4
$\begingroup$

Let $p$ be an odd prime. Show that the product of the distinct primitive roots, $\mod{p}$, is $\equiv$ $1$ or $-1$ $\pmod{p}$.

I think this can be done by viewing the primitive roots as a elements of $(\mathbb{Z}/p\mathbb{Z})^*$ and then showing that their product is either $1$ or $-1$ in $(\mathbb{Z}/p\mathbb{Z})^*$.

I have this hint but dont know how to proceed.

$\endgroup$
8
$\begingroup$

For almost all primes, the product is actually congruent to $1$ modulo $p$.

If $p=2$, then the product of the primitive roots of $p$ is congruent both to $1$ and $-1$ modulo $p$. And if $p=3$, then the product of the primitive roots of $p$ is congruent to $-1$ modulo $p$.

We now show that if $p\gt 3$ is prime, then the product of the primitive roots of $p$ is congruent to $1$ modulo $p$.

Recall (or prove) that if $g$ is a primitive root of $p$, then the modular inverse $g^{-1}$ of $g$ is a primitive root of $p$. Note also that if $p\ne 3$, and $g$ is a primitive root of $p$, then $g^{-1}\not\equiv g\pmod{p}$. For $x^{-1}\equiv x\pmod{p}$ if and only if $x^2\equiv 1\pmod{p}$, that is, if and only if $x\equiv \pm 1\pmod{p}$.

Thus if $p\gt 3$, the primitive roots of $p$ can be divided into couples $\{g,g^{-1}\}$. The product of the elements in a couple is congruent to $1$, and therefore if $p\gt 3$ then the product of all the primitive roots is congruent to $1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.