By definition, a cycle decomposition of an element of $S_n$ is a product of disjoint cycles. === Construction of Disjoint Permutations === Let $\sigma \in S_n$ be a permutation on $S_n$. Let $\mathcal R_\sigma$ be the equivalence defined in Permutation Induces Equivalence Relation. Let $\N_n$ be used to denote the (one-based) initial segment of natural numbers: :$\N_n = \closedint 1 n = \set {1, 2, 3, \ldots, n}$ Let $\N_n / \mathcal R_\sigma = \set {E_1, E_2, \ldots, E_m}$ be the quotient set of $\N_n$ determined by $\mathcal R_\sigma$. By Equivalence Class of Element is Subset: :$E \in \N_n / \mathcal R_\sigma \implies E \subseteq \N_n$ For any $E_i \in \N_n / \mathcal R_\sigma$, let $\rho_i: \paren {\N_n \setminus E_i} \to \paren {\N_n \setminus E_i}$ be the identity mapping on $\N_n \setminus E_i$. By Identity Mapping is Permutation, $\rho_i$ is a permutation. Also, let $\phi_i = \tuple {E_i, E_i, R}$ be a relation where $R$ is defined as: :$\forall x, y \in E_i: \tuple {x, y} \in R \iff \map \sigma x = y$ It is easily seen that $\phi_i$ is many to one. For all $x \in E_i$: {{begin-eqn}} {{eqn | l = x       | o = \mathcal R_\sigma       | r = \map \sigma x       | c =  }} {{eqn | ll= \leadsto       | l = \map \sigma x       | o = \in       | r = E_i       | c =  }} {{eqn | ll= \leadsto       | l = \sigma \sqbrk {E_i}       | o = \subseteq       | r = E_i       | c =  }} {{end-eqn}} which shows that $\phi_i$ is left-total. It then follows from the definition of a mapping that $\phi_i: E_i \to E_i$ is a mapping defined by: :$\map {\phi_i} x = \map \sigma x$ It is seen that $\phi_i$ is an injection because $\sigma$ is an injection. So by Injection from Finite Set to Itself is Permutation, $\phi_i$ is a permutation on $E_i$. By Intersection with Relative Complement is Empty, $E_i$ and $\N_n \setminus E_i$ are disjoint. By Union with Relative Complement: : $E_i \cup \paren {\N_n \setminus E_i} = \N_n$ So by Union of Bijections with Disjoint Domains and Codomains is Bijection, let the permutation $\sigma_i \in S_n$ be defined by: :$\map {\sigma_i} x = \map {\paren {\phi_i \cup \rho_i} } x = \begin{cases} \map \sigma x & : x \in E_i \\ x & : x \notin E_i \end{cases}$ By Equivalence Classes are Disjoint, it follows that each of the $\sigma_i$ are disjoint. {{qed|lemma}} === These Permutations are Cycles === It is now to be shown that all of the $\sigma_i$ are cycles. From Order of Element Divides Order of Finite Group, there exists $\alpha \in \Z_{\gt 0}$ such that $\sigma_i^\alpha = e$, and so: :$\map {\sigma_i^\alpha} x = \map e x = x$ By the Well-Ordering Principle, let $k = \min \set {\alpha \in \N_{\gt 0}: \map {\sigma_i^\alpha} x = x}$ Because $\sigma_i$ fixes each $y \notin E_i$, it suffices to show that: :$E_i = \set {x, \map {\sigma_i} x, \ldots, \map {\sigma_i^{k - 1} } x}$ for some $x \in E_i$. If $x \in E_i$, then for all $t \in \Z$: :$x \mathrel {\mathcal R_\sigma} \map {\sigma_i^t} x \implies \map {\sigma_i^t} x \in E_i$ It has been shown that: :$(1) \quad \set {x, \map {\sigma_i} x, \ldots, \map {\sigma_i^{k - 1} } x} \subseteq E_i$ Let $x, y \in E_i$. Then: {{begin-eqn}} {{eqn | l = x       | o = \mathcal R_\sigma       | r = y }} {{eqn | ll= \leadsto       | l = \map {\sigma_i^t} x       | r = y       | c = for some $t \in \Z$, by Permutation Induces Equivalence Relation }} {{eqn | ll= \leadsto       | l = \map {\sigma_i^{k q + r} } x       | r = y       | c = for some $q \in \Z$, and $0 \le r \lt k$ by the Division Theorem }} {{eqn | ll= \leadsto       | l = \map {\sigma_i^r \sigma_i^{k q} } x       | r = y }} {{eqn | ll= \leadsto       | l = \map {\sigma_i^r} x       | r = y       | c = Fixed Point of Permutation is Fixed Point of Power }} {{end-eqn}} It has been shown that: :$(2) \quad E_i \subseteq \set {x, \map {\sigma_i} x, \ldots, \map {\sigma_i^{k - 1} } x}$ Combining $(1)$ and $(2)$ yields: :$E_i = \set {x, \map {\sigma_i} x, \ldots, \map {\sigma_i^{k - 1} } x}$ {{qed|lemma}} === The Product of These Cycles form the Permutation === Finally, it is now to be shown that $\sigma = \sigma_1 \sigma_2 \cdots \sigma_m$. From the Fundamental Theorem on Equivalence Relations: :$x \in \N_n \implies x \in E_j$ for some $j \in \set {1, 2, \ldots, m}$. Therefore: {{begin-eqn}} {{eqn | l = \map {\sigma_1 \sigma_2 \cdots \sigma_m} x       | r = \map {\sigma_1 \sigma_2 \cdots \sigma_j} x }} {{eqn | r = \map {\sigma_j} x       | c = because $\sigma_j \sqbrk {E_j} = E_j$ }} {{eqn | r = \map \sigma x       | c = Definition of $\sigma_i$ }} {{end-eqn}} and so existence of a cycle decomposition has been shown. {{qed|lemma}} === Uniqueness of Cycle Decomposition === Take the cycle decomposition of $\sigma$, which is $\sigma_1 \sigma_2 \cdots \sigma_m$. Let $\tau_1 \tau_2 \cdots \tau_s$ be some product of disjoint cycles such that $\sigma = \tau_1 \tau_2 \cdots \tau_s$. It is assume that this product describes $\sigma$ completely and does not contain any duplicate $1$-cycles. Let $x$ be a moved element of $\sigma$. Then there exists a $j \in \set {1, 2, \ldots, s}$ such that $\map {\tau_j} x \ne x$. And so: {{begin-eqn}} {{eqn | l = \map \sigma x       | r = \map {\tau_1 \tau_2 \cdots \tau_j} x }} {{eqn | r = \map {\tau_j} x       | c = Power of Moved Element is Moved }} {{end-eqn}} It has already been shown that $x \in E_i$ for some $i \in \set {1, 2, \ldots, m}$. Therefore: {{begin-eqn}} {{eqn | l = \map {\sigma_i} x       | r = \map {\tau_j} x }} {{eqn | l = \map {\sigma_i^2} x       | r = \map {\tau_{j \prime} \tau_j} x       | c = by Power of Moved Element is Moved }} {{eqn | r = \map {\tau_j^2} x       | c = because $\map {\sigma_i^2} x \ne \map {\sigma_i} x$ and this product is disjoint }} {{eqn | l = \vdots       | o = \vdots       | r = \vdots }} {{eqn | l = \map {\sigma_i^{k - 1} } x       | r = \map {\tau_j^{k - 1} } x }} {{end-eqn}} This effectively shows that $\sigma_i = \tau_j$. Doing this for every $E_i$ implies that $m = s$ and that there exists a $\rho \in S_m$ such that: :$\sigma_{\map \rho i} = \tau_i$ In other words, $\tau_1 \tau_2 \cdots \tau_m$ is just a reordering of $\sigma_1 \sigma_2 \cdots \sigma_m$. {{qed}}