= Solution
For a <permutation> $g$, start with a letter $i$ and follow $i,g(i),g^2(i),\ldots$. Finiteness gives a repetition; since $g$ is invertible, the first return is to $i$. The visited letters form a <permutation cycle> on which $g$ acts by moving to the next letter. If there are letters left, repeat the construction starting with one of them. Its trajectory cannot meet an earlier one, because applying a suitable inverse power of $g$ would put its starting letter in that earlier trajectory. Continuing partitions the letters into disjoint cycles, including fixed points as cycles of length one.
The resulting <disjoint permutation cycles> have disjoint supports, and their product agrees with $g$ on every letter. This proves the <cycle decomposition of a permutation>. A power $g^m$ fixes a cycle of length $l$ exactly when $l\mid m$, so if the cycle lengths are $l_1,\ldots,l_t$,
$$
\boxed{\operatorname{ord}(g)=\operatorname{lcm}(l_1,\ldots,l_t).}
$$
The <least common multiple> is essential: multiplying the lengths would overcount whenever cycles share a factor.
Back to article page
