Notes on Multiplicative Groups and Order: Fermat's Theorem on Primes and Wilson's Theorem

05 Feb 2025

Remarks

Multiplicative Group	Additive Group
1 is the identity element	0 is the identity element
When the order of $x$ is $r$, $x^r = 1$	When the order of $x$ is $r$, $rx = 0$

• For a prime $p$, if information about $p - 1$ is given, consider the multiplicative group $(\mathbb{Z}/p\mathbb{Z})^\times$ of order $p - 1$.
• The multiplicative group of a finite field is cyclic.
• For a cyclic multiplicative group $G$, $nm \mid |G|$ is equivalent to the existence of an element $x$ such that $x^{nm} = 1$ but $x^n, x^m \neq 1$.
• Order is different from period. In general, (period) = (order) + 1.

Fermat’s Theorem on Primes

Theorem. A prime $p$ can be expressed in the form $p = a^2 + b^2$ for two integers $a, b$ if and only if $p \equiv 1 \pmod{4}$.

Proof.

(⇒) Trivial.

(⇐) Let $G = (\mathbb{Z}/p\mathbb{Z})^\times$. By the condition, $4 \mid |G|$. Therefore, there exists an element $n$ such that $n^4 = 1$ but $n^2 \neq 1$. That is, $n^2 = -1$. Hence, $p \mid n^2 + 1$. Now, considering in $\mathbb{Z}[i]$, we have $p \mid (n + i)(n - i)$. If $p$ were irreducible in $\mathbb{Z}[i]$, then $p \mid n + i$ or $p \mid n - i$ (note that the fact that $\mathbb{Z}[i]$ is UFD is used here) but both are impossible. Therefore, $p$ is not irreducible, and by conjugacy,

\[p = (a + bi)(a - bi) = a^2 + b^2\]

■

Wilson’s Theorem

Theorem. $p$ is prime if and only if $(p - 1)! \equiv -1 \pmod{p}$.

Proof.

(⇒) Let $G = (\mathbb{Z}/p\mathbb{Z})^\times$. The product of all elements of $G$ is $(p - 1)!$. Except for $\pm 1$, which are the roots of $x^2 = 1$, all other elements of $G$ do not have themselves as their inverse. Therefore, $(p - 1)! \equiv -1$.

(⇐) Let $N = (p - 1)! + 1$. By the condition, $p \mid N$. If $p$ were not prime, then for some prime $q < p$, we would have $q \mid N$. However, $N \equiv 1 \pmod{q}$, which is a contradiction. ■