New PDF release: Algèbre [Lecture notes]

By Antoine Chambert-Loir

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Y n ) de A telle que y1 = ∑ni=1 m i x i . 52 CHAPITRE 2. GROUPES Démonstration. — On raisonne par récurrence sur l’entier m = ∑ni=1 ∣m i ∣. Comme les m i sont premiers entre eux dans leur ensemble, ils ne sont pas tous nuls et m ⩾ 1. Si m = 1, il existe un entier i tel que m i = ±1 et m j = 0 si j ≠ i. Alors la suite (y1 , . . , y n ) obtenue en posant y1 = m i x i y i = x1 (si i ≠ 1) et y j = x j si j ∈/ {1, i} convient. Supposons maintenant m > 1, de sorte qu’au moins deux termes de la suite (m1 , .

M′n ) ∈ Zn , on a m′ m′ g((m1 , . . , m n ))g((m1′ , . . , m′n )) = x1m1 . . x nm n x1 1 . . x n n m 1 +m′1 = x1 m +m′n . . xn n = g((m1 + m1′ , . . , m n + m′n )). On a finalement f¯ ○ g(m1 , . . , m n ) = f¯(x1m1 . . x nm n ) = f (s1m1 . . s nm n ) = ∑ m i e i = (m1 , . . , m n ), donc f¯ ○ g = id. De plus, g( f¯(x i )) = g(e i ) = x i pour tout i. Comme {x1 , . . , x n } engendre G(S; R), on a donc g ○ f¯ = id. Cela conclut la preuve que f¯ est un isomorphisme de G(S; R) sur Zn .

Soit f ∶ A → C un homomorphisme de groupes. Pour a ∈ A et b ∈ Ker( f ), on a f (aba −1 ) = f (a) f (b) f (a)−1 = f (a) f (a)−1 = e, donc aba−1 ∈ Ker( f ). En particuier, le noyau d’un homomorphisme de groupes est donc un sous-groupe distingué. Plus généralement, démontrons l’image réciproque f −1 (B) d’un sous-groupe distingué B de C est un sous-groupe distingué de A. Soit en effet a ∈ A et b ∈ f −1 (B) ; on a f (aba−1 ) = f (a) f (b) f (a)−1 ∈ B puisque f (b) est un sousgroupe distingué de C ; donc aba−1 ∈ f −1 (B), ce qui démontre que f −1 (B) est un sous-groupe distingué de B.

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Algèbre [Lecture notes] by Antoine Chambert-Loir

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