Primitive Roots mod a prime number

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!

• Hint: since $h$ is non-zero mod $p$ it must be equal to $g^k$ for some positive integer $k$ (since $g$ is a primitive root). Now try to deduce something about $k$ knowing that $h$ is a primitive root and $p$ is odd. Can $g^{k+1}$ also be a primitive root? Mar 5, 2018 at 16:17

Hint:

$$g^{(p-1)/2}\equiv h^{(p-1)/2}\equiv-1\pmod p$$

$$(gh)^{(p-1)/2}\equiv?$$

• Mar 5, 2018 at 16:12
• So (gh)$^{(p-1)/2}$ $\equiv$ 1 (mod p) , but how does that prove that gh is not a primitive root of mod p? I'm sorry, I've looked through the answers to the question you sent, which was very helpful, but I'm still having trouble connecting everything. Mar 5, 2018 at 16:33
• @vfantina, So, ord$_p(gh)\le\dfrac{p-1}2<p-1$ Mar 5, 2018 at 16:39

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 $7$.