Let \(G\) be a finite group acting on a set \(X\). For \(x\in X\), let \(G_x\) be the stabilizer of \(x\) and \(G\cdot x\) be the orbit of \(x\).
Prove that there is a bijection between the left cosets \(G/G_x\) and \(G\cdot x\).
Prove that the center of every finite \(p\dash\)group \(G\) is nontrivial by considering that action of \(G\) on \(X=G\) by conjugation.
Solution
For (a), define \[ \Phi:G/G_x\longrightarrow G\cdot x, \qquad \Phi(gG_x)=g\cdot x. \] If \(gG_x=hG_x\), then \(h^{-1}g\in G_x\), so \((h^{-1}g)\cdot x=x\) and therefore \(g\cdot x=h\cdot x\). Thus \(\Phi\) is well defined. It is surjective by the definition of the orbit. If \(g\cdot x=h\cdot x\), then \(h^{-1}g\in G_x\), so \(gG_x=hG_x\); hence \(\Phi\) is injective. Therefore it is a bijection, and in particular \[ |G\cdot x|=[G:G_x]. \]
For (b), let the finite \(p\)-group \(G\) act on itself by conjugation. The fixed points are exactly the elements of \(Z(G)\). For any \(x\in G\), the stabilizer is the centralizer \(C_G(x)\), so by part (a) the conjugacy class of \(x\) has size \[ [G:C_G(x)], \] a power of \(p\). Hence every non-singleton conjugacy class has size divisible by \(p\). The class equation therefore gives \[ |G|=|Z(G)|+\sum_i |\mathcal C_i|, \] where every term in the sum is divisible by \(p\). Since \(p\mid |G|\), it follows that \(p\mid |Z(G)|\). The identity lies in \(Z(G)\), so \(|Z(G)|\ge p\), and in particular \(Z(G)\) is nontrivial.