Let $r_1$ and $r_2$ be incongruent primitive roots modulo $p$ where $p$ is an odd prime. Show that $r_1 \times r_2$ is not necessarily a primitive root modulo $p$.
We will show this with a counter-example.
We start with $r_1=2, r_2=3, p=5$, noting that 2 and 3 are both primitive roots of 5.
We now show that $r_1 \times r_2 = 6$ is not a primitive root of 5.
The following calculations show that 6 is not a primitive root of 5 because the smallest index of 6 such that the result is congruent to 1 is not $\phi(5)=4$.
| n | 2^n mod 5 | 3^n mod 5 | 6^n mod 5 |
| 1 | 2 | 3 | 1 |
| 2 | 4 | 4 | 1 |
| 3 | 3 | 2 | 1 |
| 4 | 1 | 1 | 1 |
We can also see this by noting that $6 \equiv 1 \pmod 5 \implies 6^n \equiv 1 \pmod 5$ for any positive integer $n$.