Groups of Order 6

Theorem
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:

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:

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:

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:

Hence we can calculate the powers of $x y$ in turn:

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:

and:

It remains to investigate $x^2 y$:

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.