Bourbaki

Exercise 19, § 7

Algebra, I, 164

Bourbaki marks this one as harder than the rest.

Let X be a set, G=F(X)G = F(X) the free group constructed on X and S=X∪X−1S = X \cup X^{-1}.

(a) If g∈Gg \in G, every sequence (s1,…,sn)(s_1, \ldots, s_n) of elements of S such that g=s1…sng = s_1 \ldots s_n and sisi+1≠es_i s_{i+1} \neq e for 1≤i<n1 \leq i < n is called a reduced decomposition of g. Show that every element of G admits one and only one reduced decomposition; the integer n is the length of g (cf. Exercise 26).

(b) Let Y be a subset of G containing e. Show the equivalence of the following conditions: (b1)(b_1) Y is S-connected (cf. Exercise 18). (b2)(b_2) For all g∈Yg \in Y, if (s1,…,sn)(s_1, \ldots, s_n) is the reduced decomposition of g, then s1…si∈Ys_1 \ldots s_i \in Y for 1≤i≤n1 \leq i \leq n. (c) Let Y1Y_1 and Y2Y_2 be two non-empty S-connected subsets of G such that Y1∩Y2=∅Y_1 \cap Y_2 = \varnothing. Show that there exists at most one ordered pair

(y1,y2)∈Y1×Y2 (y_1, y_2) \in Y_1 \times Y_2
such that y1y_1 and y2y_2 are S-neighbours. There exists one if and only if Y1∪Y2Y_1 \cup Y_2 is S-connected; Y1Y_1 and Y2Y_2 are then said to be neighbours.

(d) Let Y1,…,YnY_1, \ldots, Y_n be a sequence of disjoint non-empty S-connected subsets of G. Suppose that YiY_i and Yi+1Y_{i+1} are neighbours for 1≤i<n1 \leq i < n. Show that, if n≥3n \geq 3, Y1Y_1 and YnY_n are not neighbours.

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.