Every permutation is a product of disjoint cycles

Let \(\sigma\in S_n\) act on \(\{1,\dots,n\}\). Its orbits partition this finite set.

For an orbit containing \(a\), choose the least \(m>0\) with \(\sigma^m(a)=a\). Then \[ a,\sigma(a),\dots,\sigma^{m-1}(a) \] are distinct and form the entire orbit, and on this orbit \(\sigma\) acts as the cycle \[ (a\ \sigma(a)\ \dots\ \sigma^{m-1}(a)). \] Distinct orbits are disjoint, so the corresponding cycles are pairwise disjoint. Their product agrees with \(\sigma\) on every point, hence equals \(\sigma\).

The decomposition is unique up to reordering the disjoint cycles and inserting or deleting \(1\)-cycles, because the supports of the nontrivial cycles are precisely the nontrivial orbits of the cyclic group \(\langle\sigma\rangle\).