Groups of order $45$ are abelian

Let \(|G|=45=3^2\cdot5\). Sylow’s theorem gives \[ n_5\mid9,\qquad n_5\equiv1\pmod5, \] so \(n_5=1\). Likewise \[ n_3\mid5,\qquad n_3\equiv1\pmod3, \] so \(n_3=1\). Hence the Sylow subgroups \(P\) of order \(9\) and \(Q\) of order \(5\) are both normal.

Their orders are coprime, so \(P\cap Q=1\), and \[ |PQ|=|P||Q|=45, \] thus \(G=PQ\). Since both \(P\) and \(Q\) are normal, \([P,Q]\subseteq P\cap Q=1\), so \[ G\cong P\times Q. \] Every group of order \(9\) is abelian, and \(Q\cong C_5\), hence \(G\) is abelian. Consequently \[ G\cong C_9\times C_5\cong C_{45} \] or \[ G\cong (C_3\times C_3)\times C_5\cong C_3\times C_{15}. \]