State the class equation for finite groups.
Use the class equation to show that if a finite group \(G\) has order \(p^k\) for some prime \(p\) and \(k \geq 1\), then \(G\) has a nontrivial center.
Solution
1. If \(x_1,\ldots,x_r\) represent the noncentral conjugacy classesof a finite group \(G\), then its class equation is\[ |G|=|Z(G)|+\sum_{j=1}^r [G:C_G(x_j)], \]where \(C_G(x)=\{g\in G:gx=xg\}\).
Conjugacy is an equivalence relation, so its classes partition \(G\). An element has a singleton conjugacy class precisely when it commutes with every element of \(G\), that is, when it belongs to \(Z(G)\). For any \(x\in G\), the map \[ G/C_G(x)\longrightarrow\{gxg^{-1}:g\in G\},\qquad gC_G(x)\longmapsto gxg^{-1} \] is well-defined and surjective. It is injective because \(gxg^{-1}=hxh^{-1}\) is equivalent to \(h^{-1}g\in C_G(x)\), which is equivalent to equality of the two left cosets. Thus each noncentral class has the displayed index as its size. Adding the class sizes gives the formula.
2. If \(|G|=p^k\) with \(k\geq1\), then \(|Z(G)|\geq p\).
Each centralizer has order dividing \(p^k\) by Lagrange’s theorem [@DF04], so each index \([G:C_G(x_j)]\) is a power of \(p\). Since \(x_j\) is noncentral, its centralizer is proper, and that power is greater than \(1\). Hence every term in the sum in step 1 is divisible by \(p\). The same equation implies \(p\mid |Z(G)|\). The center contains the identity, so its size is a positive multiple of \(p\) and therefore at least \(p>1\). In particular it is nontrivial.