Olivier Debarre's Algèbre 1 [Lecture notes] PDF

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Additional resources for Algèbre 1 [Lecture notes]

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On note alors Z n+1 (G) G l’image inverse par la surjection canonique G → G/Z n (G) du centre de G/Z n (G), c’est-à-dire Z n+1 (G) = {g ∈ G | ∀x ∈ G g xg −1 x −1 ∈ Z n (G)}. On obtient ainsi une suite croissante de sous-groupes {e} = Z 0 (G) Z 1 (G) Z 2 (G) ··· où les quotients successifs sont abéliens. 1. — On dit qu’un groupe G est nilpotent s’il existe n ∈ N tel que Z n (G) = G. 2. — 1° Tout groupe abélien est nilpotent, puisque Z 1 (G) = G. 2° Le groupe Sn est nilpotent pour n 2, mais pas pour n 3, puisqu’on a alors Z(Sn ) = {Id} (exerc.

An {e}, CHAPITRE I. GROUPES 38 5° Le groupe Z n’a pas de suite de composition : en effet, tout sous-groupe de Z est du type mZ, et m est premier si on veut que le quotient Z/mZ soit simple. Il reste donc isomorphe à Z et on ne peut pas atteindre {0} en un nombre fini de pas. Une suite de composition G = G0 G1 · · · Gs = {e} est dite équivalente à la suite (14) si r = s et qu’il existe une permutation σ ∈ Sr telle que Gσ(i ) /Gσ(i )+1 Gi /Gi +1 . Le théorème suivant indique l’existence et l’unicité des suites de composition pour les groupes finis : il dit ainsi qu’en un certain sens tous les groupes finis sont construits à partir de ces blocs de base.

On en déduit que y est dans le centre de G/Z m (G), donc que y est dans Z m+1 (G), ce qui montre le pas de récurrence. 7. — Tout groupe nilpotent est résoluble. Démonstration. — Cela résulte du fait qu’on a Dn (G) ⊆ Cn (G) pour tout n 0. 8. — Le produit de deux groupes nilpotents est nilpotent. Démonstration. — Cela résulte du fait qu’on a Cn (G × H) ⊆ Cn (G) × Cn (H) pour tout n 0. La propriété d’être nilpotent passe aussi aux sous-groupes et aux groupes quotients. 9. — Soit G un groupe nilpotent et soit H un sous-groupe de G.

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Algèbre 1 [Lecture notes] by Olivier Debarre

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