Prove that every permutation is the product of disjoint cycles I need to prove that every permutation $\sigma \in S_n, \sigma \neq id$ can be written as a distinct product of disjoint cycles and don't really know where to start. The statement is totally clear, but I cannot think of an elegant way to proof its correctness.
 A: Let $\sigma$ be the permutation, and let $[n]=\{1,2,\ldots,n\}$ be the set that we are considering.
First, choose one element from the set, say $x$, and permute it until you reach $x$ again, and collect every element you reaches. i.e. Formulate the cycle 
$$C_x=(x\;\sigma(x)\;\sigma^2(x)\ldots\sigma^k(x))$$
And the set
$$E_x=\{x,\sigma(x),\sigma^2(x),...,\sigma^k(x)\}$$
Where $\sigma^{k+1}(x)=x$ and $\sigma^r(x)\neq x, \forall r\le k$. Now, $k$ can be found since $[n]$ is finite, and since $\sigma$ is a bijection, if $\sigma^i(x)=\sigma^j(x)$, then $\sigma^{i-j}(x)=\sigma^{j-j}(x)=x$, so all of the $\sigma^d(x)$ are distinct from each other, $\forall d\le k$.
Pick one element, say $1$, then $C_1$ is a cycle. And choose $p\in [n]\setminus E_1$, and find $C_p$ and $E_p$. Since $p\notin E_1$, $E_p\cap E_1=\varnothing$. Continue this until (can be reach since $[n]$ is finite) $[n]$ is exhausted. And multiply all $C_x$s, and we are done.
By the way, $\sigma=\mathrm{id}$ also can be written so, take $(1)(2)(3)\cdots(n)$
