I know how to prove Fermat's little theorem using binomial expansion and induction.
Question:
How can I prove it using Lagrange's theorem?
So I want to show $c^p\equiv c\pmod p$, i.e. $c^{p-1}\equiv 1\pmod p$ since $\Bbb F_p$ is a field.
We have for some $k\geq 1$ that $c^k\equiv 1\pmod p$. In partiular, $k\mid p-1$ (Lagrange) since $c$ is an element of multiplicative order $k$ in $\Bbb F_p$. Thus $k\in \{1, p-1\}$.
How do I exclude the case $k=1$ now?
Best Answer
The multiplicative group $(\mathbb{Z}/p\mathbb{Z})^\times$ has order $p-1.$ Take any $a\in\mathbb{Z}/p\mathbb{Z}$ and let $k$ be the order of $a,$ i.e. $k$ is the smallest positive integer such that $a^k=1 \bmod{p}.$ Then the group $\{1,a,a^2,\dots,a^{k-1}\}$ is a subgroup of $G$ of order $k.$ By Lagrange's theorem, $k$ divides $p-1$, i.e. $p-1=kn$ for some $n.$
Then $a^{p-1}=a^{kn}=(a^k)^n=1^n=1 \bmod{p}$