Notes - Groups TT23, Wilson’s theorem
Flashcards
Can you state Wilson’s theorem, about $(p-1)!$?
If $p$ is a prime, then
\[(p-1)! \equiv -1 \pmod p\]What gets you started when proving Wilson’s theorem, i.e.
If $p$ is a prime, then
\[(p-1)! \equiv -1 \pmod p\]
If $p$ is a prime, then
Considering the self-inverse elements in $\mathbb Z _ p$.
Can you quickly justify that the only self-inverse elements in $\mathbb Z _ p^\ast$ are $\overline 1$ and $\overline{-1}$?
\[\begin{aligned}
\overline x = \overline{x}^{-1} &\iff \overline{x}^2 = 1 \\\\
&\iff (\overline x - \overline 1)(\overline x + \overline 1) = \overline 0 \\\\
&\iff \overline x = \overline 1 \text{ or } \overline x = \overline {-1}
\end{aligned}\]
Quickly prove Wilson’s theorem, i.e.
If $p$ is a prime, then
\[(p-1)! \equiv -1 \pmod p\]
If $p$ is a prime, then
Consider the partition formed by $x \sim y \iff x = y \text{ or } x = y^{-1}$ in $\mathbb Z _ p^\ast$. Then the self inverse elements are
\[\begin{aligned} \overline x = \overline{x}^{-1} &\iff \overline{x}^2 = 1 \\\\ &\iff (\overline x - \overline 1)(\overline x + \overline 1) = \overline 0 \\\\ &\iff \overline x = \overline 1 \text{ or } \overline x = \overline {-1} \end{aligned}\]Then as the equivalence classes form a partition of $\mathbb Z _ p^\ast$,
\[\begin{aligned} (p-1)! &= \prod_{\overline k \in \mathbb Z_p^\ast} \overline k \\\\ &= \prod_\text{equiv. classes} \text{ } \prod_\text{each equiv. class} \overline k \\\\ &= \left (\overline 1 \times (-\overline 1) \right) \times \prod_\text{doubleton equiv. classes} \overline k \\\\ &= -\overline 1 \end{aligned}\]as in each doubleton equivalence class $\overline x \times \overline x^{-1} = \overline 1$.