Bourbaki

Exercise 3, Appendix 0

Tag 03GZ, permanent. Cite it as [Bourbaki, Tag 03GZ].

Theory of Sets, I, 60

Let E be a set and let ff be a mapping of E×EE \times E into E ("internal law of composition"). Let S be a set of signs which is the disjoint union of E and the set consisting of the single element ff. Put n(f)=1n(f) = 1, and n(x)=0n(x) = 0 for all x∈Ex \in E.

(a) Let M be the set of significant words of L0(S)L_0(S). Show that there is a unique mapping vv of M into E satisfying the following conditions : (1) v(x)=xv(x) = x for all x∈Ex \in E; (2) if AA and BB are two significant words, then v(fAB)=f(v(A),v(B))v(fAB) = f(v(A), v(B)).

(b) For each word A=(si)0⩽i⩽nA = (s_i)_{0 \leqslant i \leqslant n} in L0(S)L_0(S) let A∗A^* be the word (sik)(s_{i_k}), where the iki_k are the indices ii such that si≠fs_i \neq f, arranged in ascending order. Two words AA, BB of L0(S)L_0(S) are said to be similar if A∗=B∗A^* = B^*. Show that if the law of composition ff is associative, i.e., if

f(f(x,y),z)=f(x,f(y,z)),f(f(x, y), z) = f(x, f(y, z)),

then v(A)=v(B)v(A) = v(B) whenever AA and BB are similar significant words ("general theorem of associativity"). (A significant word

A=(si)0⩽i⩽2nA = (s_i)_{0 \leqslant i \leqslant 2n}

is said to be normal if si=fs_i = f for i=0,2,4,…,2n−2i = 0, 2, 4, \ldots, 2n - 2, and si≠fs_i \neq f for the other indices. Show that every significant word is similar to a unique normal word A′A', and prove that v(A)=v(A′)v(A) = v(A') by induction on the length of AA.)

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