Bourbaki marks this one as harder than the rest.
Let E be a set and f a mapping of Em into E; we write
f(x1,x2,…,xm)=x1x2…xm;
suppose that f satisfies the following conditions:
(1)
(x1x2…xm)xm+1…x2m−1=x1(x2x3…xm+1)xm+2…x2m−1 identically; (2) For all
a1,a2,…,am−1, the mappings
x↦xa1a2…am−1
x↦a1a2…am−1x
are bijections of E onto E.
(a) Show that identically
(x1x2…xm)xm+1…x2m−1=x1x2…xi(xi+1…xi+m)xi+m+1…x2m−1
for every index i such that 1≤i≤m−1 (argue by induction on i, by considering the element
((x1x2…xm)xm+1…x2m−1)a1a2…am−1).
(b) For all a1,a2,…,am−2, there exists u∈E such that for all x
x=xa1a2…am−2u=ua1a2…am−2x.
(c) In the set Ek of sequences (u1,u2,…,uk) of k elements of E (1≤k≤m−1) consider the equivalence relation Rk which states: for all x1,x2,…,xm−k,u1u2…ukx1x2…xm−k=v1v2…vkx1x2…xm−k; let Ek denote the quotient set Ek/Rk and G the sum set (Set Theory, II, § 4, no. 8) of E1=E,E2,…,Em−1. Let α∈Ei,β∈Ej; if (u1,u2,…,ui) is a sequence of class α, (v1,…,vj) a sequence of class β, consider the sequence
(u1,u2,…,ui,v1,…,vj) in
Ei+j if
i+j<m, the sequence
(u1,u2,…,ui+j−m,(ui+j−m+1…uiv1v2…vj)) in
Ei+j−m+1 if
i+j≥m; show that the class of this sequence in
Ei+j (resp.
Ei+j−m+1) depends only on the classes
α and
β; it is denoted by
α.β. Show that a group law is thus defined on
G, that
H=Em−1 is a normal subgroup of
G and that the quotient group
G/H is cyclic of order
m−1; prove finally that
E is identical with a class (mod.
H) which generates
G/H and that
x1x2…xm is just the composition, in the group
G, of the sequence
(x1,x2,…,xm).