## Group Theory and Linear Algebra

Last updated: 24 September 2014

## Lecture 25: Group actions, orbits, stabilizers

The dihedral group ${D}_{n}$ is the set $Dn= { 1,x,x2,…, xn-1,y, xy,x2y,…, xn-1y }$ with operation given by $(xiyj) (xkyℓ)= x(i-k mod n) y(j+ℓ mod 2)$ so that, in particular $y2=1,xn=1 andyx=x-1y.$

A $G\text{-set}$ $S,$ or an action of a group $G$ on a set $S$ is a function $G×S ⟶ S (g,s) ⟼ gs$ such that

 (a) If ${g}_{1},{g}_{2}\in G$ and $s\in S$ then ${g}_{1}\left({g}_{2}s\right)=\left({g}_{1}{g}_{2}\right)s,$ (b) If $s\in S$ then $1·s=s\text{.}$

Let $s\in S\text{.}$ The stabilizer of $s$ is $Gs= {g∈G | gs=s}.$ The orbit of $s$ is $Gs={gs | g∈G}.$

Let $G$ be a group and let $S$ be a $G\text{-set.}$

 (a) The orbits of $G$ acting on $S$ partition $S\text{.}$ (b) Let $s\in S\text{.}$ Then ${G}_{s}$ is a subgroup of $G,$ $φ: GGs ⟶ Gs gGs ⟼ gs$ is a well defined function and $\phi$ is a bijection.

$\text{"}{D}_{4}$ acting on a square". ${e}_{1} {e}_{2} {e}_{3} {e}_{4} c d b a b c a d a b d c d a c b 1 x x2 x3 b a c d c b d a d c a b a d b c y xy x2y x3y$ ${D}_{4}$ acts on the vertices $V=\left\{a,b,c,d\right\}\text{:}$ $xa=b, xya=a.$ ${D}_{4}$ acts on the edges $E=\left\{{e}_{1},{e}_{2},{e}_{3},{e}_{4}\right\}\text{:}$ $x2e1=e3, ye1=e1.$

$\text{"}G$ acts on itself by conjugation". $G×G ⟶ G (g,s) ⟼ gsg-1 or g·s=gsg-1,$ where $g\in G$ and $s\in G\text{.}$

Let $s\in G\text{.}$ The centraliser of $s$ in $G$ is the stabiliser of $s,$ $𝒵G(s)=Stab(s)= {g∈G | gsg-1=s}.$ The conjugacy class of $s$ in $G$ is the orbit of $s,$ $𝒞s={gsg-1 | g∈G}.$ The centre of $G$ is $𝒵(G)= {s∈G | gs=sg for all g∈G}.$

HW: Show that $s\in 𝒵\left(G\right)$ if and only if ${𝒵}_{G}\left(s\right)=G\text{.}$

HW: Show that $s\in 𝒵\left(G\right)$ if and only if ${𝒞}_{s}=\left\{s\right\}\text{.}$

${D}_{4}$ acts on itself by conjugation. $x ↺ 1 ↻ y and x ↺ x2 ↻ y$ $x ↺ x ⇄yy x3 ↺ x since yxy-1 = x3yy-1=x3 yx2y-1 = x2yy-1=x2 yx3y-1 = xyy-1=x$ $y ↺ y ⇄xx x2y ↻ y since xyx-1= xyx3=xxy= x2y$ $xy ⇄x,yy,x x3y$ So the conjugacy classes are ${1}, {x2}, {x,x3}, {y,x2y}, {xy,x3y}$ and the centralizers of elements in ${D}_{4}$ are $𝒵D4(1) = Stab(1)=D4 𝒵D4(x) = Stab(x)= {1,x,x2,x3} 𝒵D4(x2) = Stab(x2)=D4 𝒵D4(x3) = Stab(x3)= {1,x,x2,x3} 𝒵D4(y) = Stab(y)= {1,x2,y,x2y} 𝒵D4(xy) = Stab(xy)= {1,x2,xy,x3y} 𝒵D4(x2y) = Stab(x2y)= {1,x2,y,x2y} 𝒵D4(x3y) = Stab(x3y)= {1,x2,xy,x3y}$

## Notes and References

These are a typed copy of Lecture 25 from a series of handwritten lecture notes for the class Group Theory and Linear Algebra given on October 4, 2011.