Let \(G\) be a group of order \(60.\)
Then, \(G\) may or may not be simple.
For example, \(G = \mathbb{Z}/60\mathbb{Z}\) is an example of a non-simple group of order \(60.\) This can be easily verified by considering the subgroup \(H = \{0, 30\} \le G.\) This is clearly normal as \(G\) is abelian.
On the other hand, consider \(G = A_5,\) the alternating group of degree \(5.\) It is known that \(A_n\) is simple for \(n \ge 5.\) Thus, \(G\) is simple and moreover, it has order \(5!/2 = 60.\)
Interestingly, it can be shown that \(A_5\) is the only simple group (up to isomorphism) of order \(60.\)