Euler’s theorem for abelian groups

For abelian groups, that is, groups where addition is commutative, that is, $g_{1} g_{2} = g_{2} g_{1}$ for any pair $g_{1}, g_{2} \in G$ , we have the following theorem due to Euler:

Theorem: Let $G$ be an abelian group with $n$ elements. Then

g^{n} = e

for any $g \in G$ .

Proof: Label the $n$ distinct elements of $G$ as $x_{1}, x_{2}, ... x_{n}$ . Since multiplication by any element $g \in G$ is invertible, we have $g x_{1} \neq = g x_{2}$ whenever $x_{1} \neq = x_{2}$ . So the set $g G = {g x_{1}, g x_{2}, ..., g x_{n}}$ contains the same elements as $G$ . So the products $\prod_{k = 1}^{n} x_{k}$ and $\prod_{k = 1}^{n} g x_{k}$ must be the same as well. Moreover, the same product can be written as $g^{n} \prod_{k = 1}^{n} x_{k}$ . So we have

k = 1 \prod n x_{k} = k = 1 \prod n g x_{k} = g^{n} k = 1 \prod n x_{k}

It follows that $g^{n} = e$ . $□$

As a corollary, we can now easily prove Fermat’s little theorem, which is often used in number theory:

Corollary: Let $p$ be a prime, and $n$ be a nonzero integer. For any $n$ that does not divide

n^{p - 1} \equiv 1 (mod p)

Proof: The integers $1, 2, ..., p - 1$ form a commutative group of order $p - 1$ under multiplication modulo $p$ . $s q u a re$