Classify all groups of order \(99\).
Solution
Let \(G\) be a group of order \(99=3^2\cdot11\).
1. The Sylow \(11\)-subgroup \(N\) is unique and normal.
Sylow gives \(n_{11}\mid9\) and \(n_{11}\equiv1\pmod{11}\). Among \(1,3,9\), only \(1\) satisfies the congruence. Thus \(n_{11}=1\).
2. If \(P\) is a Sylow \(3\)-subgroup, then \(G=NP\) and \(N\cap P=1\).
The orders of \(N\) and \(P\) are \(11\) and \(9\), so their intersection is trivial. Since \(N\trianglelefteq G\), \[ |NP|=\frac{|N||P|}{|N\cap P|}=99=|G|. \] Hence \(G=NP\).
3. The conjugation action of \(P\) on \(N\) is trivial.
Because \(N\cong C_{11}\), \[ \operatorname{Aut}(N)\cong C_{10}. \] Conjugation gives a homomorphism \(P\to C_{10}\). Its image has order dividing both \(9\) and \(10\), hence has order \(1\). Thus \(P\) centralizes \(N\).
4. Therefore \(G\cong N\times P\).
By 2 the factors generate \(G\) and intersect trivially, and by 3 they commute elementwise.
5. There are exactly two isomorphism types:\[ C_{99} \qquad\text{and}\qquad C_{11}\times C_3\times C_3. \]
A group of order \(9\) is isomorphic either to \(C_9\) or to \(C_3\times C_3\). Combining this with 4 gives \[ C_{11}\times C_9\cong C_{99} \] or \[ C_{11}\times C_3\times C_3. \] These are not isomorphic because only the first has an element of order \(9\).