Let \(G\) be a group with order 135. We show that \(G\) is not simple.
Note that \(|G| = 3^3\cdot5\).
By Sylow Theorem (3), we see that \(n_5 \mid 27\) and \(n_5 \equiv 1 \mod 5\). Looking at the divisors of \(27\), we see that \(n_5 = 1\) is the only option and hence, we are done.