Example 4-11 The side chain of methane $\mathrm{CH}_{4}$ is , if the 4 H bonds are replaced by one of $\mathrm{H}, \mathrm{Cl}, \mathrm{CH}_{3}, \mathrm{C}_{2} \mathrm{H}_{5}$, how many different chemical structures are there?
Solving the problem is equivalent to coloring the 4 vertices of a regular tetrahedron with 4 colors, and finding the number of different schemes. The rigid motions that make the regular tetrahedron \(v_{1} v_{2} v_{3} v_{4}\) coincide are of two types: one is rotating \(120^{\circ}, 240^{\circ}\) around the central line \(x x^{\prime}\) passing through a vertex; the other is rotating \(180^{\circ}\) around the line \(y y^{\prime}\) passing through the midpoints of \(v_{1} v_{2}, v_{3} v_{4}\). The elements of the rotation group \(G\) of Figure 4-13 are:
$$
\begin{array}{l}
\left(v_{1}\right)\left(v_{2}\right)\left(v_{3}\right)\left(v_{4}\right),\left(v_{1}\right)\left(v_{2} v_{3} v_{4}\right), \\
\left(v_{1}\right)\left(v_{4} v_{3} v_{2}\right),\left(v_{2}\right)\left(v_{1} v_{3} v_{4}\right), \\
\left(v_{2}\right)\left(v_{4} v_{3} v_{1}\right),\left(v_{3}\right)\left(v_{1} v_{2} v_{1}\right), \\
\left(v_{3}\right)\left(v_{4} v_{2} v_{1}\right),\left(v_{4}\right)\left(v_{1} v_{2} v_{3}\right), \\
\left(v_{4}\right)\left(v_{3} v_{2} v_{1}\right),\left(v_{1} v_{2}\right)\left(v_{3} v_{4}\right), \\
\left(v_{1} v_{3}\right)\left(v_{2} v_{4}\right),\left(v_{1} v_{4}\right)\left(v_{2} v_{3}\right),
\end{array}
$$

Therefore, the number of different chemical structures is
$$
\frac{1}{12} \times\left[4^{4}+8 \cdot 4^{2}+3 \cdot 4^{2}\right]=\frac{1}{12} \times[256+128+48]=36
$$