 Solution

In a group of order six, element orders divide six. If there were no element of order three or six, every nonidentity element would have order two, making the group abelian by the preceding argument. Choose $a\ne e$ and $b\notin \{e,a\}$. Then $\{e,a,b,ab\}$ is a subgroup of order four: its elements are distinct, it is closed because $a,b$ commute and square to $e$, and each element is its own inverse. This contradicts Lagrange's theorem, since four does not divide six. Hence an element of order three or six exists.