Prove that $\pi(x)$ is a product of $m$ $n$-cycles. Here is the question I want to proof:
Let $G$ be a finite group and let $\pi: G \rightarrow S_G $ be the left regular representation. Prove that if $x$ is an element of $G$ of order $n$ and $|G| = mn,$ then $\pi(x)$ is a product of $m$ $n$-cycles.
I am convinced with what is required to proof, here is my justification for it:
Let $G$ be a finite group and let $\pi: G \rightarrow S_G $ be the left regular representation. So explicitly, the map is as follows:
$$a \mapsto \pi_a, \text{ where } \pi_a(g) = ag.$$
Given that $x \in G$ and $|x| = n.$ Also, given that $|G| = mn.$ Consider $\pi_x,$ which is a permutation of $G$ as defined above. Now, since $\pi_x$ is a permutation one can always write it as a product of disjoint cycles. The problem asks to prove that $\pi_x$ is a product of $m$ $n$-cycles. Let us take $g \in G.$ Then $\pi_x(g) = xg, \pi_x(xg) = x^2g,$ and so on. In general we observe that, $\pi_x(x^k g) = x^{k+1}g,$ and hence $\pi_x(x^{n-1} g) = x^{n}g.$ So, we get the following cycle $(g, xg, x^2g, \dots, x^{n-1}g).$ This is clearly an $n$-cycle. In this way one can again take an element of $G$ outside the elements of the previously made cycle and make an $n$-cycle of it. Since the order of the group $G$ is $mn$, we have that $\pi_x$ can be written as a product of $m$ $n$-cycles.
But I am unable to prove it. Could anyone help me in the proof please?
Also, I found the following proof online but I do not understand the sequence of ideas in it:
"Let $G$ be a finite group and a mapping $\pi: G \rightarrow S_G $ be the left regular representation. We know that the action of $G$ is faithful; therefore, action of $H = \langle x \rangle$ on $G$ is also faithful.
We know that, for every $g \in G$ we have that $\operatorname{stab}_H(g) = 1$ such that, $[H: \operatorname{stab}_H(g)] = n$ therefore, every $H$-orbit of $G$ is having order $n.$ We know that the $H$ orbit of an element $g \in G$ is cycle. It contains $g$ in the decomposition of $\pi(x).$ Also, $H$ is cyclic having generator $x.$ Since we know that $|G| = mn;$ there are $m$ distinct orbits. Thus $G$ is a product of $G$-disjoint $G$-cycles."
 A: Clearly all cycles of $π(x)$ have length at most $n$ (since $x^n=1$). Suppose a cycle in $π(x)$ has length $k<n$. For an element $a$ appearing in this cycle, $x^ka=π(x^k)(a)=a$. Multiplying on the right by $1/a$ we obtain $x^k=1$ contrary to the assumption that $n>k$ is the order of $x$.
A: Consider the action of the subgroup $\langle x\rangle=\{x^k,k=0,\dots,n-1\}$ on $G$ by left multiplication. For any $a\in G$, the orbit of $a$ reads:
$$O(a)=\{a,xa,\dots,x^{n-1}a\} \tag1$$
which has size $n$$^\dagger$. So, $G$ splits into $\frac{|G|}{n}=\frac{mn}{n}=m$ orbits of size $n$ each.
Now, for every $g\in O(a)$, there is $k\in\{0,\dots,n-1\}$ such that, for every $l\in\Bbb N$:
$$\pi_x^l(g)=\pi_x^l(x^ka)=x^{k+l\pmod n}a \tag2$$
(induction on $l$). Therefore, the restriction of $\pi_x$ to $O(a)$ is an $n$-cycle of $S_{O(a)}$. In order to turn it into an $n$-cycle of $S_G$, you have to extend ${\pi_x}_{\mid O(a)}$ to $G\setminus O(a)$ by the identity map, $\pi_x(g)=g$. So, named after $(1)$ the set $\{a_1,\dots,a_m\}$ of orbit representatives, let's define the map $\alpha_i$, $i=1,\dots,m$, in this way:
\begin{alignat}{1}
&{\alpha_i}_{\mid O(a_i)}:={\pi_x}_{\mid O(a_i)} \\
&{\alpha_i}_{\mid O(a_j)}:=\operatorname{id}_{O(a_j)}, \text{ for every }j=1,\dots,m \text{ such that } j\ne i\\
\tag3
\end{alignat}
Accordingly:
\begin{alignat}{2}
&g \in O(a_i) &&\Longrightarrow (\alpha_i\alpha_j)(g)=\alpha_i(\alpha_j(g))=\alpha_i(g)=\pi_x(g) \\
&g \in O(a_j) &&\Longrightarrow (\alpha_i\alpha_j)(g)=\alpha_i(\alpha_j(g))=\alpha_i(\pi_x(g))=\pi_x(g) \\
&g \in O(a_{l\ne i,j}) &&\Longrightarrow (\alpha_i\alpha_j)(g)=\alpha_i(\alpha_j(g))=\alpha_i(g)=g \\
\tag4
\end{alignat}
or, equivalently:
\begin{alignat}{2}
&g \in O(a_i)\sqcup O(a_j) &&\Longrightarrow (\alpha_i\alpha_j)(g)=\pi_x(g) \\
&g \in O(a_{l\ne i,j}) &&\Longrightarrow (\alpha_i\alpha_j)(g)=g \\
\tag5
\end{alignat}
By induction on $(5)$:
\begin{alignat}{2}
&g \in O(a_1)\sqcup\dots\sqcup O(a_r)=G &&\Longrightarrow (\alpha_1\dots\alpha_r)(g)=\pi_x(g) \\
\tag6
\end{alignat}
namely:
$$\pi_x=\alpha_1\dots\alpha_r \tag7$$
and $\pi_x$ is the product of $m$ $n$-cycles (of $S_G$).

$^\dagger$In fact, for $0\le j\le i \le n-1$ (and hence $0\le i-j \le n-1$):
\begin{alignat}{1}
&x^ia=x^ja &&\iff \\
&x^i=x^j &&\iff \\
&x^{i-j}=e &&\iff \\
&i-j=0&&\iff \\
&i=j \\
\end{alignat}
whence all the elements listed in $(1)$ are pairwise distinct.
A: The proof you found online is basically the same thing as your justification. An orbit of H is just a cycle constructed by your method. It is just using Orbit-Stabilizer Theorem to prove that each such orbit is of order n. After you find that each g can be used to construct a n-cycle in this way, it is easy to show that we can find m g's in G to get m distinct cycles.
