Prove that $ o( g) =m$ iff is a multiplication of $ \frac{n}{m}$ disjoint cycles. The problem:
Given a group $G$ or order $n$, and a Cayley embedding $\phi \ :\ G\to S_{n}$. Prove that some $g\in G$ is of order $m$ iff $\phi ( g)$ is a multiplication of $\frac{n}{m}$  disjoint cycles of length $m$.
I was able to prove that if $\phi ( g) \ $ is s a multiplication of $\displaystyle \frac{n}{m}$ disjoint cycles of order $m$, then the order of $g$ is $m$. This was fairly straightforward given the fact that:

The order of a multiplication of disjoint cycles is the $lcm$ of the length of the cycles.

In our case the $lcm$ is obviously $m$, so it was easy to prove from here that the order of $g$ is $m$.
The other direction, meaning if the order of $g$ is $m$, then $\phi ( g)$ is a multiplication of $\frac{n}{m}$ disjoint cycles of length $m$, I'm struggling to prove it.
I also want to admit that my understanding of Cayley embeddings, and Cayley's theorem in general, is very poor. What I know is simply that it's a homomorphism and also injective, not much more.
Any help?
 A: Let's consider the action by left multiplication of $\langle g\rangle$ on $G$. For $x\in G$, the stabilizer is given by:
\begin{alignat}{1}
\operatorname{Stab}(x) &= \{g^k\in\langle g\rangle\mid g^kx=x\} \\
&= \{g^k\in\langle g\rangle\mid g^k=e\} \\
&= \{e\} \\
\tag 1
\end{alignat}
Therefore, $G$ is partitioned into $r:=\frac{n}{m}$ orbits each of size $m$ (orbit-stabilizer theorem). This means that there are $r$ elements $\tilde g_i\in G$, $i=1,\dots,r$, such that $G$ reads:
$$G=\bigsqcup_{i=1}^rO(\tilde g_i) \tag 2$$
where:
$$O(\tilde g_i):=\{g^k\tilde g_i, k=1,\dots,m\} \tag 3$$
Let's now take the permutation by left multiplication by $g$, $\phi_g\in S_G$. By $(2)$ and $(3)$, $\forall x\in G, \exists i\in \{1,\dots,r\}, k\in \{1,\dots,m\}$ such that $x=g^k\tilde g_i$. Therefore, $\forall l\in \{1,\dots,m\}$:
\begin{alignat}{1}
\phi_g^l(x) &\stackrel{(*)}{=}g^lx \\
&=g^{l+k \pmod m}\tilde g_i \\
\end{alignat}
[$(*)$ induction on $l$], whence, $\forall x\in G$:
$$x\in O(\tilde g_i) \iff \phi_g^l(x)\in O(\tilde g_i), \space\forall l\in\{1,\dots,m\} \tag 4$$
Now, let's define $\alpha_i$ as the extension by the identity map of the restriction of $\phi_g$ to the orbit $O(\tilde g_i)$, namely:
\begin{alignat}{2}
&x \in O(\tilde g_i) &&\Longrightarrow \alpha_i(x) :=\phi_g(x) \\
&x \in O(\tilde g_{j\ne i}) &&\Longrightarrow \alpha_i(x) :=x \\
\tag 5
\end{alignat}
Firstly, $\alpha_i \in S_G$ for every $i=1,\dots,r$, because ${\phi_g}_{|O(\tilde g_i)}$ is a bijection on $O(\tilde g_i)$. Then, by $(4)$ and $(5)$:
\begin{alignat}{2}
& x \in O(\tilde g_i) &&\Longrightarrow \alpha_i^m(x)=\phi_g^m(x)=x\\
& x \in O(\tilde g_{j\ne i}) &&\Longrightarrow \alpha_i^m(x)=x\\
\tag 6
\end{alignat}
and finally, for every $i=1,\dots,r$:

$$\alpha_i^m=Id_G \tag 7$$

so all the $\alpha_i$'s are $m$-cycles. Moreover, again by $(4)$ and $(5)$, for $j\ne i$ we get:
\begin{alignat}{2}
&x \in O(\tilde g_i) &&\Longrightarrow (\alpha_i\alpha_j)(x)=\alpha_i(\alpha_j(x))=\alpha_i(x)=\phi_g(x) \\
&x \in O(\tilde g_j) &&\Longrightarrow (\alpha_i\alpha_j)(x)=\alpha_i(\alpha_j(x))=\alpha_i(\phi_g(x))=\phi_g(x) \\
&x \in O(\tilde g_{l\ne i,j}) &&\Longrightarrow (\alpha_i\alpha_j)(x)=\alpha_i(\alpha_j(x))=\alpha_i(x)=x \\
\tag 8
\end{alignat}
or, equivalently:
\begin{alignat}{2}
&x \in O(\tilde g_i)\sqcup O(\tilde g_j) &&\Longrightarrow (\alpha_i\alpha_j)(x)=\phi_g(x) \\
&x \in O(\tilde g_{l\ne i,j}) &&\Longrightarrow (\alpha_i\alpha_j)(x)=x \\
\tag 9
\end{alignat}
By induction on $(9)$:
\begin{alignat}{2}
&x \in O(\tilde g_1)\sqcup\dots\sqcup O(\tilde g_r)=G &&\Longrightarrow (\alpha_1\dots\alpha_r)(x)=\phi_g(x) \\
\tag {10}
\end{alignat}
namely:

$$\phi_g=\alpha_1\dots\alpha_r \tag {11}$$

Note that $\alpha_i(x)\ne x \iff x \in O(\tilde g_i)$, and hence $\alpha_i$ and $\alpha_j$ have disjoint supports as soon as $i\ne j$. Therefore, $(11)$ (along with $(7)$) reads: $\phi_g$ is the product (composition) of $\frac{n}{m}$ disjoint cycles of length $m$.
A: If $g$ has order $m$, then the orbit of $e$ under multiplication by $g$ is
$$e\mapsto g\mapsto g^2\mapsto g^3\mapsto\cdots\mapsto g^{m-1}\mapsto g^m=e.$$
That is, the cycle that contains $e$ is $(e,g,g^2,\ldots,g^{m-1})$.
Now, how about any other orbit/cycle? If we have $x\in G$, then the orbit of $x$ is given by $x\mapsto gx\mapsto g^2x\mapsto\cdots$ etc. What is the first repeat? If $g^ix = g^jx$, then $g^i=g^j$, so $i\equiv j\pmod{m}$. So the first repeat is $g^mx=x$. So the cycle/orbit of $x$ is $(x,gx,g^2x,\ldots,g^{m-1}x)$.
