14.2: The Class Equation

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

$$X_G = \{ x \in X : gx = x \text{ for all } g \in G \}\text{.} \nonumber$$

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

$$|X| = |X_G| + \sum_{i = k}^n |{\mathcal O}_{x_i}|\text{,} \nonumber$$

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 $G\text{,}$

$$Z(G) = \{x : xg = gx \text{ for all } g \in G \}\text{,} \nonumber$$

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

$$|G| = |Z(G)| + n_1 + \cdots + n_k\text{.} \nonumber$$

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:

$$|G| = |Z(G)| + [G: C(x_1) ] + \cdots + [ G: C(x_k)]\text{.} \nonumber$$

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

Example $14.12$

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

$$\{ (1) \}, \quad \{ (1 \, 2 \, 3), (1 \, 3 \, 2) \}, \quad \{(1 \, 2), (1 \, 3), (2 \, 3) \}\text{.} \nonumber$$

Solution

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

Example $14.13$

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

$$\{ (1 \, 3), (2 \, 4) \}, \quad \{ (1 \, 4 \, 3 \, 2), (1 \, 2 \, 3 \, 4) \}, \quad \{ (1 \, 2)(3 \, 4), (1 \, 4)(2 \, 3) \}\text{.} \nonumber$$

Solution

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

Example $14.14$

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 $\tau \in S_n\text{.}$

Solution

By Theorem $6.16$,

$$\tau \sigma \tau^{-1} = ( \tau( a_1), \ldots, \tau(a_k))\text{.} \nonumber$$

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.

Theorem $14.15$

Let $G$ be a group of order $p^n$ where $p$ is prime. Then $G$ has a nontrivial center.

Proof

We apply the class equation

$$|G| = |Z(G)| + n_1 + \cdots + n_k\text{.} \nonumber$$

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 $g \neq 1\text{.}$

Corollary $14.16$

Let $G$ be a group of order $p^2$ where $p$ is prime. Then $G$ is abelian.

Proof

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,

$$gh = a^m x a^n y = a^{m+n} x y = a^n y a^m x = hg\text{,} \nonumber$$

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