Exercise 18, § 7

Let S be a symmetric subset of a group G. Two elements g1, g2 of G are said to be S-neighbours if g2−1g1 belongs to S. A sequence (g1, ..., gn) of elements of G is called an S-chain if gi and gi+1 are S-neighbours for 1 ≤ i ≤ n − 1: the elements g1 and gn are called respectively the origin and the extremity of the chain. a) Let Y be a subset of G, and let R_Y\{a, b\} be the relation “there exists an S-chain contained in Y with origin a and extremity b”. Now prove that R_Y is an equivalence relation in Y (cf. E, II, p. 52, exerc. 10); the equivalence classes for this relation are called the S-connected components of Y. One says that Y is S-connected if it has at most one connected component. b) Now prove that G is S-connected if and only if S generates G. c) Suppose that no element s ∈ S satisfies the relation s² = e; choose a subset X of S such that S is the disjoint union of X and X⁻¹. Now prove the equivalence of the following conditions: (i) The family X is free. (ii) There does not exist an S-chain (g₁, ..., gₙ) in G, formed of distinct elements, in number n ≥ 3, such that g₁ and gₙ are S-neighbours. (Use exerc. 13 of I, p. 145.)