Bourbaki

Exercise 4, § 7

Algebra, I, p. 143

Bourbaki marks this one as harder than the rest.

Let XX be a set and cc an element not belonging to XX. Put Y=X∪{c}Y = X \cup \{c\}. Let MM be the magma having Mo(Y)Mo(Y) as underlying set, the composition law being given by (u,v)↦cuv(u, v) \mapsto cuv. Let LL be the submagma of MM generated by XX, and let ε\varepsilon be the mapping of YY into Z\mathbf{Z} equal to −1-1 on XX and to 11 at cc. a) Prove that LL is the subset of MM formed by the words w=y1…ypw = y_1 \ldots y_p, where yi∈Yy_i \in Y, satisfying the following two conditions: (i) ε(y1)+⋯+ε(yp)=−1\varepsilon(y_1) + \cdots + \varepsilon(y_p) = -1. (ii) ε(y1)+⋯+ε(yi)≥0\varepsilon(y_1) + \cdots + \varepsilon(y_i) \geq 0 for 1≤i≤p−11 \leq i \leq p - 1. Prove that every element of L−XL - X is written uniquely in the form cuvcuv, with u,vu, v in LL. (The elements ww satisfying (i) and (ii) form a submagma L′L' of MM containing XX. If w=y1…ypw = y_1 \ldots y_p belongs to L′L', let kk be the smallest integer such that ε(y1)+⋯+ε(yk)=0\varepsilon(y_1) + \cdots + \varepsilon(y_k) = 0; prove that one has y1=cy_1 = c and that the elements u=y2…yku = y_2 \ldots y_k and v=yk+1…ypv = y_{k+1} \ldots y_p belong to L′L'; one has w=cuvw = cuv, whence by induction on pp the relation w∈Lw \in L. Then prove that the relations u′∈L,v′∈Lu' \in L, v' \in L and w=cu′v′w = cu'v' entail u′=u,v′=v.u' = u, v' = v.) b) Prove that the injection of XX into LL extends to an isomorphism of M(X)M(X) onto LL.

Solution

No solution has been written yet.

Machine translation of the English, by gpt-5-6-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.