MATH 251 Homework 6 Solutions 1. Let G be a finite group, N ¢ G ...

P ≠ Q and P ∩ Q ≠ 1. So |P ∩ Q| = 5. Hence [P : P ∩ Q] = 5 = [Q : P ∩ Q]. So P ∩ Q ✂ P, Q. Hence|P ||Q|P, Q ≤ N G (P ∩ Q). So P Q ⊆ N G (P ∩ Q). But |P Q| =|P ∩Q| = 25·255= 125. Hence |N G (P ∩ Q)| ≥ 125and [G : N G (P ∩ Q)] ≤ 400125 < 4. Since |G| does not divide 4!, we get that N G(P ∩ Q) = G by the SmallIndex Argument. So P ∩ Q ✂ G, a contradiction. So G is not simple.4