Exercise 4.3.3 (Conjugacy classes in $Z_2\times S_3$, $S_3 \times S_3$, $Z_3 \times A_4$)

Proof. Conjugacy classes of $G$:

(a)

$G=Z_2\times S_3$.

We write $Z_2=\{1,-1\}$.

Let $h=(a,\sigma )\in Z_2\times S_3$, and $g=(b,\tau )\in Z_2\times S_3$. Since $Z_2$ is abelian,

$$\begin{array}{llll}\hfill \mathit{𝑔h}{g}^{-1}& =(\mathit{𝑏𝑎}{b}^{-1},\mathit{𝜏𝜎}{\tau }^{-1})& \hfill & \\ \hfill & =(a,\mathit{𝜏𝜎}{\tau }^{-1}).& \hfill & \end{array}$$

So the conjugacy class of $(a,\sigma )\in Z_2\times S_3$ is $\{a\}\times C$, where $C$ is the conjugacy class of $\sigma$ in $S_3$, given in Section 3. This gives $6$ conjugation classes in $G$.

$$\begin{array}{cc}\hfill (a,\sigma )\hfill & \hfill C(a,\sigma )\hfill \\ \hfill (1,())\hfill & \hfill \{(1,())\}\hfill \\ \hfill (1,(12))\hfill & \hfill \{(1,(12),(1,(23)),(1,(13)\}\hfill \\ \hfill (1,(123))\hfill & \hfill \{(1,(123),(1,(132))\}\hfill \\ \hfill (-1,())\hfill & \hfill \{(-1,())\}\hfill \\ \hfill (-1,(12))\hfill & \hfill \{(-1,(12),(-1,(23)),(-1,(13)\}\hfill \\ \hfill (-1,(123))\hfill & \hfill \{(-1,(123),(-1,(132))\}\hfill \end{array}$$

The center of $G$ is $Z(G)=\{(1,()),(-1,())\}$ and the class equation is

$$12=2+3+2+3+2.$$

(b)

$G=S_3\times S_3$.

If $h=(\sigma ,\tau )\in S_3\times S_3$ and $(\lambda ,\mu )\in S_3\times S_3$, then

$$\begin{array}{llll}\hfill \mathit{𝑔h}{g}^{-1}& =(\mathit{𝜆𝜎}{\lambda }^{-1},\mathit{𝜇𝜏}{\mu }^{-1}).& \hfill & \end{array}$$

Therefore the conjugacy class of $h$ is the cartesian product of the conjugacy class of $\sigma$ by the conjugacy class of $\tau$.

This gives $8$ conjugacy classes.

$$\begin{array}{cc}\hfill (\sigma ,\tau )\hfill & C(\sigma ,\tau )\hfill \\ \hfill ((),())\hfill & \{((),())\}\hfill \\ \hfill ((),(12))\hfill & \{((),(12),((),(23)),((),(13)\}\hfill \\ \hfill ((12),())\hfill & \{((12),()),((23),()),((13),())\}\hfill \\ \hfill ((123),())\hfill & \{(123),()),(132),())\}\hfill \\ \hfill ((),(123))\hfill & \{((),(123)),((),(132))\}\hfill \\ \hfill ((12),(12))\hfill & \{((12),(12)),((12),(23)),((12),(13)),((13),(12)),\hfill \\ \hfill \hfill & ((13),(23)),((13),(13)),((23),(12),((23),(23)),((23),(13)\}\hfill \\ \hfill ((12),(123))\hfill & \{((12),(123),((23),(123),((13),(123),\hfill \\ \hfill \hfill & ((12),(132),((23),(132)),((13),(132))\}\hfill \\ \hfill ((123),(12))\hfill & \{((123),(12)),((123),(23)),((123),(13)),\hfill \\ \hfill \hfill & ((132),(12)),((132),(23)),((132),(13))\}\hfill \\ \hfill ((123),(123))\hfill & \{((123),(123)),((123),(132)),((132),(123)),((132),(132))\}\hfill \end{array}$$

The center is trivial, and the class equation is

$$36=1+3+3+2+2+9+6+6+4.$$

(c)

$G=Z_3\times A_4$.

We know the $4$ conjugation classes of $A_4$ by Exercise 2. As in part (a), the conjugacy class of $(a,\sigma )\in Z_3\times A_4$ is $\{a\}\times {𝒪}_{\sigma }$, where ${𝒪}_{\sigma }$ is the conjugacy class of $\sigma$ in $A_4$. This gives $12$ conjugacy classes (we write $Z_3=\{1,x,x^2\}$):

$$\{x^i\}\times {𝒪}_{\sigma },i\in [[0,2]],\sigma \in \{(),(123),(213),(12)(34)\}$$

of sizes $1,4,4,3,1,4,4,3,1,4,4,3$ (class equation: $36=3+3\cdot (4+4+3)$)