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If an element has order six, it generates $G$, so $G\cong C_6$. Otherwise choose an element $x$ of order three. The preceding result excludes any further order-three elements outside $\{e,x,x^2\}$, and there are no elements of order six by assumption. Each of the other three elements must therefore have order two. ThusThe noncyclic case is the symmetric group on three points, as the classification of a group of order six further identifies.

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.

The permutation group of a three-element set has six elements: the identity, three transpositions, and two three-cycles. A three-cycle and any transposition generate it, with relations $x^3=h^2=e$, $hx{h}^{-1}=x^{-1}$. Its alternating group is the normal cyclic subgroup of order three. It is the unique nonabelian isomorphism type of a group of order six.