Groups of order 55

Let \(|G|=55=5\cdot11\). Sylow gives \[ n_{11}\equiv1\pmod{11},\qquad n_{11}\mid5, \] so \(n_{11}=1\). Hence the Sylow \(11\)-subgroup \(P\cong C_{11}\) is normal. If \(Q\cong C_5\) is a Sylow \(5\)-subgroup, then \(P\cap Q=1\) and \(PQ=G\), so \[ G\cong C_{11}\rtimes C_5. \] The action is a homomorphism \[ C_5\longrightarrow\operatorname{Aut}(C_{11})\cong C_{10}. \] There are two possibilities up to isomorphism. The action is trivial, giving \[ C_{11}\times C_5\cong C_{55}, \] or it is injective onto the unique subgroup of order \(5\) in \(C_{10}\), giving one nonabelian semidirect product. Thus there are exactly two groups of order \(55\) up to isomorphism.