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

###### Theorem.

If $a,p\in\mathbb{Z}$ with $p$ a prime and $p\nmid a$, then $a^{p-1}\equiv 1\pmod{p}$.

###### 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 $H=\{1,a,a^{2},\ldots\}$). 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. $a^{h}\equiv 1\mod p$.

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

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

as claimed. ∎

Title proof of Fermat’s little theorem using Lagrange’s theorem ProofOfFermatsLittleTheoremUsingLagrangesTheorem 2013-03-22 14:23:53 2013-03-22 14:23:53 alozano (2414) alozano (2414) 4 alozano (2414) Proof msc 11-00 LagrangesTheorem