Let \(G\) be a finite group with \(n\) distinct conjugacy classes, and let \(g_1, g_2, \dots, g_n\) be a set of representatives for the conjugacy classes of \(G\).
Prove that if \(g_i g_j = g_j g_i\) for all \(1 \le i, j \le n\), then \(G\) is abelian.
Solution
Goal: Prove that \(G\) is abelian by considering the subgroup \(H = \langle g_1, \dots, g_n \rangle\) and applying the conjugate cover theorem for finite groups.
1. Subgroup \(H = \langle g_1, \dots, g_n \rangle\) generated by representatives:
1.1. Define \(H = \langle g_1, \dots, g_n \rangle \le G\) to be the subgroup of \(G\) generated by the \(n\) conjugacy class representatives.
1.2. By hypothesis, \(g_i g_j = g_j g_i\) for all \(1 \le i, j \le n\), so all generators of \(H\) pairwise commute.
1.3. Since any group generated by pairwise commuting elements is abelian, \(H\) is an abelian subgroup of \(G\).
2. \(G\) is the union of conjugates of \(H\):
2.1. Let \(x \in G\) be an arbitrary element.
2.2. Since \(g_1, \dots, g_n\) represent all conjugacy classes of \(G\), the conjugacy classes partition \(G\):\[G = \bigsqcup_{i=1}^n \operatorname{Cl}(g_i).\]
2.3. Thus \(x \in \operatorname{Cl}(g_k)\) for some \(k \in \{1, \dots, n\}\).
2.4. By definition of conjugacy class, there exists an element \(g \in G\) such that \(x = g g_k g^{-1}\).
2.5. Since \(g_k \in H\), \(x = g g_k g^{-1} \in g H g^{-1}\).
2.6. Therefore:\[G \subseteq \bigcup_{g \in G} g H g^{-1} \subseteq G \implies G = \bigcup_{g \in G} g H g^{-1}.\]
3. Lemma: No finite group is a union of conjugates of a proper subgroup.
3.1. Let \(K < G\) be a proper subgroup with \([G : K] = m > 1\).
3.2. The number of distinct conjugate subgroups of \(K\) in \(G\) is \([G : N_G(K)] \le [G : K] = m\).
3.3. Each conjugate \(g K g^{-1}\) contains \(|K|\) elements and shares the identity element \(e\) with all other conjugates.
3.4. The union of all conjugates contains at most\[\left| \bigcup_{g \in G} g K g^{-1} \right| \le 1 + [G : N_G(K)] (|K| - 1) \le 1 + [G : K] (|K| - 1) = 1 + |G| - [G : K].\]
3.5. Since \([G : K] \ge 2\), \(1 + |G| - [G : K] \le |G| - 1 < |G|\).
3.6. Thus \(\bigcup_{g \in G} g K g^{-1} \subsetneq G\).
4. \(H = G\) and \(G\) is abelian:
4.1. From 2, \(G = \bigcup_{g \in G} g H g^{-1}\).
4.2. By the contrapositive of 3, \(H\) cannot be a proper subgroup of \(G\).
4.3. Thus \(H = G\).
4.4. From 1, \(H\) is abelian, so \(G\) is abelian.
5. Conclusion:
$G$ is abelian.