Bourbaki

Exercise 22, § 7

Algebra, I

(a) Let (yn)n∈Z(y_n)_{n \in \mathbf{Z}} be a basic family of a group FF. Let φ\varphi be the automorphism of FF which transforms yny_n into yn+1y_{n+1}; this automorphism defines an operation τ\tau of Z\mathbf{Z} on FF; let E=F×τZE = F \times_{\tau} \mathbf{Z} be the corresponding semidirect product. Now prove that the elements x=(e,1)x = (e, 1) and y=(y0,0)y = (y_0, 0) form a basic family of EE. (Let Fx,yF_{x, y} be the free group constructed on {x,y}\{x, y\}, and let ff be the canonical homomorphism of Fx,yF_{x, y} into EE. One will show that there exists a homomorphism g:E→Fx,yg : E \to F_{x, y} such that g((0,1))=xg((0, 1)) = x and g((yn,0))=xnyx−ng((y_n, 0)) = x^n y x^{-n} and that ff and gg are reciprocal to one another.) (b) Deduce that the normal subgroup of Fx,yF_{x, y} generated by yy is a free group with basic family (xnyx−n)n∈Z(x^n y x^{-n})_{n \in \mathbf{Z}}. Recover this result by applying exerc. 20 to the system of representatives formed by the xnx^n for n∈Zn \in \mathbf{Z}. (c) Extend what precedes to free groups of arbitrary rank.

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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.