Show that 7 is not a primitive root modulo 19.
We remind ourselves of Definition (6.10) which states that if $a \pmod n$ has order $\phi (n)$ then $a$ is called a primitive root modulo $n$ or just a primitive root of $n$.
We first note that $\gcd(7,19)=1$ and so the order exists. The order is a factor of $\phi(19)=18$. These factors are 1, 2, 3, 6, 9, 18 and are the only ones we need to test.
The following calculations show the order of $7 \pmod {19}$ is 3.
| n | 7^n | 7^n mod 19 |
| 2 | 49 | 11 |
| 3 | 343 | 1 |
Because 3 is not $\phi(19)=18$, we conclude that 7 is not a primitive root.