Subgroups of Symmetry Group of Regular Hexagon

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Theorem

Let $\HH = ABCDEF$ be a regular hexagon.

Let $D_6$ denote the symmetry group of $\HH$.

SymmetryGroupRegularHexagon.png

Let $e$ denote the identity mapping

Let $\alpha$ denote rotation of $\HH$ anticlockwise through $\dfrac \pi 3$ radians ($60 \degrees$)

Let $\beta$ denote reflection of $\HH$ in the $AD$ axis.


The subsets of $D_6$ which form its subgroups are as follows:


Order $1$
$\set e$


Order $2$
$\set {e, \alpha^3}$
$\set {e, \beta}$
$\set {e, \alpha \beta}$
$\set {e, \alpha^2 \beta}$
$\set {e, \alpha^3 \beta}$
$\set {e, \alpha^4 \beta}$
$\set {e, \alpha^5 \beta}$


Order $3$
$\set {e, \alpha^2, \alpha^4}$


Order $4$
$\set {e, \alpha^3, \beta, \alpha^3 \beta}$
$\set {e, \alpha^3, \alpha \beta, \alpha^4 \beta}$
$\set {e, \alpha^3, \alpha^2 \beta, \alpha^5 \beta}$


Order $6$
$\set {e, \alpha, \alpha^2, \alpha^3, \alpha^4, \alpha^5}$
$\set {e, \alpha^2, \alpha^4, \beta, \alpha^2 \beta, \alpha^4 \beta}$
$\set {e, \alpha^2, \alpha^4, \alpha \beta, \alpha^3 \beta, \alpha^5 \beta}$


Order $12$
$D_6$ itself.


Proof




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