Example 1 Find the primitive root of $p=23$.

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Example 1 Find the primitive root of $p=23$.

Since the index of $a$ modulo $p$ must be a divisor of $p-1$, to find the index, we just need to compute the residue of $a^{d}$ modulo $p$, where $d \mid p-1$.

Here $p-1=22=2 \cdot 11$, its divisors $d=1,2,11,22$. First, find the index of $a=2$ modulo 23:
$$\begin{array}{c}
2^{2} \equiv 4(\bmod 23) \\
2^{11} \equiv\left(2^{4}\right)^{2} \cdot 2^{3} \equiv(-7)^{2} \cdot 8 \equiv 3 \cdot 8 \equiv 1(\bmod 23)
\end{array}$$

So $\delta_{23}(2)=11,2$ is not a primitive root modulo 23. Next, find $\delta_{23}(3)$:
$$\begin{array}{c}
3^{2} \equiv 9(\bmod 23), \quad 3^{3} \equiv 4(\bmod 23) \\
3^{11} \equiv\left(3^{3}\right)^{3} \cdot 3^{2} \equiv 4^{3} \cdot 9 \equiv(-5) \cdot 9 \equiv 1(\bmod 23)
\end{array}$$

So, $\delta_{23}(3)=11,3$ is not a primitive root modulo 23. Next, find $\delta_{23}(4)$.
$$4^{2} \equiv-7(\bmod 23), \quad 4^{11} \equiv\left(4^{4}\right)^{2} \cdot 4^{3} \equiv 3^{2} \cdot(-5) \equiv 1(\bmod 23)$$

So $\delta_{23}(4)=11,4$ is not a primitive root modulo 23. Next, find $\delta_{23}(5)$.
$$\begin{aligned}
5^{2} & \equiv 2(\bmod 23) \\
5^{11} & \equiv\left(5^{4}\right)^{2} \cdot 5^{3} \equiv 4^{2} \cdot 10 \\
& \equiv 4 \cdot(-6) \equiv-1(\bmod 23) \\
5^{22} & \equiv 1(\bmod 23)
\end{aligned}$$

So, $\delta_{23}(5)=22,5$ is a primitive root modulo 23, and it is the smallest positive primitive root.
To further find the primitive roots modulo $p^{\alpha}, 2 p^{\alpha}$, we need to verify whether equation (8) holds when $g=5, p=23$. This is essentially finding the residue of $g^{p-1}$ modulo $p^{2}$. Here $23^{2}=529$.
$$\begin{aligned}
5^{2} & \equiv 25\left(\bmod 23^{2}\right) \\
5^{8} & \equiv(23+2)^{4} \equiv 4 \cdot 23 \cdot 2^{3}+2^{4} \equiv 10 \cdot 23-7\left(\bmod 23^{2}\right) \\
5^{10} & \equiv(10 \cdot 23-7)(23+2) \equiv 13 \cdot 23-14 \\
& \equiv 12 \cdot 23+9\left(\bmod 23^{2}\right) \\
5^{20} & \equiv(12 \cdot 23+9)^{2} \equiv 216 \cdot 23+81 \\
& \equiv 13 \cdot 23-11\left(\bmod 23^{2}\right) \\
5^{22} & \equiv(13 \cdot 23-11)(23+2) \equiv 15 \cdot 23-22 \\
& \equiv 1+14 \cdot 23\left(\bmod 23^{2}\right)
\end{aligned}$$

Since $23 \nmid 14$ and 5 is odd, it proves that 5 is a primitive root for all moduli $23^{a}, 2 \cdot 23^{a}$.