Let \(G\) be a group with order 150. We show that \(G\) is not simple.
Note that \(|G| = 2\cdot3\cdot5^2\).
By Sylow Theorem (3), we get that \(n_5 \in \{1, 6\}.\)
Then, we are done.
Let \(G\) act on Syl\(_5(G) = \{P_1, \ldots, P_6\}\) by conjugation.
Consider the corresponding natural homomorphism \(m:G\to S_6\).
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 \(150\) elements. However \(m(G) \le S_6\) and \(|S_6| = 6!\), a contradiction as \(150\not\mid6!.\) This shows us that ker \(m\) cannot be \((1).\)
Thus, ker \(m\) is a proper nontrivial subgroup of \(G.\)
As kernels of homomorphisms are always normal, we are done!