Dummit Foote Solutions Chapter 4 Now

Problem: If ( |G| = p^2 ) for ( p ) prime, prove ( G ) is abelian.

Solution: Recall the class equation: ( |G| = |Z(G)| + \sum [G : C_G(g_i)] ).

Each term ( [G : C_G(g_i)] > 1 ) divides ( |G| = p^2 ), so can be ( p ) or ( p^2 ). But ( [G : C_G(g_i)] = p^2 ) would imply ( C_G(g_i) = e ), impossible for non-identity ( g_i ) since ( G ) is finite. So each non-central term = ( p ). dummit foote solutions chapter 4

Thus ( p^2 = |Z(G)| + kp ), where ( k ) = number of non-central conjugacy classes.

Hence ( |Z(G)| = p(p - k) ). Since ( |Z(G)| \ge 1 ) and divides ( p^2 ), possibilities: Problem : If ( |G| = p^2 )

Thus ( |Z(G)| = p^2 ), so ( G ) is abelian. QED.

Note: This exercise is standard in any "Dummit Foote solutions Chapter 4" PDF. Understand this proof thoroughly—it reapplies in Sylow theory. Thus ( |Z(G)| = p^2 ), so ( G ) is abelian

Chapter 4 builds the action framework for:

Key Theorem (Orbit-Stabilizer): For a finite group ( G ), ( |\mathcalO_a| = [G : G_a] ).

Chapter 4 is titled: Group Actions, Sylow Theorems, and Applications
But in many syllabi, Chapter 4 covers Group Actions (after Ch. 3 on subgroups & quotients).

Core topics: