Theorem: Groups of Order 6

There exist exactly $2$ groups of order $6$, up to isomorphism:
:$C_6$, the cyclic group of order $6$
:$S_3$, the symmetric group on $3$ letters.

Proof:
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 = {{Group-axiom|1}}
}}
{{eqn | r = x \paren {x y} y
      | c = as $x y = y x$ {{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 $x y = y x$ {{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}}