Bourbaki

Exercise 38, § 7

Algebra, I, p. 150

Bourbaki marks this one as harder than the rest.

Let XX be a finite set, F(X)F(X) the free group constructed on XX, and S=X∪X−1S = X \cup X^{-1}. a) Let SnS_n be the set of products s1…sms_1 \ldots s_m for si∈S,m≤ns_i \in S, m \leq n. Now prove that every SS-connected component (cf. I, p. 146, exer. 18) of F(X)−SnF(X) - S_n contains one and only one element of the form s1…sn+1s_1 \ldots s_{n+1}, with si∈Ss_i \in S and sisi+1≠es_i s_{i+1} \neq e for 1≤i≤n1 \leq i \leq n. b) Deduce from a) a bijection of the set BB of ends of F(X)F(X) (cf. exer. 37) onto the set of infinite sequences (s1,…,sn,…)(s_1, \ldots, s_n, \ldots) of elements of SS such that sisi+1≠es_i s_{i+1} \neq e for every ii. In particular, the number of ends of F(X)F(X) is equal to 2N02^{N_0} if Card(X)≥2\mathrm{Card}(X) \geq 2.

EXERCISES

Solution

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Machine translation of the English, by gpt-5-6-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.