I was trying to prove the following statement, however when I finished I realized that the proof of my book (A first course in abstract algebra by Fraleigh) actually holds in 3 lines.
Hence I doubt of the coherence of my proof since I had to prove an external claim I introduced myself in it, which made it ridiculously long.
I did so because I didn't know if I could assume this apparently obvious claim (since obvious doesn't mean that it doesn't deserves to be proved).
9.8 Theorem Every permutation $\sigma$ of a finite set is a product of disjoint cycles.
Let $A$ be a finite set and $\sigma\in S_A$ a permutation.
We firstly claim that the fact that A is finite implies that for every $a\in A$ there exists a $n\in \mathbb{N}$ s.t. $\sigma^n a = a$.
Suppose that it isn't the case, i.e. $\exists a\in A:\forall n\in\mathbb{N}:\sigma^n a \neq a$.
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Let's consider the case when the images of $\sigma$ do not repeat themselves in $A$, i.e. $(i,j)\in\mathbb{N}^2, x\in A$ \begin{gather} \sigma^i x = \sigma^j x \Leftrightarrow i=j \end{gather} Then, for every $n\in\mathbb{N}$, $\sigma^n x$ is unique in $A$.
Let $\phi_x:A\rightarrow \mathbb{N}$ s.t. $\phi_x a = n \Leftrightarrow a = \sigma^n x$.
Therefore, \begin{gather} \phi_x a = \phi_x b \Leftrightarrow a=b \\ \forall n \in\mathbb{N}:\exists a\in A, a=\sigma^n x \\ \Rightarrow \phi_x[A] = \mathbb{N} \end{gather} We observe that $\phi$ is consequently injective and surjective, so it is an isomorphism between $A$ and $\mathbb{N}$.
Therefore $A$ is not finite which is absurd.
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Now let's consider the case when there's at least one repetition of the images of $\sigma$ in $A$.
Thus, \begin{gather} \exists \Gamma \subseteq A:\Gamma = \lbrace \sigma^n a \mid n\in\mathbb{N} \rbrace \\ \Rightarrow \exists x \in\Gamma : x=\sigma^i a \wedge x=\sigma^j a,\text{ with }(i,j)\in\mathbb{N}^2:i\neq j \end{gather} Without loss of generality, let $j > i$ and $n\in\mathbb{N}$. \begin{gather} \sigma^i a = \sigma^j a \\ \Rightarrow a = \sigma^{-i}\sigma^j a = \sigma^n a \end{gather} Which goes against our hypothesis.
Therefore, $\forall a \in A: \exists n\in\mathbb{N}$ s.t. $\sigma^n a = a$.
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Pick an $a\in A$.
We know that $\exists n\in\mathbb{N}$ s.t. $\sigma^n a = a$.
Observe that $C_0=(a,\sigma a,...,\sigma^{n-1}a)$ is a cycle.
If there is a $b$ in $A$ such that $b$ isn't in $C_0$, repeat the operation until every element in $A$ belong in a unique cycle.
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A is finite so there's a finite number $m\in\mathbb{N}$ of disjoint cycles.
Observe that \begin{gather} \sigma=C_0C_1...C_m \end{gather}
Hence, every permutation $\sigma$ of a finite set is a product of disjoint cycles.