Up to isomorphism, the groups of order \(14\) are the cyclic group \(\ZZ/14\ZZ\), which is abelian, and the dihedral group \(D_7\) of order \(14\), which is nonabelian.
Let \(\abs G=14\). By Sylow's theorems, the number of Sylow \(7\)-subgroups divides \(2\) and is congruent to \(1\) modulo \(7\), so there is a unique, hence normal, Sylow \(7\)-subgroup \(N\cong\ZZ/7\ZZ\). A Sylow \(2\)-subgroup \(H\cong\ZZ/2\ZZ\) meets \(N\) trivially and \(NH=G\), so \(G\cong N\rtimes_\theta H\) for a homomorphism \(\theta\colon H\to\Aut(N)\cong(\ZZ/7\ZZ)^\times\). The elements of order dividing \(2\) in \((\ZZ/7\ZZ)^\times\) are \(\pm1\). If \(\theta\) is trivial, \(G\cong\ZZ/7\ZZ\times\ZZ/2\ZZ\cong\ZZ/14\ZZ\); if the generator of \(H\) acts by inversion, \(G\cong D_7\).