Bourbaki

Exercise 35, § 6

Algebra, I, p. 140

Let GG be a group such that the derived group D=D(G)D = D(G) is contained in the center C of G. a) Prove that the mapping (x,y)↦(x,y)(x, y) \mapsto (x, y) induces a mapping φ:(G/C)×(G/C)→D\varphi : (G/C) \times (G/C) \to D such that, on writing these groups additively, one has:

φ(α+β,γ)=φ(α,γ)+φ(β,γ)φ(α,β)=−φ(β,α)φ(α,β+γ)=φ(α,β)+φ(α,γ)φ(α,α)=0 \begin{align*} \varphi(\alpha + \beta, \gamma) &= \varphi(\alpha, \gamma) + \varphi(\beta, \gamma) & \varphi(\alpha, \beta) &= -\varphi(\beta, \alpha) \\ \varphi(\alpha, \beta + \gamma) &= \varphi(\alpha, \beta) + \varphi(\alpha, \gamma) & \varphi(\alpha, \alpha) &= 0 \end{align*}
for all α,β,γ\alpha, \beta, \gamma in G/CG/C. b) Prove that, for every integer nn, one has
(yx)n=ynxn(x,y)n(n−1)2 (yx)^n = y^n x^n (x, y)^{\frac{n(n-1)}{2}}

for x,yx, y in GG. Deduce that, if nn is odd and if dn=1d^n = 1 for every d∈Dd \in D, the mapping x↦xnx \mapsto x^n defines, by passing to the quotient, a homomorphism θ:G/D→C\theta : G/D \to C.

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.