2
$\begingroup$

Let $p$ be prime. Prove that if $a$ is a primitive root modulo $p$ and $ab\equiv1\bmod p$, then $b$ is a primitive root modulo $p$.

I understand the definition of primitive roots. I am having trouble starting the proof and where to go afterwards.

Thank you in advance!

$\endgroup$
0
$\begingroup$

Since $p$ is a prime, the powers of the primitive root $a$ modulo $p$ generate all the positive integers less than $p$ in some sequence, including 1. Since $b\equiv a^{-1}\bmod p$, the powers of $b$ modulo $p$ must generate the same sequence of positive integers less than $p$, but in reverse. Therefore the order of $b$ modulo $p$ is $p-1$ and $b$ is a primitive root modulo $p$.

$\endgroup$
0
$\begingroup$

In general, in any group you have that the order of $a$ is equal to the order of its inverse. This is just the special case in which $a$ is a generator of the multiplicative group of $\mathbb Z_p$.

$\endgroup$

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.