State the class equation.
Suppose \(|G| = p^k\) and \(H\) is a normal subgroup of \(G\) with \(|H| > 1\). Show that \(H\) contains a non-identity element of the center \(Z(G)\) of \(G\).
Solution
1. If \(x_1,\dots,x_r\) represent the noncentral conjugacy classes of a finite group \(G\), then the class equation is\[ |G|=|Z(G)|+\sum_{i=1}^r [G:C_G(x_i)]. \]
The conjugacy class of \(x\in G\) has cardinality \([G:C_G(x)]\) by orbit-stabilizer for the conjugation action. The central elements are exactly the singleton conjugacy classes. Partitioning \(G\) into its conjugacy classes therefore gives the displayed equation.
2. Let \(G\) act on the normal subgroup \(H\) by conjugation. The fixed-point set of this action is\[ H^G=H\cap Z(G). \]
Normality of \(H\) ensures that conjugation by every element of \(G\) preserves \(H\). An element \(h\in H\) is fixed by every conjugation map exactly when \(ghg^{-1}=h\) for every \(g\in G\), equivalently \(h\in Z(G)\).
3. Every nontrivial orbit of the conjugation action of \(G\) on \(H\) has cardinality divisible by \(p\).
For \(h\in H\), orbit-stabilizer gives \[ |G\cdot h|=[G:C_G(h)]. \] Since \(|G|=p^k\), every subgroup index is a power of \(p\). If the orbit is not a singleton, its size is therefore a positive power of \(p\), hence divisible by \(p\).
4. The cardinality \(|H\cap Z(G)|\) is divisible by \(p\).
Partitioning \(H\) into conjugation orbits gives \[ |H|=|H\cap Z(G)|+\sum_j |\mathcal O_j|, \] where the \(\mathcal O_j\) are the nontrivial orbits. Because \(H\) is a nontrivial subgroup of the finite \(p\)-group \(G\), its order is a positive power of \(p\), hence divisible by \(p\). By 3 every term in the sum is divisible by \(p\). Therefore \(|H\cap Z(G)|\equiv0\pmod p\).
5. Hence \(H\) contains a nonidentity element of \(Z(G)\).
The intersection \(H\cap Z(G)\) contains the identity, so it is nonempty. By 4 its cardinality is a positive multiple of \(p\), hence at least \(p\ge2\). Therefore it contains an element other than the identity.