The incongruent primitive roots modulo 19 are 2, 3, 10, 13, 14, and 15.
Determine the order of
(a) $-2 \pmod {19} $
(b) $-3 \pmod {19} $
(c) $-10 \pmod {19} $
(d) $-13 \pmod {19} $
(e) $-14 \pmod {19} $
(f) $-15 \pmod {19} $
We note that the all the values are negations of known primitive roots. So we'll first prove a general lemma that.
Lemma. For prime $p$ where $\frac{p-1}{2}$ is odd, if $a$ is a primitive root of $p$, then the order of $-a$ is $\frac{p-1}{2}$.
Since $a$ is a primitive root, $a^{p-1} \equiv 1 \pmod p$. Since $\frac{p-1}{2}$ is odd, it is an integer, so by Proposition (3.14)(b)
$$ a^{\frac{p-1}{2}} \equiv 1 \pmod p \quad \lor \quad a^{\frac{p-1}{2}} \equiv -1 \pmod p $$
The first case is not possible because $a$ is a primitive root, and the smallest index $j$ such that $a^j \equiv 1 \pmod p$ is $j=p-1$. This leaves
$$ a^{\frac{p-1}{2}} \equiv -1 \pmod p $$
Multiplying both sides by $-1 = (-1)^{\frac{p-1}{2}}$ gives
$$ (-a)^{\frac{p-1}{2}} \equiv 1 \pmod p $$
This doesn't tell us the order of $-a$ as there could be a smaller positive integer $k$ such that $(-a)^k \equiv 1 \pmod p$.
Imagine $z$ is the order of $(-a)$. Then
$$ (-a)^z \equiv 1 \pmod p $$
Squaring
$$ a^{2z} \equiv 1 \pmod p $$
Since $a$ is a primitive root, $2z \ge p-1$, that is $z \ge \frac{p-1}{2}$.
We have two facts:
- The order of $-a$ is greater than or equal to $\frac{p-1}{2}$.
- $ (-a)^{\frac{p-1}{2}} \equiv 1 \pmod p $
These two facts tell us the order of $-a \pmod p$ is $\frac{p-1}{2}$.
(a) Since 2 is a primitive root of prime 19, and since $\frac{19-1}{2}=9$ is odd, then by the Lemma, the order of $-2$ is 9.
(b) Since 3 is a primitive root of prime 19, and since $\frac{19-1}{2}=9$ is odd, then by the Lemma, the order of $-3$ is 9.
(c) Since 10 is a primitive root of prime 19, and since $\frac{19-1}{2}=9$ is odd, then by the Lemma, the order of $-10$ is 9.
(d) Since 13 is a primitive root of prime 19, and since $\frac{19-1}{2}=9$ is odd, then by the Lemma, the order of $-13$ is 9.
(e) Since 14 is a primitive root of prime 19, and since $\frac{19-1}{2}=9$ is odd, then by the Lemma, the order of $-14$ is 9.
(f) Since 15 is a primitive root of prime 19, and since $\frac{19-1}{2}=9$ is odd, then by the Lemma, the order of $-15$ is 9.