I'm reading the proof for the following statement:
Every permutation is a product of pairwaise disjoint cycles, and this descomposition is unique up to the order of terms.
At the beginning of the proof, it says:
Given $\sigma\in S_n$ (symmetric group), let $\mathbb{Z}$ act on $X=\{1,\ldots,n\}$ by $m\bullet x=\sigma^m x$ $\left(\sigma x=\sigma(x)\right)$. It is indeed a group action. Then $X$ partitions $X$ into orbits. We see that $\sigma x=x$ the orbit of $x$ is trivial.
Just in case: the orbit of $x\in X$ is the set $G\bullet x=\mathcal O_x=\{g\bullet x:g \in G\}$ (if $\bullet$ is a left group action of $G$ on $X$).
That last statement is my doubt: $\sigma x=x$ iff l the orbit of $ x$ is trivial.
I thought about the following: If $\sigma x=x$, $x$ is a fixed point of the permutation $\sigma$. Then, $\sigma^nx=x~\forall n\in\mathbb{N}$ which implies $\mathbb{Z}\bullet x=\{x\}$. Conversely, if the orbit is trivial, $\sigma^n x=x~\forall n\in\mathbb{N}$, in particular, $\sigma^1 x=\sigma x=x$.
Was my reasoning right? I mean, a fixed point of a permutation will be a fixed point "forever"? (In the sense of applying over and over the same permutation). Thanks in advance.
You are correct, though I'd have found neater to saying that the cycle structure of $\sigma\in S_n$ pops up when considering the natural action of $\langle\sigma\rangle$ on $X$, which makes your argument even better fitting in. And the fixed points of $\sigma$ (and hence of all the elements of $\langle\sigma\rangle$) correspond to the "break points" between adjacent round brackets in the cycle notation of $\sigma$, namely: "$\dots)(\dots$".