Exercise 13.4.10

Proof.

(a)

For an element $g$ of a subgroup $G$, its conjugacy class is the set of elements conjugate to it: $$\{\mathit{xg}{x}^{-1}:x\in G\}.$$

Suppose the conjugacy classes of $g$ and $h$ overlap, i.e., $\mathit{xg}{x}^{-1}=\mathit{yh}{y}^{-1}$ for some $x$ and $y$ in the subgroup. Therefore

$$g=x^{-1}\mathit{yh}{y}^{-1}x=(x^{-1}y)h{(x^{-1}y)}^{-1},$$

which shows each element of $G$ that is conjugate to $g$ is also conjugate to $h$. In the other way, from $\mathit{xg}{x}^{-1}=\mathit{yh}{y}^{-1}$ write $h=(y^{-1}x)h{(y^{-1}x)}^{-1},$ and similar calculation shows that each element of $G$ that is conjugate to $h$ is also conjugate to $g$.

Hence the conjugacy classes are equal, which means that subgroup $G$ is the set of distinct conjugacy classes $\{S_1,S_2,...,S_l\}$ and any element $x\in G$ belongs to one of the conjugacy class.

Let $S=\{\sigma \in G\mid \sigma \mathit{has}\mathit{cycle}\mathit{type}{d}_1,...,d_r\}$. If $G$ does not have element with cycle type $d_1,...,d_r$, then $S$ is empty.

If $S$ is not empty, then any $\sigma \in S$ belongs to some conjugacy class. Let the subset $\{S_{k_1},S_{k_2},...,S_{k_m}\}\subset \{S_1,S_2,...,S_l\}$ is such that any element of $S$ belongs to one of subset in $\{S_{k_1},S_{k_2},...,S_{k_m}\}$.

In Ex. 13.4.9 has been showed that all elements of conjugate class have the same cycle type. It means that if $\tau \in S_{k_i}$, then $\tau$ has cycle type $d_1,...,d_r$, i.e., $\tau \in S$, and this is valid for all $S_{k_i},k_i\in (k_1,...,k_m)$.

Therefore, $S={\text{∪}}_{i=1}^m{S}_{k_i}$, i.e., the set of elements of $G$ with the same cycle type is either empty or a union of conjugacy classes of $G$.

(b)

Consider subgroup $A_3\subset S_3$. Since $A_3$ doesn’t have transpositions, the cycle type $(d_1,d_2)=(1,2)$ is not possible and set of $A_3$ elements with cycle type (1,2) is empty.

Consider subgroup $A_4\subset S_4$.

The 3-cycle (123) and its inverse (132) are not conjugate in A4. To see this, let’s determine all possible $\sigma \in S_4$ that conjugate (123) to (132). For $\sigma \in S_4$, the condition $\sigma (123){\sigma }^{-1}=(132)$ is the same as $(\sigma (1)\sigma (2)\sigma (3))=(132)$. There are three possibilities:

$\text{∙}$

$\sigma (1)=1$, so $\sigma (2)=3$ and $\sigma (3)=2$, and necessarily $\sigma (4)=4$. Thus $\sigma =(23)$.

$\text{∙}$

$\sigma (1)=3$, so $\sigma (2)=2$ and $\sigma (3)=1$, and necessarily $\sigma (4)=4$. Thus $\sigma =(13)$.

$\text{∙}$

$\sigma (1)=2$, so $\sigma (2)=1$ and $\sigma (3)=3$, and necessarily $\sigma (4)=4$. Thus $\sigma =(12)$.

Therefore the only possible $\sigma$’s are transpositions, which are not in $A_4$. It is obvious that all other 3-cycles are conjugate to (123) or (132).

Hence, all elements of $A_4$ with cycle type $(d_1,d_2)=(1,3)$ are in the union of two conjugacy classes with the representatives $\{(123),(132)\}$.