Groups of order $p^2$ and of order $pq$

For \(|G|=p^2\), the center of the \(p\)-group \(G\) is nontrivial. If \(|Z(G)|=p\), then \(G/Z(G)\) is cyclic, which forces \(G\) abelian; hence in fact \(Z(G)=G\). Therefore \[ G\cong C_{p^2}\quad\text{or}\quad C_p\times C_p. \]

Now let \(|G|=pq\) with primes \(p<q\). Sylow gives \[ n_q\equiv1\pmod q,\qquad n_q\mid p, \] so \(n_q=1\). Thus the Sylow \(q\)-subgroup \(Q\cong C_q\) is normal. If \(P\cong C_p\) is a Sylow \(p\)-subgroup, then \[ G\cong C_q\rtimes_\theta C_p, \qquad \theta:C_p\to\operatorname{Aut}(C_q)\cong C_{q-1}. \]

If \(p\nmid q-1\), the homomorphism \(\theta\) is trivial, hence \[ G\cong C_q\times C_p\cong C_{pq}. \] If \(p\mid q-1\), then \(C_{q-1}\) has a unique subgroup of order \(p\). Thus there are exactly two isomorphism classes: the cyclic group \(C_{pq}\), and one nonabelian semidirect product \[ \langle x,y\mid x^q=y^p=1,\ yxy^{-1}=x^r\rangle, \] where \(r\in(\mathbb Z/q\mathbb Z)^\times\) has order \(p\).