Every permutation is the product of disjoint cycles.

For a given permutation $α \in S_{n}$ , we can unambiguously write

α = (a_{1}, a_{2}, \dots, a_{m}) (b_{1}, b_{2}, \dots, b_{k}) \dots (c_{1}, c_{2}, \dots, c_{s})

where each different letter signifies a disjoint cycle.

Proof

Let $α \in Perm (A)$ , with $A = {1, 2, \dots, n}$

Choose any $a_{1}$ in $A$
Let $a_{2} = α (a_{1})$
Let $a_{3} = α (a_{2}) = α (α (a_{1}))$
and so on, where
$a_{k} = α^{k - 1} (a_{1})$

Since $A$ is finite, $α^{i} (a_{1}) = α^{j} (a_{1})$ for some $i, j$ (let $i < j$ without loss of generality)
Then $a_{1} = α^{m} (a_{1})$ for $m = j - i$

Thus, there is a cycle of $m$ elements of $A$ , and we express this as

α = (a_{1}, a_{2}, \dots, a_{m}) \dots

where the centered dots reflect the fact that it is possible for $m < n$ , so some elements may be uncaptured; in such a case, we choose a $b_{1}$ not appearing in the first cycle and repeat the process to generate a second cycle, i.e. until $b_{1} = α^{k} (b_{1})$ for some $k$ .
#Lemma 1 This second cycle will have no elements in common with the first cycle.

α = (a_{1}, a_{2}, \dots, a_{m}) (b_{1}, b_{2}, \dots, b_{k}) \dots

If $(m + k) < n$ , repeat the process for another uncaptured element of $A$ , say $c_{1}$ .
And so on, until all elements of $A$ appear in the disjoint cycle notation.

Lemma 1: This second cycle will have no elements in common with the first cycle.

Suppose for the sake of contradiction that some element $x$ is contained in both the first and second cycles.
Then $$\alpha^i(a_1)=x=\alpha^j(b_1)$$
However, this would entail $α^{i - j} (a_{1}) = b_{1}$ , i.e. that $b_{1} = a_{t}$ for some $t$ . This contradicts the way $b_{1}$ was chosen.