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."
Best Answer
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.