proof of Fermat’s little theorem using Lagrange’s theorem

Theorem.

If $a\mathrm{,}p\mathrm{\in }\mathrm{Z}$ with $p$ a prime and $p\mathrm{\nmid }a$, then $a^{p\mathrm{-}\mathrm{1}}\mathrm{\equiv }\mathrm{1}\mathrm{(}\mathrm{mod}p\mathrm{)}$.

Proof.

We will make use of Lagrange’s Theorem: Let $G$ be a finite group and let $H$ be a subgroup of $G$. Then the order of $H$ divides the order of $G$.

Let $G={(\mathbb{Z}/p\mathbb{Z})}^{\times }$ and let $H$ be the multiplicative subgroup of $G$ generated by $a$ (so $H=\{1,a,a^2,\mathrm{\dots }\}$). Notice that the order of $H$, $h=|H|$ is also the order of $a$, i.e. the smallest natural number $n>1$ such that $a^n$ is the identity in $G$, i.e. $a^h\equiv 1modp$.

By Lagrange’s theorem $h\mid |G|=p-1$, so $p-1=h\cdot m$ for some $m$. Thus:

$$a^{p-1}={(a^h)}^m\equiv 1^m\equiv 1modp$$

as claimed.∎