I haven't fully wrapped my head around primitive roots yet and I have a question with them:
Let $p$ be an odd prime and $g$, $h$ be two primitive roots modulo $p$. Show that $gh$ is not a primitive root modulo $p$.
I think I'll need to use the fact that if $g$ is a primitive root modulo p then a reduced residue system modulo $p$ is $g$, $g^2$,..., $g^{p-1}$
Any help would be much appreciated!
On
ANSWER AND COMMENT.-It is known that if $g$ is a primitive root modulo $p$ the other possible primitive roots are given by $g^n=h$ where $(n, \phi(p))=(n,p-1)=1$. Therefore if $$gh=g^{n+1}$$ is another primitive root, then $$(n+1,p-1)=1$$ It is not possible to have both $$(n, p-1)=1\text{ and } (n+1,p-1)=1\space \space \text {(why?) }$$.
Example of searching another primitive root.
$3$ is a primitive root modulo $7$ and $\phi(7)=6$. Thus $3^5=5$ modulo $7$ is the only other p.r. because $2,3,4,6$ are not coprime with $6$ (exponent $1$ corresponds to the p. r. $3$ itself). There are just two primitive roots modulo $7$.
Hint:
$$g^{(p-1)/2}\equiv h^{(p-1)/2}\equiv-1\pmod p$$
$$(gh)^{(p-1)/2}\equiv?$$