Fixed Elements form 1-Cycles

Theorem
Let $$S_n$$ denote the symmetric group on $n$ letters.

Let $$\pi \in S_n$$.

Let $$\operatorname{Fix} \left({\pi}\right)$$ be the set of elements fixed by $\pi$.

For any $$\pi \in S_n$$, all the elements of $$\operatorname{Fix} \left({\pi}\right)$$ form 1-cycles.

Proof
Let $$\pi$$ be a permutation, and let $$x \in \operatorname{Fix} \left({\pi}\right)$$.

From the definition of a fixed element, $$\pi \left({x}\right) = x$$.

From the definition of a $k$-cycle, we see that $$1$$ is the smallest $$k \in \mathbb{Z}: k > 0$$ such that $$\pi^k \left({x}\right) = x$$.

The result follows.