Cycle type of elements in left regular presentation I am interested in the proof of the following statement:
Let $G$ be a finite group, and let $\pi : G \to S_{\vert G \vert}$ be the left regular presentation. If $x \in G$ has order $n$, and $\vert G \vert = mn$, then $\pi(x)$ decomposes into $m$ disjoint $n$-cycles.
The question is coming from here: https://crazyproject.wordpress.com/2010/05/03/characterization-of-parity-in-the-left-regular-representation-of-a-finite-group/. I understand the proof offered until they say "the $H$-orbit of an element $g \in G$ is precisely the cycle containing $g$ in the decomposition of $\pi(x)$..." This part is just totally unclear to me. An orbit is a set... how can it be a cycle? My understanding of the statement is that since $x$ is not the identity, it fixes no elements of $G$. Therefore $\pi(x)$ cannot fix any elements any of the symbols $\{1,2,...,\vert G \vert\}$ since the map $\pi$ is injective. Thanks!
 A: Their proof relies on looking at the orbit decomposition of $G$ as the cyclic subgroup of $x$ acts on $G$. By appropriately ordering each orbit you get a cycle and then the result follows. Here's a bunch of details...
Consider $H = \langle x \rangle = \{ 1,x,x^2,\dots,x^{n-1} \}$ the cyclic subgroup generated by $x$. 
By Lagrange's theorem there are $\dfrac{|G|}{|H|} = \dfrac{mn}{n}=m$ right cosets of $H$ in $G$. Suppose $H, Hy_2, \dots, Hy_m$ are the distinct right cosets (let $y_1=1$ for convenience). Then $Hy_j = \{ y_j,xy_j,\dots,x^{n-1}y_j\}$ for each $j=1,\dots,m$. Each of these right cosets, $Hy_j$, is one of the orbits of the action of $H$ on $G$.
To help represent left multiplication by $x$ as a permutation let's fix an order for the elements of $G$. List $G$'s elements in the following order: $1,x,\dots,x^{n-1},y_1,xy_1,\dots,x^{n-1}y_1,\dots,y_m,xy_m,\dots,x^{n-1}y_m$.
Now multiply by $x$ and get $x^iy_j$ maps to $x^{i+1}y_j$ where we can reduce $i+1$ modulo $n$ (the order of $x$). So multiplying by $x$ on the left yields the following cycle: $y_j \mapsto xy_j \mapsto x^2y_j \mapsto \cdots \mapsto x^{n-1}y_j \mapsto y_j$ for each $j=1,\dots,m$. This is an $n$-cycle (It can't loop back early since: If $x^ky_j=x^\ell y_j$ for $k<\ell$ we'd have $x^k=x^\ell$ and so $x^{\ell-k}=1$ contradicting the order of $x$ being $n$).
Thus left multiplication by $x$ decomposes into an $n$-cycle for each of the $m$ right cosets of $H=\langle x \rangle$. That says $x$ is represented by a permutation consisting of $m$ cycles each of length $n$. :)
