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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?

MyNameIs
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    Because the onle element with order 1 is the identity – Martín Vacas Vignolo Jul 26 '16 at 14:42
  • I don't follow. If $k=1$ then $c\equiv 1\pmod p$. Hence $c\in {1,p+1,2p+1,\cdots}$. ? – MyNameIs Jul 26 '16 at 14:49
  • Your deduction that $k\in{1,p-1}$ is wrong. What if $p=7?$ Then $k$ can be, for example, $2.$ (In fact, $6^2=36\equiv 1.$) What Fermat's Little Theorem says is that $c^{p-1}$ is always congruent to 1 no matter what the $c$ is, not that $p-1$ is the only power such that $c^{p-1}\equiv 1.$ – quasicoherent_drunk Jul 27 '16 at 01:30

2 Answers2

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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 $a^{p-1}=a^{kn}=(a^k)^n=1^n=1 \bmod{p}$

quasicoherent_drunk
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  • Somehow that's simpler than I thought it would be, thanks! – MyNameIs Jul 27 '16 at 07:06
  • Yup, it's really just a direct corollary of Lagrange's. A generalization would be that if $G$ is a finite group containing $g,$ then $g^{\lvert G\rvert}=e$ because the subgroup generated by $g$ has order $o(g),$ which divides $\lvert G\rvert,$ and $g^{o(g)}=e.$ This also explains why Euler's theorem (with the totient function) works. – quasicoherent_drunk Jul 28 '16 at 05:15
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The order of the group $(\mathbb{Z}/p)^\times$ is $p-1$. To use Lagrange means to know, that if $a$ is not devided by $p$ then there exists $b\in (\mathbb{Z}/p)^\times$ defined by $a$ with $b^{p-1}=1$ and therefore $a^{p-1}\equiv 1\mod p$. And so we get Fermat's Little Theorem.

richarddedekind
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user90369
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