Algèbre [Lecture notes]] by Antoine Chambert-Loir

By Antoine Chambert-Loir

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On a la relation : Card(X) = ∑ Card(O) = ∑ (A ∶ AO ). O∈X/A O∈X/A Démonstration. — Comme les orbites de A dans X sont les classes d’équivalence d’une relation d’équivalence, elles définissent une partition de X, ce qui démontre la première égalité. D’autre part, l’application de A/AO dans X donnée par aAO ↦ a ⋅ xO est une injection d’image O, donc Card(O) = Card(A/AO ) = (A ∶ AO ). La deuxième égalité en découle. 14). — Soit G un groupe fini et soit H un sous-groupe de G. On a Card(G) = (G ∶ H) Card(H).

Supposons inversement que Ker( f ) = {eA } et démontrons que f est injectif. Soit a, b des élements de A tels que f (a) = f (b). Alors, f (ab−1 ) = f (a) f (b)−1 = eB , donc ab −1 = eA et a = b. Cela prouve que f est injectif. 5. 1. — Soit (Ai )i∈I une famille de monoïdes ; pour tout i, notons e i l’élément neutre de Ai . Soit A le produit de cette famille ; un élément de A est une famille (a i )i∈I , où a i ∈ Ai pour tout i ∈ I. On définit une loi de composition dans A en posant (a i )i∈I ⋅ (b i )i∈I = (a i b i )i∈I .

De groupes) f ∶ B → A tel que f i = p i ○ f pour tout i ∈ I. Démonstration. — Définissons une application f ∶ B → A par la formule f (b) = ( f i (b))i∈I , pour b ∈ B. C’est un morphisme de monoïdes (resp. de groupes) et l’on a f i (b) = p i ( f (b)) pour tout b ∈ B, c’est-à-dire f i = p i ○ f . Inversement, soit f ′ ∶ B → A une application telle que p i ○ f ′ = f i pour tout i ∈ I ; pour b ∈ B, on a f ′ (b) = (p i ○ f ′ (b))i∈I = ( f i (b)) = f (b), ce qui démontre que f = f ′ . 3. — Soit A un groupe et soit B et C des sous-groupes de A tels que B ∩ C = {e} et B ⋅ C = A.

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