Groups of order 15

Let \(|G|=15=3\cdot5\). By Sylow, \[ n_5\mid3,\qquad n_5\equiv1\pmod5, \] so \(n_5=1\). Likewise \[ n_3\mid5,\qquad n_3\equiv1\pmod3, \] and since \(5\not\equiv1\pmod3\), we get \(n_3=1\).

Thus the Sylow subgroups \(P\) of order \(5\) and \(Q\) of order \(3\) are both normal. Their intersection is trivial, and \[ |PQ|=\frac{|P||Q|}{|P\cap Q|}=15, \] so \(G=PQ\). Since both factors are normal and intersect trivially, they commute and \[ G\cong P\times Q\cong \mathbb Z/5\times\mathbb Z/3\cong\mathbb Z/15. \] Therefore there is exactly one group of order \(15\) up to isomorphism.