E3 - application of cosets to permutation groups

stabiliser

let $G$ be a group of permutations of a set $S$ , then for each $i$ in $S$ , the stabiliser of $i$ in $G$ is defined as:

{stab}_{G} (i) = {ϕ \in G ∣ ϕ (i) = i}

so, stabiliser of an element is any permutation that keeps that element unchanged
note: ${stab}_{G} (i) \leq G$

orbit

${orb}_{G} (i) = {ϕ (i) ∣ ϕ \in G}$

so, it is the set of all locations a point can move to
eg: let $G$ be the following subgroup of $S_{8} :$ ${(1), (132) (465) (78), (132) (465), (123) (456) (78), (78)}$ , then:

\begin{matrix} {orb}_{G} (1) = {1, 3, 2}, & {stab}_{G} (1) = {(1), (78)}, \\ {orb}_{G} (2) = {2, 1, 3}, & {stab}_{G} (2) = {(1), (78)}, \\ {orb}_{G} (4) = {4, 5, 6}, & {stab}_{G} (4) = {(1), (78)}, \\ {orb}_{G} (5) = {7, 8}, & {stab}_{G} (7) = {(1), (132) (465)}, \end{matrix}

orbit-stabiliser theorem

let $G$ be a finite group of permutations of a set $S$ , then for any $i$ from $S :$

| G | = | {orb}_{G} (i) | | {stab}_{G} (i) |

by lagrange's theorem, $| {orb}_{G} (i) | / | {stab}_{G} (i) |$ is the number of distinct left cosets of ${stab}_{G} (i)$ in $G$
so, it will suffice to show a one-to-one relation between the left cosets and the elements in the orbit of $i$ , suppose $T : ϕ {stab}_{G} (i) \to ϕ (i)$
for $T$ to be a well defined function, need to show: $α {stab}_{G} (i) = β {stab}_{G} (i)$ $⟹ α (i) = β (i)$
this also means that $α^{- 1} β \in {stab}_{G} (i)$ , so $(α^{- 1} β) (i) = i$ , thus, $β (i) = α (i)$
this reversal shows that $T$ is one-to-one
to show that it is onto, let $j \in {orb}_{G} (i)$ , then $α (i) = i$ for some $α \in G$
clearly, $T (α {stab}_{G} (i)) = α (i) = j$ , so $T$ is onto
example: the rotation group of a cube $G$
the six faces are labelled from 1 to 6, so different rotations induce different permutations of the faces, and $G$ is group of permutations on the set ${1, 2, 3, 4, 5, 6}$
there will be some rotation about a central, horizontal or vertical axis that carries 1 to any other, so $| {orb}_{G} (1) | = 6$
${stab}_{G} (1)$ corresponds to all rotations that keep face 1 fixed, of which there are 4, about the central axis perpendicular to the face
this, $| G | = 6 \times 4 = 24$
after the number of rotations is known, determining the actual structure of the rotation group is simple

the rotation grop of a cube

the order of the rotation group of a cube is the same as $S_{4}$
the cube has four diagonals, which are permuted by the rotations, so the consecutive diagonals are labelled 1 to 4

image: J. Gallian

a $90 °$ rotation yields $α = (1234)$
another one, perpendicular to the first axis yields $β = (1423)$
so, the permutation group induced by the rotations contains the 8-element subgroup ${ϵ, α, α^{2}, α^{3} β^{2}, β^{2} α, β^{2} α^{2}, β^{2} α^{3}}$ and $α β$ of order $3$
thus, the rotations yield all 24 permutations as the order of the rotation group must be divisible by both 8 and 3
eg: the rotation group of a traditional soccer ball (20 regular-hexagonal and 12 regular-pentagonal faces), ie. truncated icosahedron
$S$ may chosen to be the set of 20 hexagons or the 12 pentagons
say 12 pentagon set is chosen, by some rotation any one can be carried to any other, the orbit of any pentagon is $S$
there are five rotations that fix any particular pentagon
therefore, there are 60 rotational symmetries
the rotation group is isomorphic to $A_{5}$