Group of order six
= Group of order six
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 $\langle x\rangle$ 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 $hxh^{-1}=x^{-1}$. The generators therefore obey $x^3=h^2=e$, $hxh^{-1}=x^{-1}$, the relations realized by $(123)$ and $(12)$ in $S_3$.