Up to isomorphism, the groups of order \(10\) are the cyclic group \(\ZZ/10\ZZ\), which is abelian, and the dihedral group \(D_5\) of order \(10\), which is nonabelian.
Let \(\abs G=10\). By Sylow's theorems, the number of Sylow \(5\)-subgroups divides \(2\) and is congruent to \(1\) modulo \(5\), so there is a unique, hence normal, Sylow \(5\)-subgroup \(N\cong\ZZ/5\ZZ\). 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 \(\theta\colon H\to\Aut(N)\cong(\ZZ/5\ZZ)^\times\). The elements of order dividing \(2\) in \((\ZZ/5\ZZ)^\times\) are \(\pm1\). If \(\theta\) is trivial, \(G\cong\ZZ/5\ZZ\times\ZZ/2\ZZ\cong\ZZ/10\ZZ\); if the generator of \(H\) acts by inversion, \(G\cong D_5\).