Class

A class consists of elements of a group $G$ that are conjugate to one another. Two elements $a,b\in G$ are conjugate to each other if $b=g^{-1}ag$ , where $g\in G$ . If $a$ is conjugate to $b$ , then $b$ is conjugate to $a$ , as

$b=g^{-1}ag\implies gb=ag\implies gbg^{-1}=a\implies a=g'^{-1}bg'$

where $g'=g^{-1}$ (note that $g^{-1}\in G$ according to the inverse property of a group). The identity element $e$ is a class by itself since $e=g^{-1}eg$ .

If the elements of $G$ are represented by matrices, $b=g^{-1}ag$ is called a similarity transformation. Furthermore, if $a$ and $b$ are conjugate to each other, and $b$ and $c$ are conjugate to each other, then $a$ and $c$ are conjugate to each other. This is because $b=g_1^{-1}ag_1$ and $c=g_2^{-1}bg_2\implies b=g_2cg_2^{-1}$ and therefore, $g_1^{-1}ag_1=g_2cg_2^{-1}\implies c=g_3^{-1}ag_3$ , where $g_3=g_1g_2$ .

Question

Show that the symmetry operators $\sigma_v$ , $\sigma_v'$ and $\sigma_v''$ of the $C_{3v}$ point group belong to the same class.

Answer

With reference to the $C_{3v}$ multiplication table,

we have

$\sigma_v''^{-1}\sigma_v\sigma_v''=\sigma_v''\sigma_v\sigma_v''=\sigma_v''C_3=\sigma_v'$

Similarly, $\sigma_v^{-1}\sigma_v'\sigma_v=\sigma_v\sigma_v'\sigma_v=\sigma_vC_3=\sigma_v''$ . Since $\sigma_v$ and $\sigma_v'$ are conjugate to each other, and $\sigma_v'$ and $\sigma_v''$ are conjugate to each other, then $\sigma_v$ and $\sigma_v''$ are conjugate to each other. Therefore, $\sigma_v,\sigma_v',\sigma_v''$ form a class. Using the same logic, we find that $C_3'$ and $C_3^2$ form another class.

All elements of the same class in a group $G$ have the same order, which is defined as the smallest value of $n$ such that $a^n=e$ , where $a\in G$ . This is because if $a$ is conjugate to $b$ , we have

$b^n=(g^{-1}ag)(g^{-1}ag)\cdots (g^{-1}ag)=g^{-1}a(gg^{-1})a(gg^{-1})ag\cdots\\=g^{-1}a^ng=g^{-1}eg=e$

The above equation of $b^n=g^{-1}a^ng=e$ is valid if and only if $a^n=e$ . This means that the smallest value of $n$ in $a^n=e$ and in $b^n=e$ must be the same. Therefore, elements $a$ and $b$ of the same class in a group have the same order and is denoted by $\vert a\vert=\vert b\vert$ .

Question

Verify that the symmetry operators $\sigma_v$ , $\sigma_v'$ and $\sigma_v''$ of the $C_{3v}$ point group have the same order of 2.

Answer

It is clear that when the reflection operator $\sigma_v$ acts on a shape twice, it sends the shape into itself. The same goes for $\sigma_v'$ and $\sigma_v''$ . Hence, $\vert\sigma_v\vert=\vert\sigma_v'\vert=\vert\sigma_v''\vert=2$ .

As mentioned in an earlier article, the similarity transformation of a matrix $A$ to a matrix $B$ leaves the trace of $A$ , which is defined as $Tr(A)=\sum_ia_{ii}$ , invariant. This implies that elements of the same class in a group have the same trace.

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