Proof

First we show that every g in S_n can be written as a product of disjoint cycles (the existence). Then we prove the uniqueness of this product. Both parts are proved by induction.

Existence.

Uniqueness.

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Existence
We use induction with respect to the number of elements in the support of the permutation g.

If the support of g is empty, then g is e, the identity element, a 0-cycle. We regard this as an empty product of cycles.

Now assume that for some number k > 0 any element g with |supp(g)| < k can be written as a product of disjoint cycles. Let g be an element with k elements in its support. Fix an element x in supp(g). We try to `split off' a cycle containing x.

We set a₁ = x and a_{i + 1} = g(a_i) for i > 0. Let m denote the smallest positive integer for which a_{m + 1} = x and consider the cycle c = (a₁, ..., a_m). Its support is a subset of supp(g). So the permutation h = gc^-1 fixes all points of fix(g) as well as the points a_i, 0 < i < m + 1. Indeed, h(a_i) = gc^-1 (a _i) = g(a_{i - 1}) = a_i, where we set a₀ = a_m. This implies that the support of h is contained in supp(g)\ {a₁, ..., a_m}. By the induction assumption we may write h as a product of disjoint cycles c₁, c₂, ... , c_t. The support of these cycles is contained in supp(h) and therefore disjoint from {a₁, ..., a_m}. But then

g = hc = c₁ ··· c_t · c

is a product of disjoint cycles. By induction we have finished the first part of the proof.

Uniqueness
Assume that g is the product of the disjoint cycles c₁, ..., c_t and at the same time of the disjoint cycles d₁, ..., d_s, all of length at least 2. We prove the uniqueness by induction on t.

The case t = 0 is trivial. So assume that t > 0. Then supp(g) is not empty and we can find an element x in supp(g). As x is not fixed by g, there exist cycles c_i and d_j which do not fix x. Without loss of generality we may suppose that x supp(c_t) and x supp(d_s). For every m N, we have

c_t^m(x) = g^m(x) = d_s^m(x).

In particular c_t = d_s. But then also c₁ ··· c_{t - 1} = gc_t ^{- 1} = gd_s ^{- 1} = d₁ ··· d_{s - 1}. The induction hypothesis yields that t - 1 = s - 1 and, possibly after renumbering of the indices, c_i = d_i for all i from 0 to t. This proves the proposition.