Alain Kraus's Cours d’arithmétique PDF

By Alain Kraus

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De Lagrange), il est donc ´egal ` a p. En particulier, tous les ´el´ements autres que l’´el´ement neutre sont g´en´erateurs. Exercice 14. Montrer que tout sous-groupe fini de (C∗ , ×) est cyclique. 4. Supposons G cyclique d’ordre n. 1) Tout sous-groupe de G est cyclique. 2) Pour tout diviseur d ≥ 1 de n, l’ensemble Hd = x ∈ G | x d = e est un sous-groupe de G d’ordre d. 3) L’application qui ` a d associe Hd est une bijection entre l’ensemble des diviseurs positifs de n et l’ensemble des sous-groupes de G.

Le groupe additif (Z/nZ, +) est d’ordre n et ses ´el´ements sont les classes des entiers compris entre 0 et n − 1. Exercice 7. Montrer que 0, 3, 6, 9 est un sous-groupe de Z/12Z, + . Expliciter un sous-groupe d’ordre 6 de Z/12Z, + . Exercice 8. D´eterminer tous les sous-groupes de Z/6Z, + . 5. Sous-groupe engendr´e par un ´el´ement - Ordre d’un ´el´ement Soit G un groupe multiplicatif, d’´el´ement neutre e. 1). Pour tout x ∈ G, il existe donc un plus petit sous-groupe de G qui contient x (au sens de l’inclusion), `a savoir l’intersection de tous les sous-groupes de G qui contiennent le singleton x .

Posons t = ϕ(n). Notons b1 , · · · , bt les entiers compris entre 1 et n et premiers avec n. Pour tout i = 1, · · · , t, il existe des entiers qi et ri tels que l’on ait abi = nqi + ri avec 0 ≤ ri < n. On a la congruence t t abi ≡ i=1 ri mod. n. i=1 Par ailleurs, les entiers abi ´etant premiers avec n, on a pgcd(ri , n) = 1 et 1 ≤ ri ≤ n (on a ri = 0 car n ≥ 2). V´erifions alors que l’on a b1 , · · · , bt = r1 , · · · , rt . 17 Citons-en deux : le probl`eme de Lehmer. On a vu que si p premier, on a ϕ(p) = p−1.

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Cours d’arithmétique by Alain Kraus


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