No group of order 36 is simple

Let \(|G|=36\). Sylow gives \[ n_3\mid4,\qquad n_3\equiv1\pmod3, \] so \(n_3=1\) or \(4\).

If \(n_3=1\), the Sylow \(3\)-subgroup is a nontrivial proper normal subgroup.

Suppose \(n_3=4\). Conjugation on the four Sylow \(3\)-subgroups gives \[ \rho:G\to S_4. \] If \(\ker\rho=1\), then \(G\) embeds in \(S_4\), impossible because \(36\nmid24\). Thus the kernel is nontrivial. It is also proper because the conjugation action is transitive and hence nontrivial. Therefore \(\ker\rho\) is a nontrivial proper normal subgroup of \(G\).

So no group of order \(36\) is simple.