Groups of order 10

Let \(|G|=10=2\cdot5\). Sylow gives \[ n_5\mid2,\qquad n_5\equiv1\pmod5, \] so \(n_5=1\). Thus the Sylow \(5\)-subgroup \(P\cong C_5\) is normal.

If \(Q\cong C_2\) is a Sylow \(2\)-subgroup, then \(P\cap Q=1\) and \(PQ=G\), so \[ G\cong C_5\rtimes C_2. \] The action is a homomorphism \[ C_2\to\Aut(C_5)\cong C_4. \] There are exactly two possibilities: the trivial action and the unique nontrivial action of order \(2\), namely inversion.

The trivial action gives \[ C_5\times C_2\cong C_{10}. \] The inversion action gives \[ \langle r,s\mid r^5=s^2=1,\ srs=r^{-1}\rangle\cong D_5. \] Hence the groups of order \(10\) are exactly \[ \boxed{C_{10}\quad\text{and}\quad D_5}. \]