Let \(G\) be a group with order \(2^n\cdot3\) where \(n\) is a positive integer strictly greater than 1. We show that \(G\) is not simple.
By Sylow Theorem (3), we see that \(n_2 \mid 3\) and thus, we have the following two cases:
Then, we are done.
Then, \(G\) acts on Syl\(_2(G) = \{P_1, P_2, P_3\}\) by conjugation.
Consider the corresponding natural homomorphism \(m:G\to S_3\).
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 \(2^n\cdot 3 \ge 12\) elements. However \(m(G) \le S_3\) and \(|S_3| = 6\), a contradiction.
Thus, ker \(m\) is a proper nontrivial subgroup of \(G\).
As kernels of homomorphisms are always normal, we are done.