##### Example14.12

It is easy to check that the conjugacy classes in \(S_3\) are the following: \begin{equation*}\{ (1) \}, \quad \{ (123), (132) \}, \quad \{(12), (13), (23) \}.\end{equation*} The class equation is \(6 = 1+2+3\).

Let \(X\) be a finite \(G\)-set and \(X_G\) be the set of fixed points in \(X\); that is, \begin{equation*}X_G = \{ x \in X : gx = x \text{ for all } g \in G \}.\end{equation*} Since the orbits of the action partition \(X\), \begin{equation*}|X| = |X_G| + \sum_{i = k}^n |{\mathcal O}_{x_i}|,\end{equation*} where \(x_k, \ldots, x_n\) are representatives from the distinct nontrivial orbits of \(X\).

Now consider the special case in which \(G\) acts on itself by conjugation, \((g,x) \mapsto gxg^{-1}\). The *center* of \(G\),
\begin{equation*}Z(G) = \{x : xg = gx \text{ for all } g \in G \},\end{equation*}
is the set of points that are fixed by conjugation. The nontrivial orbits of the action are called the *conjugacy classes* of \(G\). If \(x_1, \ldots, x_k\) are representatives from each of the nontrivial conjugacy classes of \(G\) and \(|{\mathcal O}_{x_1}| = n_1, \ldots, |{\mathcal O}_{x_k}| = n_k\), then
\begin{equation*}|G| = |Z(G)| + n_1 + \cdots + n_k.\end{equation*}
The stabilizer subgroups of each of the \(x_i\)'s, \(C(x_i) = \{ g \in G: g x_i = x_i g \}\), 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)].\end{equation*}
One of the consequences of the class equation is that the order of each conjugacy class must divide the order of \(G\).

It is easy to check that the conjugacy classes in \(S_3\) are the following: \begin{equation*}\{ (1) \}, \quad \{ (123), (132) \}, \quad \{(12), (13), (23) \}.\end{equation*} The class equation is \(6 = 1+2+3\).

The center of \(D_4\) is \(\{ (1), (13)(24) \}\), and the conjugacy classes are \begin{equation*}\{ (13), (24) \}, \quad \{ (1432), (1234) \}, \quad \{ (12)(34), (14)(23) \}.\end{equation*} Thus, the class equation for \(D_4\) is \(8 = 2 + 2 + 2 + 2\).

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 \(\tau \in S_n\). By Theorem 6.16, \begin{equation*}\tau \sigma \tau^{-1} = ( \tau( a_1), \ldots, \tau(a_k)).\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 \(r_i\). 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. The problem of finding the number of such partitions for any positive integer \(n\) is 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.

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

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