Suppose the group \(G\) acts on a set \(X\). Show that the stabilizers of elements in the same orbit are conjugate: for any \(x \in X\) and \(g \in G\), \(G_{g \cdot x} = g G_x g^{-1}\).
Let \(G\) be a finite group and let \(H < G\) be a proper subgroup. Show that the union of all conjugates of \(H\) is strictly smaller than \(G\): \[ \bigcup_{g \in G} g H g^{-1} \subsetneq G. \]
Suppose a finite group \(G\) acts transitively on a set \(S\) with \(|S| \ge 2\). Show that there exists an element \(g \in G\) having no fixed points in \(S\) (i.e. \(g \cdot s \ne s\) for all \(s \in S\)).
Solution
Goal: Prove conjugacy of stabilizers in (a), prove that conjugates of a proper subgroup do not cover \(G\) in (b), and deduce the existence of a fixed-point-free element (derangement) for transitive actions in (c).
1. Part (a): Conjugacy of stabilizers in an orbit.
1.1. Let \(x \in X\) and \(g \in G\), and let \(y = g \cdot x \in X\).
1.2. An element \(h \in G\) stabilizes \(y\) if and only if \(h \cdot y = y\).
1.3. Substituting \(y = g \cdot x\):\[h \cdot (g \cdot x) = g \cdot x \iff (g^{-1} h g) \cdot x = x \iff g^{-1} h g \in G_x.\]
1.4. Multiplying on the left by \(g\) and on the right by \(g^{-1}\):\[g^{-1} h g \in G_x \iff h \in g G_x g^{-1}.\]
1.5. Therefore \(G_y = G_{g \cdot x} = g G_x g^{-1}\).
2. Part (b): Conjugates of a proper subgroup do not cover \(G\).
2.1. Let \(N_G(H) = \{g \in G : g H g^{-1} = H\}\) be the normalizer of \(H\) in \(G\).
2.2. The distinct conjugates of \(H\) in \(G\) are parameterized by the cosets of \(N_G(H)\), so there are precisely \(k = [G : N_G(H)]\) distinct conjugates \(H_1, H_2, \dots, H_k\).
2.3. Since \(H \le N_G(H) \le G\), by the tower law of indices \([G : H] = [G : N_G(H)] [N_G(H) : H] \ge [G : N_G(H)] = k\).
2.4. Every conjugate \(g H g^{-1}\) contains the identity element \(e\).
2.5. Each of the \(k\) distinct conjugates contains \(|H| - 1\) non-identity elements.
2.6. Bound the size of the union of all conjugates:\[\left| \bigcup_{g \in G} g H g^{-1} \right| = \left| \{e\} \cup \bigcup_{i=1}^k (H_i \setminus \{e\}) \right| \le 1 + \sum_{i=1}^k (|H_i| - 1) = 1 + k (|H| - 1).\]
2.7. Since \(k \le [G : H]\), we have\[\left| \bigcup_{g \in G} g H g^{-1} \right| \le 1 + [G : H](|H| - 1) = 1 + [G : H]|H| - [G : H] = |G| + 1 - [G : H].\]
2.8. Since \(H\) is a proper subgroup (\(H \ne G\)), the index \([G : H] \ge 2\).
2.9. Thus:\[\left| \bigcup_{g \in G} g H g^{-1} \right| \le |G| + 1 - [G : H] \le |G| - 1 < |G|.\]
2.10. Therefore \(\bigcup_{g \in G} g H g^{-1} \subsetneq G\).
3. Part (c): Transitive actions on \(|S| \ge 2\) have a fixed-point-free element.
3.1. Method 1 (via Part (b)):
3.1.1. Choose an arbitrary point \(s_0 \in S\), and let \(H = G_{s_0} \le G\) be its stabilizer.
3.1.2. By the Orbit-Stabilizer Theorem, since the action is transitive, \([G : H] = |G \cdot s_0| = |S| \ge 2\).
3.1.3. Thus \(H\) is a proper subgroup of \(G\).
3.1.4. By Part (a), every stabilizer is of the form \(G_{g \cdot s_0} = g H g^{-1}\).
3.1.5. An element \(g \in G\) fixes at least one point \(s \in S\) if and only if \(g \in G_s = h H h^{-1}\) for some \(h \in G\).
3.1.6. Thus the set of elements in \(G\) that fix at least one point of \(S\) is precisely \(\bigcup_{h \in G} h H h^{-1}\).
3.1.7. By Part (b), \(\bigcup_{h \in G} h H h^{-1} \subsetneq G\).
3.1.8. Therefore there exists an element \(g \in G \setminus \bigcup_{h \in G} h H h^{-1}\), which fixes no points of \(S\).
3.2. Method 2 (via Burnside’s Lemma):
3.2.1. Let \(X^g = \{s \in S : g \cdot s = s\}\) denote the fixed point set of \(g \in G\).
3.2.2. By Burnside’s Lemma, the number of orbits is the average number of fixed points:\[1 = \frac{1}{|G|} \sum_{g \in G} |X^g| \implies |G| = \sum_{g \in G} |X^g| = |X^e| + \sum_{g \ne e} |X^g|.\]
3.2.3. Since the identity fixes all of \(S\), \(|X^e| = |S| \ge 2\).
3.2.4. If every \(g \ne e\) had at least one fixed point (\(|X^g| \ge 1\)), then\[\sum_{g \in G} |X^g| = |X^e| + \sum_{g \ne e} |X^g| \ge 2 + (|G| - 1) = |G| + 1,\]a contradiction.
3.2.5. Thus there exists at least one \(g \in G\) with \(|X^g| = 0\).
4. Conclusion:
Stabilizers in an orbit are conjugate, conjugates of a proper subgroup cannot cover a finite group, and transitive actions on sets of size $\ge 2$ always contain derangements.