Classify all groups of order \(99 = 3^2 \cdot 11\) up to isomorphism.
Solution
1. The Sylow \(11\)-subgroup of \(G\) is unique and normal.
Let \(n_{11}\) be the number of Sylow \(11\)-subgroups. Sylow’s theorem gives \[ n_{11}\mid 9, \qquad n_{11}\equiv 1\pmod{11}. \] The divisors of \(9\) are \(1,3,9\), and only \(1\) is congruent to \(1\) modulo \(11\). Hence \[ n_{11}=1. \] Write \(Q\) for this unique Sylow \(11\)-subgroup. Then \(Q\trianglelefteq G\), and since \(|Q|=11\), \[ Q\cong C_{11}. \]
2. The Sylow \(3\)-subgroup of \(G\) is unique and normal.
Let \(n_3\) be the number of Sylow \(3\)-subgroups. Again by Sylow’s theorem, \[ n_3\mid 11, \qquad n_3\equiv 1\pmod 3. \] The only divisors of \(11\) are \(1\) and \(11\), while \(11\equiv2\pmod3\). Thus \[ n_3=1. \] Write \(P\) for the unique Sylow \(3\)-subgroup. Then \(P\trianglelefteq G\) and \(|P|=9\).
3. The group \(G\) is the internal direct product \(P\times Q\).
Since \(|P|=9\) and \(|Q|=11\) are coprime, \[ P\cap Q=1. \] For \(p\in P\) and \(q\in Q\), normality of \(P\) and \(Q\) implies \[ [p,q]=pqp^{-1}q^{-1}\in P\cap Q=1. \] Hence \(P\) and \(Q\) commute elementwise. Moreover, \[ |PQ|=\frac{|P||Q|}{|P\cap Q|}=9\cdot11=99=|G|, \] so \(PQ=G\). Therefore \[ G\cong P\times Q. \]
4. There are exactly two possibilities for \(P\).
Every group of order \(p^2\) is abelian. Therefore a group of order \(9\) is isomorphic to exactly one of \[ C_9, \qquad C_3\times C_3. \] Thus \[ P\cong C_9 \quad\text{or}\quad P\cong C_3\times C_3. \]
5. Consequently there are exactly two groups of order \(99\) up to isomorphism:\[ C_{99} \qquad\text{and}\qquad C_3\times C_{33}. \]
If \(P\cong C_9\), then by 3, \[ G\cong C_9\times C_{11}\cong C_{99}, \] because \(9\) and \(11\) are coprime. If \(P\cong C_3\times C_3\), then \[ G\cong C_3\times C_3\times C_{11} \cong C_3\times C_{33}. \] These two groups are not isomorphic: the first contains an element of order \(99\), whereas every element of the second has order dividing \(33\).