Let \(G\) be a finite group.
Prove that if \(H < G\) is a proper subgroup, then \(G\) is not the union of conjugates of \(H\).
Suppose that \(G\) acts transitively on a set \(X\) with \(|X| > 1\). Prove that there exists an element of \(G\) with no fixed points in \(X\).
Solution
1. (1) \(G \neq \bigcup_{g \in G} gHg^{-1}\) for \(H < G\) proper.
1.1. The number of distinct conjugates of \(H\) is \([G : N_G(H)] \le [G : H]\).
The conjugates of \(H\) are indexed by \(G/N_G(H)\), and \(N_G(H) \supseteq H\), so \([G : N_G(H)] \le [G : H]\).
1.2. Each conjugate has \(|H|\) elements, and all contain the identity.
\(|gHg^{-1}| = |H|\), and \(1 \in gHg^{-1}\) for all \(g\).
1.3. Hence \(\abs{\bigcup_g gHg^{-1}} \le 1 + [G:H](|H| - 1) = 1 + |G| - [G:H] < |G|\).
The union has at most \(1 + (\text{number of conjugates})(|H| - 1)\) elements (counting the identity once), and \([G:H] > 1\) since \(H\) is proper.
1.4. Hence the union is a proper subset of \(G\).
It has fewer than \(|G|\) elements.
2. (2) A transitive action on \(|X| > 1\) has a fixed-point-free element.
2.1. Suppose every \(g \in G\) fixes some point of \(X\).
Assume this for contradiction.
2.2. Then \(G = \bigcup_{x \in X} G_x\), where \(G_x\) is the stabilizer of \(x\).
Every element fixes some point, so every element lies in some stabilizer.
2.3. The stabilizers \(G_x\) are all conjugate (since the action is transitive).
\(G_{gx} = g G_x g^{-1}\).
2.4. Hence \(G\) is a union of conjugates of the proper subgroup \(G_x\) (proper since \(|X| > 1\) and the action is transitive).
\(G_x\) is proper because the orbit of \(x\) is all of \(X\) with \(|X| > 1\), so \(G_x \neq G\).
2.5. This contradicts
1.
1 says \(G\) is not a union of conjugates of a proper subgroup.
3. Q.E.D.
1 proves (1); 2 proves (2).