Prove: if |G| = p^2, G must be abelian. (Use the preceding Exercise F.)

Let G be a group whose order is a power of a prime p, say |G| = p^k.

Let C denote the center of G.

Conclude that if G/C is cyclic, then G is abelian.

Prove: if |G| = p^2, G/C must be cyclic.

|C| is a multiple of p      (1)

|G| is a multiple of |C|

|G| = p*p               (2)

|C| = p

|C| = p^2

Number of cosets of C in G =
Index of C in G =
(G:C) = |G|/|C| = p^2/p = p

|G/C| = p

If G is a group with a prime number p of elements, then G is a cyclic group

|G| = p^2   ->   G/C is a cyclic group   ->   G is abelian