Groups of Order 6

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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:

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


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 $P$.

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:

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

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.

$\blacksquare$


Sources

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