Algebra Prelim Fall 2013
نویسنده
چکیده
First we will prove a small lemma. Lemma 1. Let P be a p-Sylow subgroup of G. Let N be a normal subgroup of G such that P ⊂ N . Then all p-Sylow subgroups are in N . Proof. Let P̃ be a p-Sylow subgroup that is not P . Then by the Sylow Theorems, we know that gPg−1 = P̃ for some g ∈ G. Since P ⊂ N , and N is normal, we know that gPg−1 ⊂ N and hence P̃ ⊂ N . Thus all p-Sylow subgroups are in N . Now let us begin the problem. Proof. Let G be a group such that |G| = 105 = 3 ·5 ·7. Let np denote the number of p-Sylow subgroups. By the Sylow Theorems, we know the following: n3 = 1 mod 3 n3 ∣∣ 35 =⇒ n3 = 1, 7 n5 = 1 mod 5 n5 ∣∣ 21 =⇒ n5 = 1, 21 n7 = 1 mod 7 n7 ∣∣ 15 =⇒ n7 = 1, 15
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