1$. I don't see how to get to $g^m = 1$ without invoking Lagrange's theorem, though. $\endgroup$ – LSpice May 10 '18 at 1:35
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    $\begingroup$ This is my pet question. See math.stackexchange.com/questions/28332/… $\endgroup$ – lhf May 10 '18 at 2:11
  • $\begingroup$ Possible duplicate of math.stackexchange.com/questions/28317/… $\endgroup$ – lhf May 10 '18 at 2:13
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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.

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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.

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