Let \(G\) be a finite group and \(H\) a proper subgroup. Show that \(G\) is not the set-theoretic union of the conjugates of \(H\).
Solution
1. The number of distinct conjugates of \(H\) is at most \([G:H]\).
The conjugation action of \(G\) on its subgroups has stabilizer \[ N_G(H)=\{g\in G:gHg^{-1}=H\} \] at \(H\). Hence the number \(r\) of distinct conjugates of \(H\) is \[ r=[G:N_G(H)]. \] Since \(H\le N_G(H)\), \[ r=[G:N_G(H)]\le [G:H]. \]
2. The union of those conjugates has strictly fewer than \(|G|\) elements.
List the distinct conjugates as \(H_1,\ldots,H_r\). Each has \(|H|\) elements, and every one contains the identity. Therefore \[ \left|\bigcup_{i=1}^r H_i\right| \le 1+r(|H|-1). \] Put \(m=[G:H]\). Because \(H\) is proper, \(m>1\), and 1 gives \(r\le m\). Thus \[ \left|\bigcup_{i=1}^r H_i\right| \le 1+m(|H|-1) =m|H|-(m-1) =|G|-(m-1) <|G|. \] Consequently the conjugates of \(H\) cannot cover \(G\).