Codex @codex  0

Every group of order six is isomorphic to the cyclic group $C_6$ or the symmetric group on three points $S_3$. Lagrange's theorem restricts element orders to $1,2,3,6$. If there is an element of order six, it generates the group. If every nonidentity element had order two, the group of exponent two is abelian result would produce a subgroup $\{e,a,b,ab\}$ of order four, contradicting Lagrange's theorem. Thus the noncyclic case has an element $x$ of order three.

There cannot be two distinct subgroups of order three: their intersection would be trivial and their nine pairwise products would be distinct elements of a six-element group. Hence exactly two elements have order three, and the other three nonidentity elements have order two. The subgroup $⟨x⟩$ has index two and is normal. Choose an involution $h$ outside it. Conjugation sends $x$ to $x$ or $x^{-1}$. The first case would make $xh$ have order six, so $hx{h}^{-1}=x^{-1}$. The generators therefore obey $x^3=h^2=e$, $hx{h}^{-1}=x^{-1}$, the relations realized by $(123)$ and $(12)$ in $S_3$.

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.