The following is what I took from Dummit and foote on page 115. I understood everything except why we can deduce for each $\sigma\in S_n$ there exists a cycle decomposition from the bijection? is it because if we take $\sigma(x)\in \mathscr{O}$ then there exists $\tau\in G=\langle\sigma\rangle$ such that $\tau(x)=\sigma(x)$ but then $\tau=\sigma^{k}$ for some $0\leq k\leq d-1$. Which is just a iterate of $\sigma$.
Let $\sigma\in S_n$ and set $G=\langle\sigma\rangle$ then $G$ acting on $A= \{1,\ldots,n\}$ partitions $A$ into unique disjoint sets of orbits. Let $\mathscr{O}$ be one of those orbits and let $x\in \mathscr{O}$ we can define a bijection from $\mathscr{O}$ to $G/G_x$ where $G_x$ is the stablizer of $x$ in $G$. $$\sigma^i x\mapsto \sigma^i G_x$$ Because $G$ is cyclic, $G_x$ normal subgroup of $G$ and $G/G_x$ is cyclic of order $d$, where $d$ is the smallest positive integer $\sigma^d\in G_x$ And $d=|G:G_x|=|\mathscr{O}|$. Thus the distinct cosets of $G_x$ are $$G_x,\sigma G_x,\cdots, \sigma^{d-1}G_x$$ and the distinct element of $\mathscr{O}$ are $$x,\sigma(x),\cdots , \sigma^{d-1}(x)$$ On an orbit of size $d$, $\sigma$ acts as $d$-cycle. This proves the existence of a cycle decomposition for each $\sigma\in S_n$
Thanks for explaining!