Let \(G\) be a finite group and let \(H\subsetneq G\) be a proper subgroup.
Show that the number of subgroups of \(G\) of the form \(xHx^{-1}\), for \(x\in G\), is at most the index \([G:H]\).
Prove that some element of \(G\) is not contained in any subgroup of the form \(xHx^{-1}\), \(x\in G\).
Solution
Let \[ N_G(H)=\{g\in G:gHg^{-1}=H\} \] be the normalizer of \(H\).
1. The number of conjugates of \(H\) is at most \([G:H]\).
The group \(G\) acts by conjugation on its set of subgroups. The stabilizer of \(H\) is exactly \(N_G(H)\), so the orbit-stabilizer theorem gives \[ \#\{xHx^{-1}:x\in G\}=[G:N_G(H)]. \] Since every \(h\in H\) normalizes \(H\), one has \[ H\le N_G(H). \] Therefore \[ [G:N_G(H)]\le [G:H]. \]
2. The union of all conjugates of \(H\) is a proper subset of \(G\).
Let \[ m=\#\{xHx^{-1}:x\in G\}. \] By step 1, \[ m\le [G:H]. \] Every conjugate \(xHx^{-1}\) has exactly \(\abs{H}\) elements, and every one contains the identity element \(e\). Each conjugate therefore contributes at most \(\abs{H}-1\) elements other than \(e\) to the union, and \[ \abs{\bigcup_{x\in G}xHx^{-1}} \le 1+m(\abs{H}-1) \le 1+[G:H](\abs{H}-1). \] Since \(H\) is proper, \([G:H]>1\). Using \(\abs{G}=[G:H]\abs{H}\), \[ 1+[G:H](\abs{H}-1) =\abs{G}-[G:H]+1 <\abs{G}. \]
3. Q.E.D.
Step 1 proves part 1. By step 2, some element of \(G\) lies in no subgroup \(xHx^{-1}\), which proves part 2.