From Example 1 in §2, we know that 5 is a primitive root modulo 23, and $\varphi(23)=22$. First, list Table 1 in the order of indices, as it is easier to compute the absolute least residues of $5^{j}$ modulo 23. Arranging according to the size order of the absolute least reduced residue system, Table 1 becomes Table 2.

Table 1
$$\begin{array}{|c|rrrrrrrrrrrr|} \hline\gamma_{23,5}(a) & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hlinea & 1 & 5 & 2 & 10 & 4 & -3 & 8 & -6 & -7 & 11 & 9 \\ \hline\delta_{23}(a) & 1 & 22 & 11 & 22 & 11 & 22 & 11 & 22 & 11 & 22 & 11 \\ \hline \hline\gamma_{23,5}(a) & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 \\ \hlinea & -1 & -5 & -2 & -10 & -4 & 3 & -8 & 6 & 7 & -11 & -9 \\ \hline\delta_{23}(a) & 2 & 11 & 22 & 11 & 22 & 11 & 22 & 11 & 22 & 11 & 22 \\ \hline \end{array}$$

Table 2
$$\begin{array}{|c|rrrrrrrrrrr|} \hlinea & -11 & -10 & -9 & -8 & -7 & -6 & -5 & -4 & -3 & -2 & -1 \\ \hline\gamma_{23.5}(a) & 20 & 14 & 21 & 17 & 8 & 7 & 12 & 15 & 5 & 13 & 11 \\ \hline\delta_{23}(a) & 11 & 11 & 22 & 22 & 11 & 22 & 11 & 22 & 22 & 22 & 2 \\ \hline \hlinea & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 \\ \hline\gamma_{23.5}(a) & 0 & 2 & 16 & 4 & 1 & 18 & 19 & 6 & 10 & 3 & 9 \\ \hline\delta_{23}(a) & 1 & 11 & 11 & 11 & 22 & 11 & 22 & 11 & 11 & 22 & 22 \\ \hline \end{array}$$

From the table, we know that the primitive roots modulo 23 (i.e., elements with order 22) are 10, and they are:
$$-9,-8,-6,-4,-3,-2,5,7,10,11$$

Elements with order 11 are 10, and they are:
$$-11,-10,-7,-5,2,3,4,6,8,9$$

The element with order 2 is one: -1. The element with order 1 is one: 1.
Regarding the index sets $\gamma_{a,-1, g_{0}}^{(-1)}(a), \gamma_{a,-1, g_{0}}^{(0)}(a)$ for $m=2^{a}(\alpha \geqslant 3)$, we only discuss the case $g_{0}=5$, and denote them simply as $\gamma^{(-1)}(a), \gamma^{(0)}(a)$.