Exercise 4.3.2 (Conjugation classes in $D_8, Q_8, A_4$)

Proof. Conjugacy classes of $G$:

(a)

$G=D_8$ (see Example 3 p. 124).

• If $x\in Z(D_8)=⟨r^2⟩$, then the conjugacy class has size $1$: this gives the conjugacy classes $\{1\}$ and $\{r^2\}$.

• If $x\notin Z(D_8)$, then the orbit of $x$ has size $|G:C_G(x)|$, where the centralizer $C_G(x)$ satisfies $⟨x⟩\le C_G(x)\le G$. Since $x\notin Z(D_8)$, $C_G(x)\ne G$, so

  $$\begin{array}{lll}\hfill ⟨x⟩\le C_G(x)\ne G.& & \hfill \end{array}$$

  The lattice of subgroups of $D_8$ given p. 69 shows that $D_8=⟨s,r⟩$ has three subgroups of index $2$. Since these subgroups have order $4$, they are abelian. Every element $x$ of $G$ is in one of these three maximal subgroups, so $x\in H$ where $|H|=4$ and $H$ abelian, so $H\le C_G(x)\ne G$, thus $|H|=4\mid |C_G(x)|\ne 8$, so

  $$\begin{array}{lll}\hfill |C_G(x)|=4(\text{if }x\notin Z(D_8)=\{1,r^2\}).& & \hfill \end{array}$$

  Then the conjugacy class of $x$ has $|G:C_G(x)|=2$ elements. Since

  $$\begin{array}{lll}\hfill r^3=\mathit{𝑠𝑟}{s}^{-1},s{r}^2=\mathit{𝑟𝑠}{r}^{-1},s{r}^3=s(\mathit{𝑠𝑟})s^{-1},& & \hfill \end{array}$$

  we obtain all the conjugacy classes

  $$\begin{array}{lll}\hfill \{1\},\{r^2\},\{r,r^3\},\{s,s{r}^2\},\{\mathit{𝑠𝑟},s{r}^3\}.& & \hfill \end{array}$$

(b)

$G=Q_8$ (see Example 2 p. 124).

Starting from $⟨i⟩\le C_G(i)\le G$, since $i\notin Z(Q_8)=\{1,-1\}$, then $C_G(i)\ne G$, thus

$$\begin{array}{lll}\hfill ⟨i⟩=\{1,i,-1,-i\}\le C_G(i)\ne G.& & \hfill \end{array}$$

Thus $4$ divides $|C_G(i)|$ and $|C_G(i)|\ne 8$, thus $|C_G(i)|=4$, so

$$\begin{array}{lll}\hfill C_G(i)=⟨i⟩.& & \hfill \end{array}$$

Therefore the conjugacy class of $i$ has $2=|G:C_G(i)|$ elements. Moreover $-i=\mathit{𝑘𝑖}{k}^{-1}\ne i$, so the conjugacy class of $i$ is $\{i,-i\}$. We obtain similarly the conjugacy classes of $j$ and $k$. So the conjugacy classes in $Q_8$ are

$$\begin{array}{lll}\hfill \{1\},\{-1\},\{i,-i\},\{j,-j\},\{k,-k\}.& & \hfill \end{array}$$

(c)

$G=A_4$.

Conjugate permutations in $A_4$ are also conjugate in $S_4$, so have the same cycle type. The representative of cycle type in $A_4$ are

$$(),(123),(12)(34).$$

The conjugacy class of $\text{()}$ is $\text{{()}}$.

Let $\lambda =(abc)$ be any $3$-cycle in $A_4$. We know by the results on conjugacy in $S_n$ that $C_{S_4}(\lambda )=⟨\lambda ⟩$ has order $3=3(4-3)!$. Since these three elements are in $A_4$,

$$C_{A_4}(\lambda )=⟨\lambda ⟩$$

has order $3$. Therefore the conjugacy class of $(abc)$ in $A_4$ has $|A_4:C_{A_4}(\lambda )|=12∕3=4$ elements.

Since there are eight $3$-cycles in $A_4$, there are two classes of conjugation in the set of $3$-cycles.

Explicitly, consider a cycle $(abc)\in A_4$. Then $a,b,c$ are distinct elements of $\{1,2,3,4\}$. Let $d$ be the fourth element in this set ( $d=10-a-b-c$). We define the permutations $\sigma ,\tau \in S_4$ by

$$\sigma =\left(\begin{array}{cccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill \\ \hfill a\hfill & \hfill b\hfill & \hfill c\hfill & \hfill d\hfill \end{array}\right),\tau =\left(\begin{array}{cccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill \\ \hfill b\hfill & \hfill a\hfill & \hfill c\hfill & \hfill d\hfill \end{array}\right).$$

Then Proposition 10 shows that

$$\begin{array}{lll}\hfill \sigma (123){\sigma }^{-1}=(abc),& & \hfill \text{(1)}\\ \hfill \tau (213){\tau }^{-1}=(abc).& & \hfill \text{(2)}\end{array}$$

So $(abc)$ is conjugate to $(123)$ or $(132)$, and there are two conjugacy classes of $3$-cycles, so the classes of conjugation are the class of $(123)$ and the class of $(213)$. So $(123)$ and $(213)$ are not conjugate in $A_4$.

It remains $3$ elements in $A_4$:

$$(12)(34),(13)(24),(14)(23),$$

which form a class of conjugation, since

$$\begin{array}{llll}\hfill (13)(24)=(31)(24)& =\xi (12)(34){\xi }^{-1}\text{for }\xi =(132),& \hfill & \\ \hfill (14)(23)& =\eta (12)(34){\eta }^{-1}\text{for }\eta =(243).& \hfill & \end{array}$$

In conclusion, there are $4$ classes of conjugation in $A_4$:

$$\begin{array}{llll}\hfill {𝒪}_{\text{()}}& =\{()\},& \hfill & \\ \hfill {𝒪}_{(123)}& =\{(123),(134),(142),(243)\}& \hfill & \\ \hfill {𝒪}_{(213)}& =\{(132),(143),(124),(234)\},& \hfill & \\ \hfill {𝒪}_{(12)(34)}& =\{(12)(34),(13)(24),(14)(23)\}.& \hfill & \end{array}$$

The class equation is $12=1+4+4+3$.

Note: The subgroup $H=⟨(12)(34),(13)(24)⟩$ is the union of two conjugation classes, so is normal in $A_4$.