Bourbaki

Exercise 7, § 7

Algebra, I, p. 144

Bourbaki marks this one as harder than the rest.

Let MxM_x be the free magma constructed on a set {x}\{x\} with one element. (a) For every y∈Mxy \in M_x, now prove that there exists a unique endomorphism fyf_y of MxM_x such that fy(x)=yf_y(x) = y; it is an isomorphism of MxM_x onto the submagma MyM_y generated by y (cf. exerc. 6). (b) If y∈Mx,z∈Mxy \in M_x, z \in M_x, put y∘z=fy(z)y \circ z = f_y(z). Now prove that the law of composition (y,z)↦y∘z(y, z) \mapsto y \circ z makes MxM_x a monoid, with identity element x, isomorphic to the monoid End(MxM_x) of endomorphisms of MxM_x. (c) Let N be the set of monogenic submagmas of MxM_x distinct from MxM_x. An element y of MxM_x is said to be primitive if MyM_y is a maximal element of N. Now prove that xx,x(x(xx))xx, x(x(xx)) are primitive whereas x,(xx)(xx)x, (xx)(xx) are not. (d) Let P be the set of primitive elements of MxM_x. Now prove that, if y∈P,z∈P,y≠zy \in P, z \in P, y \neq z, one has My∩Mz=∅M_y \cap M_z = \varnothing (use exerc. 6). (e) Let Mo(P) be the free monoid constructed on P. If one endows MxM_x with the monoid structure defined in b), the injection P→MxP \to M_x extends to a homomorphism p:Mo(P)→Mxp : Mo(P) \to M_x. Now prove that pp is an isomorphism. (If z∈Mxz \in M_x, one will prove by induction on l(z)l(z) that one has z∈Im⁡(p)z \in \operatorname{Im}(p). On the other hand, if y1,…,yn,z1,…,zmy_1, \ldots, y_n, z_1, \ldots, z_m are elements of P such that p(y1…yn)=p(z1…zm)p(y_1 \ldots y_n) = p(z_1 \ldots z_m), one has My1∩Mz1≠∅M_{y_1} \cap M_{z_1} \neq \varnothing, whence y1=z1y_1 = z_1 and fy1(p(y2…yn))=fy2(p(z2…zm))f_{y_1}(p(y_2 \ldots y_n)) = f_{y_2}(p(z_2 \ldots z_m)); whence p(y2…yn)=p(z2…zm)p(y_2 \ldots y_n) = p(z_2 \ldots z_m), and one deduces that y1…yn=z1…zmy_1 \ldots y_n = z_1 \ldots z_m by reasoning by induction on sup⁡(n,m)\sup(n, m)).

Solution

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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.