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:
$\quad \begin{array}{r|rrrrrr} \struct {\Z_6, +_6} & \eqclass 0 6 & \eqclass 1 6 & \eqclass 2 6 & \eqclass 3 6 & \eqclass 4 6 & \eqclass 5 6 \\ \hline \eqclass 0 6 & \eqclass 0 6 & \eqclass 1 6 & \eqclass 2 6 & \eqclass 3 6 & \eqclass 4 6 & \eqclass 5 6 \\ \eqclass 1 6 & \eqclass 1 6 & \eqclass 2 6 & \eqclass 3 6 & \eqclass 4 6 & \eqclass 5 6 & \eqclass 0 6 \\ \eqclass 2 6 & \eqclass 2 6 & \eqclass 3 6 & \eqclass 4 6 & \eqclass 5 6 & \eqclass 0 6 & \eqclass 1 6 \\ \eqclass 3 6 & \eqclass 3 6 & \eqclass 4 6 & \eqclass 5 6 & \eqclass 0 6 & \eqclass 1 6 & \eqclass 2 6 \\ \eqclass 4 6 & \eqclass 4 6 & \eqclass 5 6 & \eqclass 0 6 & \eqclass 1 6 & \eqclass 2 6 & \eqclass 3 6 \\ \eqclass 5 6 & \eqclass 5 6 & \eqclass 0 6 & \eqclass 1 6 & \eqclass 2 6 & \eqclass 3 6 & \eqclass 4 6 \\ \end{array}$
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
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Subgroups and Cosets: $\S 41 \zeta$
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $5$: Cosets and Lagrange's Theorem: Groups with six elements
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $11$: The Sylow Theorems: Example $11.13$