Groups of order 99

Let \(|G|=99=3^2\cdot11\).

For a Sylow \(11\)-subgroup, \[ n_{11}\mid9,\qquad n_{11}\equiv1\pmod{11}, \] so \(n_{11}=1\). Thus the Sylow \(11\)-subgroup \(Q\cong C_{11}\) is normal.

For a Sylow \(3\)-subgroup, \[ n_3\mid11,\qquad n_3\equiv1\pmod3. \] The only possibilities dividing \(11\) are \(1\) and \(11\), and \(11\equiv2\pmod3\), so \(n_3=1\). Hence the Sylow \(3\)-subgroup \(P\) of order \(9\) is also normal.

Since \(P\cap Q=1\) and \(|P||Q|=99\), we have \[ G=P\times Q. \] Every group of order \(9\) is abelian and is isomorphic either to \(C_9\) or to \(C_3\times C_3\). Therefore the two possibilities are \[ C_9\times C_{11}\cong C_{99} \] and \[ (C_3\times C_3)\times C_{11}\cong C_3\times C_{33}. \] Thus there are exactly two groups of order \(99\) up to isomorphism: \[ \boxed{C_{99}\quad\text{and}\quad C_3\times C_{33}.} \]