Let \(G\) be a finite group and let \(H \le G\) be a subgroup. Prove that if \(H\) meets every conjugacy class of \(G\) (i.e. \(\bigcup_{g \in G} g H g^{-1} = G\)), then \(H = G\) (Jordan’s Theorem on permutation groups).
Solution
Suppose \(H<G\) is proper and put \(n=[G:H]\ge2\). Let \(k\) be the number of distinct conjugates of \(H\). Since \[ k=[G:N_G(H)]\le [G:H]=n, \] and every conjugate contains the identity, \[ \left|\bigcup_{g\in G}gHg^{-1}\right| \le 1+k(|H|-1) \le 1+n(|H|-1). \] Using \(n|H|=|G|\), \[ 1+n(|H|-1)=|G|-(n-1)<|G|. \] Therefore the union of the conjugates of a proper subgroup cannot be all of \(G\).
Hence, if \[ \bigcup_{g\in G}gHg^{-1}=G, \] then \(H\) cannot be proper, so \(H=G\). Equivalently, a proper subgroup of a finite group misses at least one conjugacy class.