Aryaman ยท I wasn't

Order 198

Let \(G\) be a group with order 198. We show that \(G\) is not simple.
Note that \(|G| = 2\cdot3^2\cdot11\).
By Sylow Theorem (3), we see that \(n_{11} \mid 18\) and \(n_{11} \equiv 1 \mod 11\). Looking at the divisors of \(18\), we see that \(n_{11} = 1\) is the only option and hence, we are done.