## Section14.2The Class Equation

Let $$X$$ be a finite $$G$$-set and $$X_G$$ be the set of fixed points in $$X\text{;}$$ that is,

\begin{equation*} X_G = \{ x \in X : gx = x \text{ for all } g \in G \}\text{.} \end{equation*}

Since the orbits of the action partition $$X\text{,}$$

\begin{equation*} |X| = |X_G| + \sum_{i = k}^n |{\mathcal O}_{x_i}|\text{,} \end{equation*}

where $$x_k, \ldots, x_n$$ are representatives from the distinct nontrivial orbits of $$X\text{.}$$

Now consider the special case in which $$G$$ acts on itself by conjugation, $$(g,x) \mapsto gxg^{-1}\text{.}$$ The center of $$G\text{,}$$

\begin{equation*} Z(G) = \{x : xg = gx \text{ for all } g \in G \}\text{,} \end{equation*}

is the set of points that are fixed by conjugation. The nontrivial orbits of the action are called the conjugacy classes of $$G\text{.}$$ If $$x_1, \ldots, x_k$$ are representatives from each of the nontrivial conjugacy classes of $$G$$ and $$|{\mathcal O}_{x_1}| = n_1, \ldots, |{\mathcal O}_{x_k}| = n_k\text{,}$$ then

\begin{equation*} |G| = |Z(G)| + n_1 + \cdots + n_k\text{.} \end{equation*}

The stabilizer subgroups of each of the $$x_i$$'s, $$C(x_i) = \{ g \in G: g x_i = x_i g \}\text{,}$$ are called the centralizer subgroups of the $$x_i$$'s. From Theorem 14.11, we obtain the class equation:

\begin{equation*} |G| = |Z(G)| + [G: C(x_1) ] + \cdots + [ G: C(x_k)]\text{.} \end{equation*}

One of the consequences of the class equation is that the order of each conjugacy class must divide the order of $$G\text{.}$$

### Example14.12.

It is easy to check that the conjugacy classes in $$S_3$$ are the following:

\begin{equation*} \{ (1) \}, \quad \{ (1 \, 2 \, 3), (1 \, 3 \, 2) \}, \quad \{(1 \, 2), (1 \, 3), (2 \, 3) \}\text{.} \end{equation*}

The class equation is $$6 = 1+2+3\text{.}$$

### Example14.13.

The center of $$D_4$$ is $$\{ (1), (1 \, 3)(2 \, 4) \}\text{,}$$ and the conjugacy classes are

\begin{equation*} \{ (1 \, 3), (2 \, 4) \}, \quad \{ (1 \, 4 \, 3 \, 2), (1 \, 2 \, 3 \, 4) \}, \quad \{ (1 \, 2)(3 \, 4), (1 \, 4)(2 \, 3) \}\text{.} \end{equation*}

Thus, the class equation for $$D_4$$ is $$8 = 2 + 2 + 2 + 2\text{.}$$

### Example14.14.

For $$S_n$$ it takes a bit of work to find the conjugacy classes. We begin with cycles. Suppose that $$\sigma = ( a_1, \ldots, a_k)$$ is a cycle and let $$\tau \in S_n\text{.}$$ By Theorem 6.16,

\begin{equation*} \tau \sigma \tau^{-1} = ( \tau( a_1), \ldots, \tau(a_k))\text{.} \end{equation*}

Consequently, any two cycles of the same length are conjugate. Now let $$\sigma = \sigma_1 \sigma_2 \cdots \sigma_r$$ be a cycle decomposition, where the length of each cycle $$\sigma_i$$ is $$r_i\text{.}$$ Then $$\sigma$$ is conjugate to every other $$\tau \in S_n$$ whose cycle decomposition has the same lengths.

The number of conjugate classes in $$S_n$$ is the number of ways in which $$n$$ can be partitioned into sums of positive integers. In the case of $$S_3$$ for example, we can partition the integer $$3$$ into the following three sums:

\begin{align*} 3 & = 1 + 1 + 1\\ 3 & = 1 + 2\\ 3 & = 3; \end{align*}

therefore, there are three conjugacy classes. There are variations to problem of finding the number of such partitions for any positive integer $$n$$ that are what computer scientists call NP-complete. This effectively means that the problem cannot be solved for a large $$n$$ because the computations would be too time-consuming for even the largest computer.

We apply the class equation

\begin{equation*} |G| = |Z(G)| + n_1 + \cdots + n_k\text{.} \end{equation*}

Since each $$n_i \gt 1$$ and $$n_i \mid |G|\text{,}$$ it follows that $$p$$ must divide each $$n_i\text{.}$$ Also, $$p \mid |G|\text{;}$$ hence, $$p$$ must divide $$|Z(G)|\text{.}$$ Since the identity is always in the center of $$G\text{,}$$ $$|Z(G)| \geq 1\text{.}$$ Therefore, $$|Z(G)| \geq p\text{,}$$ and there exists some $$g \in Z(G)$$ such that $$g \neq 1\text{.}$$

By Theorem 14.15, $$|Z(G)| = p$$ or $$p^2\text{.}$$ Suppose that $$|Z(G)| = p\text{.}$$ Then $$Z(G)$$ and $$G / Z(G)$$ both have order $$p$$ and must both be cyclic groups. Choosing a generator $$aZ(G)$$ for $$G / Z(G)\text{,}$$ we can write any element $$gZ(G)$$ in the quotient group as $$a^m Z(G)$$ for some integer $$m\text{;}$$ hence, $$g = a^m x$$ for some $$x$$ in the center of $$G\text{.}$$ Similarly, if $$hZ(G) \in G / Z(G)\text{,}$$ there exists a $$y$$ in $$Z(G)$$ such that $$h = a^n y$$ for some integer $$n\text{.}$$ Since $$x$$ and $$y$$ are in the center of $$G\text{,}$$ they commute with all other elements of $$G\text{;}$$ therefore,

\begin{equation*} gh = a^m x a^n y = a^{m+n} x y = a^n y a^m x = hg\text{,} \end{equation*}

and $$G$$ must be abelian. Hence, $$|Z(G)| = p^2\text{.}$$