An algebraic proof of Fermat's Little Theorem

Let $G$ be an abelian group. Define a scalar multiplication over $\mathbb{Z}$ as follows: $$n \cdot g = \underbrace{g+g+\cdots+g}_{n \ \text{times}}.$$ Note that in this case $|G| \ g=0$. ( We turned $G$ into a $\mathbb{Z}$-module. )

For primes, the multiplicative group $\mathbb{Z}_p$ is abelian, $| \mathbb{Z}_p | = p-1$ and the identity element is $1$. Let $a \in \mathbb{Z}_p$ and the multiplicative notation of $|G| \ g=0$ becomes $a^{p-1} \equiv 1 \bmod{p}$. But this is just Fermat's Little Theorem!

Fermat