Let \(G\) be a nonabelian group of order \(p^3\), where \(p\) is prime. Prove that \[ G'=Z(G). \]
Solution
1. One has\[ |Z(G)|=p. \]
Every finite \(p\)-group has nontrivial center, so \(|Z(G)|\) is \(p\), \(p^2\), or \(p^3\). Since \(G\) is nonabelian, \(|Z(G)|\ne p^3\).
If \(|Z(G)|=p^2\), then \(G/Z(G)\) has order \(p\) and is therefore cyclic. A group with cyclic central quotient is abelian, contradiction. Hence \(|Z(G)|=p\).
2. The quotient \(G/Z(G)\) is abelian.
By 1 it has order \[ |G/Z(G)|=p^2, \] and every group of order \(p^2\) is abelian.
3. Therefore\[ G'\le Z(G). \]
The derived subgroup \(G'=[G,G]\) is the smallest normal subgroup \(N\) such that \(G/N\) is abelian. Since \(G/Z(G)\) is abelian by 2, this universal property gives \[ G'\le Z(G). \]
4. The subgroup \(G'\) is nontrivial.
If \(G'=1\), then \(G\) itself is abelian, contrary to hypothesis.
5. Hence\[ G'=Z(G). \]
By 1, the center has prime order \(p\). By 3 and 4, \(G'\) is a nontrivial subgroup of \(Z(G)\). The only nontrivial subgroup of a group of prime order is the whole group.