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:

$\begin{array}{r|rrrrrr} \struct {\Z_6, +_6} & \eqclass 0 6 & \eqclass 1 6 & \eqclass 2 6 & \eqclass 3 6 & \eqclass 4 6 & \eqclass 5 6 \\ \hline \eqclass 0 6 & \eqclass 0 6 & \eqclass 1 6 & \eqclass 2 6 & \eqclass 3 6 & \eqclass 4 6 & \eqclass 5 6 \\ \eqclass 1 6 & \eqclass 1 6 & \eqclass 2 6 & \eqclass 3 6 & \eqclass 4 6 & \eqclass 5 6 & \eqclass 0 6 \\ \eqclass 2 6 & \eqclass 2 6 & \eqclass 3 6 & \eqclass 4 6 & \eqclass 5 6 & \eqclass 0 6 & \eqclass 1 6 \\ \eqclass 3 6 & \eqclass 3 6 & \eqclass 4 6 & \eqclass 5 6 & \eqclass 0 6 & \eqclass 1 6 & \eqclass 2 6 \\ \eqclass 4 6 & \eqclass 4 6 & \eqclass 5 6 & \eqclass 0 6 & \eqclass 1 6 & \eqclass 2 6 & \eqclass 3 6 \\ \eqclass 5 6 & \eqclass 5 6 & \eqclass 0 6 & \eqclass 1 6 & \eqclass 2 6 & \eqclass 3 6 & \eqclass 4 6 \\ \end{array}$

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:

$\begin{array}{c|ccc|ccc} \circ & e & p & q & r & s & t \\ \hline e & e & p & q & r & s & t \\ p & p & q & e & s & t & r \\ q & q & e & p & t & r & s \\ \hline r & r & t & s & e & q & p \\ s & s & r & t & p & e & q \\ t & t & s & r & q & p & e \\ \end{array}$

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 $P$.

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:

 $\ds y^{-1} x y$ $=$ $\ds e$ $\ds y^{-1} x y$ $=$ $\ds x$ $\ds y^{-1} x y$ $=$ $\ds x^2$

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:

 $\ds y^{-1} x y$ $=$ $\ds x$ $\ds \leadsto \ \$ $\ds x y$ $=$ $\ds y x$

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

 $\ds \paren {x y}^2$ $=$ $\ds x \paren {y x} y$ Group Axiom $\text G 1$: Associativity $\ds$ $=$ $\ds x \paren {x y} y$ as $x y = y x$ by hypothesis $\ds$ $=$ $\ds x^2 y^2$ $\text {(1)}: \quad$ $\ds$ $=$ $\ds x^2$ as $y^2 = e$

 $\ds \paren {x y}^3$ $=$ $\ds \paren {x y}^2 \paren {x y}$ $\ds$ $=$ $\ds x^2 x y$ as $\paren {x y}^2 = x^2$ by $(1)$ above $\ds$ $=$ $\ds x^3 y$ $\text {(2)}: \quad$ $\ds$ $=$ $\ds y$

 $\ds \paren {x y}^4$ $=$ $\ds \paren {x y} \paren {x y}^3$ $\ds$ $=$ $\ds x y y$ as $\paren {x y}^3 = y$ by $(2)$ above $\ds$ $=$ $\ds x y^2$ $\text {(3)}: \quad$ $\ds$ $=$ $\ds x$ as $y^2 = e$

 $\ds \paren {x y}^5$ $=$ $\ds \paren {x y}^4 \paren {x y}$ $\ds$ $=$ $\ds x x y$ as $\paren {x y}^4 = x$ by $(3)$ above $\text {(4)}: \quad$ $\ds$ $=$ $\ds x^2 y$

 $\ds \paren {x y}^6$ $=$ $\ds \paren {x y}^5 \paren {x y}$ $\ds$ $=$ $\ds x^2 y x y$ as $\paren {x y}^5 = x^2 y$ by $(4)$ above $\ds$ $=$ $\ds x^2 y y x$ as $x y = y x$ by hypothesis $\ds$ $=$ $\ds x^3$ as $y^2 = e$ $\ds$ $=$ $\ds e$ as $x^3 = e$

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:

 $\ds y^{-1} x y$ $=$ $\ds x^{-1}$ $\ds \leadsto \ \$ $\ds x y$ $=$ $\ds y x^{-1}$

and:

 $\ds \paren {x y}^2$ $=$ $\ds x y y x^{-1}$ $\ds$ $=$ $\ds x x^{-1}$ as $y^2 = e$ $\ds$ $=$ $\ds e$

It remains to investigate $x^2 y$:

 $\ds \paren {x^2 y}^2$ $=$ $\ds \paren {x y x^{-1} } \paren {x^2 y}$ as $x y = y x^{-1}$ $\ds$ $=$ $\ds x y x y$ $\ds$ $=$ $\ds e$ from above: $\paren {x y}^2 = e$

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.

$\blacksquare$