Let \(G\) be a group with order 84. We show that \(G\) is not simple.
Note that \(|G| = 2^2\cdot5\cdot7\).
By Sylow Theorem (3), we see that \(n_7 \mid 20\) and \(n_7 \equiv 1 \mod 20\). Looking at the divisors of \(20\), we see that \(n_7 = 1\) is the only option and hence, we are done.