We take the reduced residue system modulo $2^{\circ}$ given by formula (4) $\left(g_{0}=5\right)$. From formula (17), we know that the sufficient and necessary condition for $\delta_{2^{a}}(a)=1$ is $\gamma^{(0)}(a)=\gamma^{(-1)}(a)=0$, i.e., $a \equiv 1\left(\bmod 2^{a}\right)$. Therefore, formula (19) holds. From formula (17), we know that the sufficient and necessary condition for $\delta_{2^{a}}(a)=2$ is either
$$\begin{array}{cc} \gamma^{(0)}(a)=0, & \gamma^{(-1)}(a)=1 \\ \gamma^{(0)}(a)=2^{a-3}, & \gamma^{(-1)}(a)=0,1 \end{array}$$

Thus, there are three such elements in a reduced residue system, so formula (19) also holds. When $d>2, d \mid 2^{a-2}$, we can set $d=2^{j}, 1<j \leqslant 2^{a-2}$. From formula (17), we know that the sufficient and necessary condition for $\delta_{2^{a}}(a)=d=2^{j}$ is
$$\left(\gamma^{(0)}(a), 2^{a-2}\right)=2^{a-2-j}, \quad 0<\gamma^{(0)}(a)<2^{a-2} .$$

Let $\gamma^{(0)}(a)=2^{a-2-j} \cdot t$, the above equation becomes
$$\left(t, 2^{j}\right)=1, \quad 0<t<2^{j}$$

There are $\varphi\left(2^{j}\right)=\varphi(d)$ such $t$ values, and since $\gamma^{(-1)}(a)$ can take 0 or 1, there are exactly $2 \varphi(d)$ elements with index $d\left(2<d \mid 2^{a-2}\right)$ in a reduced residue system.

Similar to the index table for a modulus $m$ with a primitive root, we can list the index group table for modulus $2^{a}(\alpha \geqslant 3)$, where the index $\delta(a)=\delta_{2^{e}}(a)$ is derived from formula (17).