Bourbaki

Exercise 37, § 7

Algebra, I, 170

Bourbaki marks this one as harder than the rest.

Let GG be a group and SS a finite symmetric subset of GG generating GG. Let L\mathfrak{L} denote the set of finite subsets of GG, ordered by inclusion; it is a directed set.

(a) Let T∈LT \in \mathfrak{L} and let BTB_T be the set of connected SS-components of G−TG - T (cf. Exercise 18); let BT∞=BT−(L∩BT)B_T^\infty = B_T - (\mathfrak{L} \cap B_T). Show that BTB_T is finite (note that, if T≠∅T \neq \varnothing and Z∈BTZ \in B_T, there exists t∈Tt \in T and z∈Zz \in Z which are SS-neighbours).

(b) Let T,T′∈LT, T' \in \mathfrak{L} with T⊂T′T \subset T'. If Z′∈BT′Z' \in B_{T'}, show that there exists a unique element Z∈BTZ \in B_T such that Z′⊂ZZ' \subset Z; the mapping fTT′:BT′→BTf_{TT'} : B_{T'} \to B_T such that fTT′(Z′)=Zf_{TT'}(Z') = Z maps BT′∞B_{T'}^\infty onto BT∞B_T^\infty. The inverse limit of the BTB_T, with TT running through L\mathfrak{L}, is denoted by BB. Show that B=lim⁡←BT∞B = \lim \leftarrow B_T^\infty and that the image of BB in BTB_T is BT∞B_T^\infty. The set BB is called the set of ends of GG; it is non-empty if and only if GG is infinite.

(c) Let TT be a non-empty SS-connected finite subset of GG and let Z∈BT∞Z \in B_T^\infty. Show that there exist g∈Gg \in G such that gT∩T=∅gT \cap T = \varnothing and gT∩Z≠∅gT \cap Z \neq \varnothing and that then gT⊂ZgT \subset Z. Show that there exists Z1∈BTZ_1 \in B_T such that gZ1gZ_1 contains TT and all the Z′∈BTZ' \in B_T such that Z′≠ZZ' \neq Z (observe that T∪⋃Z′≠ZZ′T \cup \bigcup_{Z' \neq Z} Z' is SS-connected and does not meet gTgT); deduce that, if Z′∈BTZ' \in B_T is different from Z1Z_1, then gZ′⊂ZgZ' \subset Z. Show that, writing T′=T∪gTT' = T \cup gT, the inverse image of ZZ in BT′∞B_{T'}^\infty consists of at least n−1n - 1 elements, where n=Card(BT∞)n = \mathrm{Card}(B_T^\infty).

(d) Let S′S' be a symmetric finite subset of GG generating GG and B′B' the set of ends of GG relative to S′S'. Define a bijection of BB onto B′B' (reduce it to the case where S⊂S′S \subset S'). The cardinal of BB is thus independent of the choice of SS; it is called the number of ends of GG.

(e) Show, using (c), that the number of ends of GG is equal to 0, 1, 2 or 2k02^{k_0}.

(f) Show that the number of ends of Z\mathbf{Z} is 2 and that that of Zn\mathbf{Z}^n (n≥2n \geq 2) is 1.

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.