Bourbaki

Exercise 13, § 10

Algebra, II, p. 207

Bourbaki marks this one as harder than the rest.

Let EE be a right vector space of dimension ⩾2\geqslant 2, H0H_0 a hyperplane of E.

a) Now prove that every automorphism u belonging to F(E) (Exercise 12) is the product of an automorphism of SL(E)\mathbf{SL}(E) and possibly a hyperplane dilatation H0H_0 (proceed by induction on the codimension of the kernel of 1E−u1_E - u, using Exercises 12 a) and 12 f)).

b) Now prove that SL(E)\mathbf{SL}(E) contains the group of commutators of F(E) and is identical with this group unless E is a space of dimension 2 over the field with 2 elements. (To show that SL(E)\mathbf{SL}(E) contains the group of commutators of F(E)F(E), use a) and Exercise 12 ff). To see that SL(E)\mathbf{SL}(E) is contained in the group of commutators of F(E)F(E) except in the indicated exceptional case, show that for every homomorphism of F(E)F(E) into a commutative group, the image of a transvection is the identity element, using Exercises 11 b) (II, p. 206) and 12 c).

c) Now prove that SL(E)\mathbf{SL}(E) is equal to its group of commutators unless dim⁡E=2\dim E = 2 and that KK has 2 or 3 elements. (If dim⁡E⩾3\dim E \geqslant 3, remark that every transvection uu can be written vwv−1w−1v w v^{-1} w^{-1}, where vv is a transvection and w∈SL(E)w \in \mathbf{SL}(E), using Exercise 12 d). If dim⁡E=2\dim E = 2, remark first that every automorphism uu whose matrix with respect to a base of EE is of the form (λ00λ−1)\begin{pmatrix} \lambda & 0 \\ 0 & \lambda^{-1} \end{pmatrix} is a product of transvections by proceeding as in a); then consider the commutator uvu−1v−1uvu^{-1}v^{-1}, where vv is the transvection whose matrix with respect to the same base is (1101)\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}).

Solution

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Machine translation of the English, by gpt-5-6. Not checked by a person.