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 $⟨x⟩$ 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 $S_3$.