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, \(|Z(G)| = p^2\text{.}\)