Let \(G\) be a finite group. For any \(x \in G\) \[Z_G(x) = \{g \in G : gxg^{-1} = x\}\] is the centralizer of \(x\) in \(G\) and \[x^G = \{gxg^{-1} : g \in G\}\] is the conjugacy class of \(x\) in \(G\).
Show that \(|x^G| = [G : Z_G(x)]\).
If \(H \le G\) and \(x \in H\), prove that \(Z_H(x) = H \cap Z_G(x)\).
If \(H\) is a subgroup of index 2 in \(G\) and \(x \in H\), prove that either \(|x^H| = |x^G|\) or \(|x^H| = \frac{1}{2}|x^G|\).
Solution
1. In part (a), \(|x^G|=[G:Z_G(x)]\).
The map \(\Phi\colon G/Z_G(x)\to x^G\), \(gZ_G(x)\mapsto gxg^{-1}\), is well defined and injective, because \[ g_1Z_G(x)=g_2Z_G(x)\iff g_2^{-1}g_1\in Z_G(x)\iff (g_2^{-1}g_1)x(g_2^{-1}g_1)^{-1}=x\iff g_1xg_1^{-1}=g_2xg_2^{-1}. \] It is surjective by the definition of \(x^G\). Hence \(|x^G|=\abs{G/Z_G(x)}=[G:Z_G(x)]\).
2. In part (b), \(Z_H(x)=H\cap Z_G(x)\).
\(Z_H(x)=\{h\in H: hxh^{-1}=x\}=\{g\in G: g\in H\text{ and }gxg^{-1}=x\}=H\cap Z_G(x)\).
Assume for part (c) that \([G:H]=2\) and \(x\in H\).
3. \(2\,|x^H|=|x^G|\,[Z_G(x):Z_H(x)]\).
By step 1 applied in \(G\) and in \(H\), \(|x^G|=[G:Z_G(x)]\) and \(|x^H|=[H:Z_H(x)]\). Since \(Z_H(x)\le H\le G\) and \(Z_H(x)\le Z_G(x)\le G\) by step 2, multiplicativity of indices gives \[ [G:Z_H(x)]=[G:H]\,[H:Z_H(x)]=2\,|x^H|,\qquad [G:Z_H(x)]=[G:Z_G(x)]\,[Z_G(x):Z_H(x)]=|x^G|\,[Z_G(x):Z_H(x)]. \]
4. \([Z_G(x):Z_H(x)]\in\{1,2\}\).
A subgroup of index \(2\) is normal, so \(HZ_G(x)\) is a subgroup of \(G\) containing \(H\). By step 2 and the second isomorphism theorem, \([Z_G(x):Z_H(x)]=[Z_G(x):H\cap Z_G(x)]=[HZ_G(x):H]\), which divides \([G:H]=2\).
5. Q.E.D.
By steps 3 and 4, if \([Z_G(x):Z_H(x)]=2\) then \(|x^H|=|x^G|\), and if \([Z_G(x):Z_H(x)]=1\) then \(|x^H|=\tfrac12|x^G|\). Steps 1 and 2 answer parts (a) and (b).