Quote:

b) if r' is any other primitive root of p , then rr' is not primitive root of p.

We can write $\displaystyle r' \equiv r^k (\bmod p)$ where $\displaystyle 1\leq k \leq p-1$. For $\displaystyle r'$ be a primitive root it is necessary and sufficient that $\displaystyle \gcd (k, p-1) = 1$. Now $\displaystyle rr' \equiv rr^k \equiv r^{k+1} (\bmod p)$, now for $\displaystyle rr'$ to be a primitive root it is necessary and sufficient that $\displaystyle \gcd(k+1, p-1) = 1$. But that is impossible because $\displaystyle p-1$ is an even number and among $\displaystyle k,k+1$ one is even by pigeonholing. Thus it impossible for $\displaystyle rr'$ to be a primitive root canal.