: Every group of order ( p^2 ) is abelian. Solution idea : From 4.3.6, ( |Z(G)| = p ) or ( p^2 ). If ( |Z(G)| = p ), then ( G/Z(G) ) cyclic ⇒ ( G ) abelian (contradiction unless ( Z(G) = G )).
When working through Chapter 4 solutions, keep these strategies in mind: Identify the Action: dummit foote solutions chapter 4
Abstract Algebra, 3rd Edition - Answers & Solutions | Brainly : Every group of order ( p^2 ) is abelian