I am trying to prove the non-trivial assertion of Wilson's theorem,
If $p$ is a prime, $p$ divides $(p-1)! - (p-1)$.
I'm going to consider the set of $p$-cycles on $\mathbb{S}(\mathbb{Z}_p)$, which I'll denote $C_p$. Now, $C_p$ has size $(p-1)!$ for the following reason: it suffices to choose the images of $1,,...,p$ without repeating values or having fixed points, therefore we have $p-1$ options for the image of $1$, $p-2$ for the image of $2$, and so on until we get to the image of $p$, which is at this point already determined. Now, let's observe that if $a \in \mathbb{Z}_p^{\times}$, the translation given by summing $a$,
$$ \tau_a : \mathbb{Z}_p \rightarrow \mathbb{Z}_p \\ x \ \mapsto a + x $$
is an element of $C_p$ with inverse $\tau_{-a}$.
We can then define the set $X_p = C_p - \{\tau_a : a \in \mathbb{Z}_p^{\times}\}$ with $|X| = (p-1)! - (p-1)$, and the following equivalence relation on $X_p$:
$$ f \sim g \iff f = \tau_{-a}\cdot g\cdot \tau_a, \ \text{ for some } a \in \mathbb{Z}_p $$
If we can conclude that each equivalence class on $X_p/\sim$ has size $p$, this will imply that $p \ | \ |X_p| = (p-1)! - (p-1)$ as we claim. It is clear that if $f \in \mathbb{S}(\mathbb{Z}_p)$,
$$ [f] \subseteq \{\tau_{-a}\cdot f\cdot \tau_a : a \in \mathbb{Z}_p\} $$
To finish, it will be sufficient to see the other inclusion, that is, that for any $a$ in $\mathbb{Z}_p$ the element $\tau_{-a}\cdot f\cdot \tau_a$ is not a translation and therefore is in $X_p$ (the fact that $\tau_{-a}\cdot f\cdot \tau_a \sim f$ is immediate). Let's suppose that, on the contrary, there exists $b \in \mathbb{Z}_p^{\times}$ such that
$$ \tau_{-a}\cdot f\cdot \tau_a \equiv \tau_b $$
and therefore
$$ f \equiv \tau_a\cdot \tau_b \cdot \tau_{-a} \equiv \tau_b $$
since translation commute. We've reached then a contradiction, since $f$ cannot be a translation, so in effect each class is as previously described and has size $p$, which concludes the proof.
Is this correct? I'd also appreciate any suggestions to simplify the argument. Thanks in advance!
Edit: I've noticed that it is also necessary to see that the set $\{\tau_{-a}\cdot f\cdot \tau_a : a \in \mathbb{Z}_p\}$ has in fact $p$ distinct elements. Any ideas?
This looks correct, and the set $\{\tau_{-a}\cdot f\cdot\tau_{a}:a\in\Bbb{Z}/(p)\}$ does have $p$ distinct elements, since if $\tau_{-a}\cdot f\cdot\tau_{a} = \tau_{-b}\cdot f\cdot\tau_{b}$ then $f = \tau_{b-a}\cdot f\cdot\tau_{a-b}$ implying $a-b=0$ and thus $a=b$.
This last bit is because when looking at only $p$ letters the only permutations on $p$ letters that $p$ cycle commutes with are its powers. And the powers of $\tau_{a-b}$ are all the translations, and $f$ is not a translation.
To see the claim: conjugation by $\sigma$ takes a cycle $(abc...)$ to $(\sigma(a)\sigma(b)\sigma(c)...)$. You can see that there are only $p$ ways this can be the original cycle, so only $p$ possible $\sigma$s do the trick, and we already know the $p$ powers of the original cycle are among them, so nothing else can work.