Exercise 37, § 7
Algebra, I, 170
Bourbaki marks this one as harder than the rest.
Let be a group and a finite symmetric subset of generating . Let denote the set of finite subsets of , ordered by inclusion; it is a directed set.
(a) Let and let be the set of connected -components of (cf. Exercise 18); let . Show that is finite (note that, if and , there exists and which are -neighbours).
(b) Let with . If , show that there exists a unique element such that ; the mapping such that maps onto . The inverse limit of the , with running through , is denoted by . Show that and that the image of in is . The set is called the set of ends of ; it is non-empty if and only if is infinite.
(c) Let be a non-empty -connected finite subset of and let . Show that there exist such that and and that then . Show that there exists such that contains and all the such that (observe that is -connected and does not meet ); deduce that, if is different from , then . Show that, writing , the inverse image of in consists of at least elements, where .
(d) Let be a symmetric finite subset of generating and the set of ends of relative to . Define a bijection of onto (reduce it to the case where ). The cardinal of is thus independent of the choice of ; it is called the number of ends of .
(e) Show, using (c), that the number of ends of is equal to 0, 1, 2 or .
(f) Show that the number of ends of is 2 and that that of () is 1.
Solution
No solution has been written yet.