Let \(p\) be a prime and \(\abs{G} = p^3\). Prove that \(G\) has a normal subgroup \(N\) of order \(p^2\).
Suppose \(N = \gens{h}\) is cyclic and classify all possibilities for \(G\) if:
\(\abs h = p^3\)
\(\abs h = p\).
Hint: Sylow and semidirect products.
Solution
First, every group \(G\) of order \(p^3\) has a normal subgroup of order \(p^2\). Since \(G\) is a nontrivial finite \(p\)-group, \(Z(G)\ne1\).
If \(|Z(G)|\ge p^2\), then \(Z(G)\) contains a subgroup of order \(p^2\), and every subgroup of the center is normal.
If \(|Z(G)|=p\), then \(G/Z(G)\) has order \(p^2\) and is abelian. Choose any subgroup \[ L/Z(G)\le G/Z(G) \] of order \(p\). Because \(G/Z(G)\) is abelian, \(L/Z(G)\) is normal, hence its inverse image \(L\) is normal in \(G\); moreover \[ |L|=p^2. \] Thus a normal subgroup of order \(p^2\) always exists.
The remaining bullets in the problem, read literally, are incompatible with the preceding hypothesis. If the same subgroup is assumed to satisfy \[ N=\langle h\rangle,\qquad |N|=p^2, \] then necessarily \[ |h|=p^2. \] Therefore neither case \(|h|=p^3\) nor case \(|h|=p\) can occur for that \(N\).
If instead one merely asks what happens when an element of \(G\) has order \(p^3\), then that element generates all of \(G\), so \[ G\cong C_{p^3}. \] The condition that \(G\) merely contain an element of order \(p\) gives no classification at all: every group of order \(p^3\) contains such an element by Cauchy’s theorem.