From Existence of Cyclic Group of Order n we have that one such group of order $6$ is $C_6$ the cyclic group of order $6$: This is exemplified by the additive group of integers modulo $6$, whose Cayley table can be presented as: {{:Modulo Addition/Cayley Table/Modulo 6}} Then we have the symmetric group on $3$ letters. From Order of Symmetric Group, this has order $6$. It can be exemplified by the symmetry group of the equilateral triangle, whose Cayley table can be presented as: {{:Symmetry Group of Equilateral Triangle/Cayley Table}} It remains to be shown that these are the only $2$ groups of order $6$. Let $G$ be a group of order $6$ whose identity is $e$. By the First Sylow Theorem, $G$ has at least one Sylow $3$-subgroup. By the Third Sylow Theorem, the number of Sylow $3$-subgroups is a divisor of $6$. By the Fourth Sylow Theorem, the number of Sylow $3$-subgroups is equivalent to $1 \pmod p$. Combining these results, this number is therefore $1$. Call this Sylow $3$-subgroup $P$. By Sylow $p$-Subgroup is Unique iff Normal, $P$ is normal. From Prime Group is Cyclic, $P = \gen x$ for some $x \in G$ for $x^3 = e$. By the First Sylow Theorem, $G$ also has at least one Sylow $2$-subgroup of order $2$. Thus $G$ has an element $y$ such that $y^2 = e$. We have that $P$ is normal. Therefore: :$y^{-1} x y \in P$ Therefore one of the following applies: {{begin-eqn}} {{eqn | l = y^{-1} x y       | r = e }} {{eqn | l = y^{-1} x y       | r = x }} {{eqn | l = y^{-1} x y       | r = x^2 }} {{end-eqn}} If $y^{-1} x y = e$ then it follows that $x = 1$, which is contrary to our deduction that $x^3 = e$. Hence there remain two possibilities. First suppose $y^{-1} x y = x$. Then: {{begin-eqn}} {{eqn | l = y^{-1} x y       | r = x       | c =  }} {{eqn | ll= \leadsto       | l = x y       | r = y x       | c =  }} {{end-eqn}} Hence we can calculate the powers of $x y$ in turn: {{begin-eqn}} {{eqn | l = \paren {x y}^2       | r = x \paren {y x} y       | c = {{GroupAxiom|1}} }} {{eqn | r = x \paren {x y} y       | c = as $x y = y x$ by hypothesis }} {{eqn | r = x^2 y^2       | c =  }} {{eqn | n = 1       | r = x^2       | c = as $y^2 = e$ }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \paren {x y}^3       | r = \paren {x y}^2 \paren {x y}       | c =  }} {{eqn | r = x^2 x y       | c = as $\paren {x y}^2 = x^2$ by $(1)$ above }} {{eqn | r = x^3 y       | c =  }} {{eqn | n = 2       | r = y       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \paren {x y}^4       | r = \paren {x y} \paren {x y}^3       | c =  }} {{eqn | r = x y y       | c = as $\paren {x y}^3 = y$ by $(2)$ above }} {{eqn | r = x y^2       | c =  }} {{eqn | n = 3       | r = x       | c = as $y^2 = e$ }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \paren {x y}^5       | r = \paren {x y}^4 \paren {x y}       | c =  }} {{eqn | r = x x y       | c = as $\paren {x y}^4 = x$ by $(3)$ above }} {{eqn | n = 4       | r = x^2 y       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \paren {x y}^6       | r = \paren {x y}^5 \paren {x y}       | c =  }} {{eqn | r = x^2 y x y       | c = as $\paren {x y}^5 = x^2 y$ by $(4)$ above }} {{eqn | r = x^2 y y x       | c = as $x y = y x$ by hypothesis }} {{eqn | r = x^3       | c = as $y^2 = e$ }} {{eqn | r = e       | c = as $x^3 = e$ }} {{end-eqn}} Thus the order of $x y$ is $6$, and so $G$ is cyclic of order $6$. Now suppose $y^{-1} x y = x^2$. Then: {{begin-eqn}} {{eqn | l = y^{-1} x y       | r = x^{-1}       | c =  }} {{eqn | ll= \leadsto       | l = x y       | r = y x^{-1}       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = \paren {x y}^2       | r = x y y x^{-1}       | c =  }} {{eqn | r = x x^{-1}       | c = as $y^2 = e$ }} {{eqn | r = e       | c =  }} {{end-eqn}} It remains to investigate $x^2 y$: {{begin-eqn}} {{eqn | l = \paren {x^2 y}^2       | r = \paren {x y x^{-1} } \paren {x^2 y}       | c = as $x y = y x^{-1}$ }} {{eqn | r = x y x y       | c =  }} {{eqn | r = e       | c = from above: $\paren {x y}^2 = e$ }} {{end-eqn}} Thus we have $6$ elements: :$e, x, x^2$ which form a cyclic group of order $3$ :$y, x y, x^2 y$ all of which are self-inverse. Thus in this case $G$ is the symmetric group on $3$ letters $S_3$. The possibilities are exhausted. Hence the result. {{qed}}