Let \(G\) be a group with order 72. We show that \(G\) is not simple.
Note that \(|G| = 2^3\cdot3^2\).
By Sylow Theorem (3), we see that \(n_3 \mid 8\) and \(n_3 \equiv 1 \mod 3\). Looking at the divisors of \(8,\) we see that \(n_3 = 1\) or \(n_3 = 4.\)
Then, we are done.
Thus, there are \(4\) Sylow-\(3\) subgroups of \(G\), each of size \(9.\)
Syl\(_3(G) = \{P_1, P_2, P_3, P_4\}.\)
Let \(G\) act on Syl\(_3(G)\) by conjugation.
Consider the corresponding natural homomorphism \(m:G\to S_4\).
By Sylow Theorem (2), ker \(m \neq G\) as the action is transitive.
Assume ker \(m = (1)\). Then, \(m\) is an injective map and thus, \(m(G)\) has \(72\) elements. However \(m(G) \le S_4\) and \(|S_4| = 24\), a contradiction.
Thus, ker \(m\) is a proper nontrivial subgroup of \(G\).
As kernels of homomorphisms are always normal, we are done.