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Section 1.19 Class equation
This section will essentially be a long format example of computing stabilizers and orbits for one particular type of group action: namely, the action of a group on itself by conjugation. The
class equation of a group is a one-line summary of the number and sizes of the orbits of
\(G\) under the action of conjugation.
Definition 1.19.1 . Conjugacy class.
The conjugacy class of an element \(h\) of the group \(G\text{,}\) denoted \(C_h\) is the set of all conjugates of \(h\) in: i.e.,
\begin{equation}
C_h=\{ghg^{-1}\mid g\in G\}\text{.}\tag{1.19.1}
\end{equation}
Equivalently, the conjugacy class of \(h\) is the orbit \(O_h\) of \(h\) under the action of conjugation.
Theorem 1.19.2 . Conjugacy and the class equation.
Let \(G\) be a finite group and consider the action of \(G\) on itself by conjugation.
For all \(h\in G\text{,}\) we have
\begin{align*}
O_h \amp =C_h \\
G_h \amp =C_G(h) \text{.}
\end{align*}
It follows from the orbit-stabilizer theorem that
\begin{equation}
\abs{C_h}=[G\colon C_G(h)]\text{.}\tag{1.19.2}
\end{equation}
We have
\begin{equation}
Z(G)=\{g\in G\mid C_g=\{g\}\}\text{.}\tag{1.19.3}
\end{equation}
Let \(C_i=C_{g_i}\text{,}\) \(1\leq i\leq s\) be the distinct conjugacy classes of \(G\text{,}\) and assume further that \(\{C_i\mid 1\leq i\leq r\}\text{,}\) \(r\leq s\text{,}\) is the set of conjugacy classes consisting of a single element. We have a partition
\begin{equation}
G=\bigcup_{i=1}^{s}C_i\text{,}\tag{1.19.4}
\end{equation}
from which it follows that
\begin{equation}
\abs{G}=\sum_{i=1}^r1+\sum_{i=r+1}^s\abs{C_i}=\abs{Z(G)}+\sum_{i=r+1}^s[G\colon C_G(g_i)]\text{.}\tag{1.19.5}
\end{equation}
Proof.
Definition 1.19.3 . Class equation.
Let \(G\) be a finite group, and let \(C_1,C_2,\dots, C_s\) be the distinct conjugacy classes of \(G\text{.}\) The class equation of \(G\) is the equation
\begin{equation*}
\abs{G}=\sum_{i=1}^s\abs{C_i}\text{.}
\end{equation*}
Theorem 1.19.2 is useful both for extracting information about a group from its class equation, as well as taking some shortcuts for computing the class equation of a group.
Example 1.19.4 . Class equation: abelian groups.
Prove: a finite group \(\abs{G}\) is abelian if and only if its class equation is of the form
\begin{equation*}
\abs{G}=1+1+\cdots +1\text{.}
\end{equation*}
Example 1.19.5 . Class equation: \(D_8\) .
Compute the class equation of
\(D_8\text{.}\)
Theorem 1.19.6 . Center of a \(p\) -group.
Assume
\(G\) is a group of cardinality
\(p^n\text{,}\) where
\(p\) is a prime integer. The center of
\(G\) is nontrivial: i.e.,
\(Z(G)\ne \{e\}\text{.}\)
Proof.
Corollary 1.19.7 . Groups of cardinality \(p^2\) .
If
\(G\) is a group of cardinality
\(p^2\text{,}\) where
\(p\) is a prime integer, then
\(G\) is abelian.
Definition 1.19.8 . Partition of \(n\) .
Let \(n\) be a positive integer. A partition of \(n\) is an \(r\) -tuple \((n_1,n_2,\dots, n_r)\) for some \(1\leq r\leq n\) satisfying
\begin{align*}
n_1 \amp \leq n_2\leq \dots \leq n_r\\
n \amp =n_1+n_2+\cdots +n_r\text{.}
\end{align*}
Given a partiion \(p=(n_1,n_2,\dots, n_r)\) we define its type \(T(p)\) to be the \(n\) -tuple \((m_1,m_2,\dots, m_n)\text{,}\) where \(m_i\) is the number of entries of \(p\) equal to \(i\) for all \(1\leq i\leq n\text{.}\)
Theorem 1.19.9 . Class equation of \(S_n\) .
Let \(n\) be a positive integer, let \(\mathcal{P}\) be the set of all partitions of \(n\text{,}\) and let \(T(\mathcal{P})\) be the set of all types of these partitions. The class equation of \(S_n\) is
\begin{equation}
n!=\sum_{(m_1,m_2,\dots, m_n)\in T(\mathcal{P})}\dfrac{n!}{m_1!m_2
!\cdots m_n!1^{m_1}2^{m_2}\cdots n^{m_n}}\text{.}\tag{1.19.6}
\end{equation}
Proof.
Corollary 1.19.10 . Stabilizers of cycles.
Let \(\sigma=(a_1\ a_2\ \dots a_k)\) be a \(k\) -cycle in \(S_n\text{.}\) Let \(H=\angvec{\sigma}\) and let
\begin{equation*}
K=\{\tau\in S_n\mid \tau(a_i)=a_i \text{ for all } 1\leq i\leq k\}\text{.}
\end{equation*}
We have
\begin{align*}
C_{S_n}(\sigma) \amp =HK \amp \abs{C_{S_{n}}(\sigma)}=k(n-k)!\text{.}
\end{align*}
Example 1.19.11 . Class equation: \(A_5\) .
Determine the class equation of
\(A_5\text{.}\)
Solution .
First we describe the different cycle types of elements of \(A_5\text{.}\)
\begin{equation*}
\begin{array}{c|c}
\text{Type} \amp \text{Number of elements}\\
\hline
(a\ b\ c\ d\ e) \amp 24 \\
(a\ b\ c) \amp 20 \\
(a\ b)(c\ d) \amp 15\\
\id \amp 1
\end{array}\text{.}
\end{equation*}
Given \(\sigma\in A_n\) of one of these types, we know that its conjugacy class in \(S_5\) consists of all elements of that same type. However, we are computing conjugacy classes in \(A_5\text{!}\) Since we are conjugating \(\sigma\) by fewer elements, its conjugacy class in \(A_5\) will be a subset of its conjugacy class in \(S_5\text{.}\) It follows that each conjugacy class in \(S_5\) breaks ups into a disjoint union of conjugacy classes in \(A_5\text{.}\) As you will show in your homework, some of these conjugacy classes remain intact, while the others split into exactly two conjugacy classes.
Theorem 1.19.12 . Simplicity of \(A_n\) .
For all
\(n\geq 5\) the alternating group
\(A_n\) is simple.
Proof.
