Wilson's Theorem/Necessary Condition/Proof 3

Theorem

Let $p$ be a prime number.

Then:

$\paren {p - 1}! \equiv -1 \pmod p$

Proof

Let $p$ be prime.

Consider $\struct {\Z_p, +, \times}$, the ring of integers modulo $m$.

From Ring of Integers Modulo Prime is Field, $\struct {\Z_p, +, \times}$ is a field.

Hence, apart from $\eqclass 0 p$, all elements of $\struct {\Z_p, +, \times}$ are units

As $\struct {\Z_p, +, \times}$ is a field, it is also by definition an integral domain, we can apply:

From Product of Units of Integral Domain with Finite Number of Units, the product of all elements of $\struct {\Z_p, +, \times}$ is $-1$.

But the product of all elements of $\struct {\Z_p, +, \times}$ is $\paren {p - 1}!$

The result follows.

$\blacksquare$

Source of Name

This entry was named for John Wilson.

Historical Note

The proof of Wilson's Theorem was attributed to John Wilson by Edward Waring in his $1770$ edition of Meditationes Algebraicae.

It was first stated by Ibn al-Haytham ("Alhazen").

It appears also to have been known to Gottfried Leibniz in $1682$ or $1683$ (accounts differ).

It was in fact finally proved by Lagrange in $1793$.

Sources

• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $9$: Rings: Exercise $12$