Group Actions

Kenneth Levasseur
Mathematical Sciences
UMass Lowell
Kenneth_Levasseur@uml.edu

Definition:  Let G be a group and X a set.  G acts on X if there is a function α:G×X ->X  such that 
	(a)   α(e,x)=x   and
	(b)  α((Subscript[g, 1])α((Subscript[g, 2]),x))=α((Subscript[g, 1])(Subscript[g, 2]),x)
α is called an action on X.    w

α(g,x) is sometimes abbreviated g x.

Examples. (a) G action on itself by left translation:  
	(Subscript[τ, g])(x)= g x

Cayley's Theorem says any group G is isomorphic to a subgroup of (Subscript[S, G]):
	G ≅ {(Subscript[τ, g])| g∈ G}  with an isomorpism being g|->(Subscript[τ, g]). 
	
(b)  G acts on itself by conjugation.  γ(g,x)=g x (g^(-1)).
	γ(e,x)=e x (e^(-1))= e x e = x
	γ((Subscript[g, 1])γ((Subscript[g, 2]),x)) = γ((Subscript[g, 1]),(Subscript[g, 2])x ((Subscript[g, 2])^(-1)))
		= (Subscript[g, 1])((Subscript[g, 2])x ((Subscript[g, 2])^(-1)))((Subscript[g, 1])^(-1))
		= (Subscript[(g, 1])(Subscript[g, 2]))x (((Subscript[g, 2])^(-1))((Subscript[g, 1])^(-1)))
		         =  ((Subscript[g, 1])(Subscript[g, 2]))x ((((Subscript[g, 1])(Subscript[g, 2])))^(-1))
		         =  γ((Subscript[g, 1])(Subscript[g, 2]),x)

(c)  Consider the fabrication of circular bracelets with four jewels (Rubies, Emeralds, and Saphires) evenly spaced on the bracelet.  For example one bracelet could look like this:

jr=0.05;
{Circle[{0,0},1],Red,Disk[{1,0},jr],Disk[{-1,0},jr],Blue,Disk[{0,1},jr],Green,Disk[{0,-1},jr]}//Graphics

One of 81 placements of jewels

Note: Not all of the different jewels must be used.  For example we could have a bracelet with four rubies.

For the time being imagine that the positions of the jewels cannot be changed.  Let X be the set of possible arrangement of jewels in a bracelet.   |X| = (3^4)= 81.    The group of symmetries of a square, (Subscript[D, 4]) acts on X in a natural way.

The stablizer of x:   (Subscript[G, x])= {g∈G|  α(g,x) = x}⊆G

For the case of conjugation:  
	(Subscript[G, x])= {g∈G | g x (g^(-1))= x}= {g∈G | g x = x g}
	= the centralizer of x.

For the bracelet above, the stablizer is the cyclic subgroup of  (Subscript[D, 4])  generated by the vertical reflection.

Theorem.  (Subscript[G, x])≤G. 
Proof:  First note that e∈(Subscript[G, x]) since  α(e,x)=x.  So (Subscript[G, x]) is never empty.  If g∈(Subscript[G, x]), then  x = e x= ((g^(-1))g)x=(g^(-1))(g,x)=(g^(-1))x => ∈ (Subscript[G, x]).   Finally, if g,h∈(Subscript[G, x]),
	(g h) x=g(h x)= g x=x  =>   g h ∈(Subscript[G, x])

The orbit of x:   𝒪(x) = {g x | g∈G} ⊆ X

For the case of conjugation:  
	𝒪(x)= { g x (g^(-1))| g∈G}
	= the conjugate class of x

The orbit of the bracelet above is

jr=0.05;
Graphics/@{{Circle[{0,0},1],Red,Disk[{1,0},jr],Disk[{-1,0},jr],Blue,Disk[{0,1},jr],Green,Disk[{0,-1},jr]},{Circle[{0,0},1],Red,Disk[{1,0},jr],Disk[{-1,0},jr],Blue,Disk[{0,1},jr],Green,Disk[{0,-1},jr]}/.{{a_,b_}->RotationMatrix[Pi/2].{a,b}},{Circle[{0,0},1],Red,Disk[{1,0},jr],Disk[{-1,0},jr],Blue,Disk[{0,1},jr],Green,Disk[{0,-1},jr]}/.{{a_,b_}->RotationMatrix[Pi].{a,b}},{Circle[{0,0},1],Red,Disk[{1,0},jr],Disk[{-1,0},jr],Blue,Disk[{0,1},jr],Green,Disk[{0,-1},jr]}/.{{a_,b_}->RotationMatrix[-Pi/2].{a,b}}}//Partition[#,2]&//TableForm

The Center of G:   
	Z(G)= {x∈G | g x = x g for all g∈G}
             ={x ∈ G | the conjugate class of x is {x}}  
	If n≥3,  Z((Subscript[S, n]))={i}

The Class Equation: If G is a finite group and {(Subscript[g, 1]),(Subscript[g, 2]),…,(Subscript[g, m])} are selected as representatives of the  conjugate classes containing more than one element, then
	(Abs[G])=(Abs[Z(G)]) + (∑_(k=1)^m)(Abs[𝒪((Subscript[g, k]))])

Theorem.   The number of elements in the conjugate class of x, |𝒪(x)| is [G:(Subscript[G, x])], the number of distinct cosets of (Subscript[G, x]) in G.

Proof:  Let G/(Subscript[G, x]) be  the set of all cosets of (Subscript[G, x]) in G.  Define ϕ:𝒪(x)-> G/(Subscript[G, x]) as follows:  If y ∈𝒪(x), then   y = g x  for some g.   Then define   ϕ(y)= g (Subscript[G, x]).
	ϕ  is well-defined.
	ϕ  is one-to-one
	ϕ is onto

The Class Equation, revised:  (Abs[G])=(Abs[Z(G)]) + (∑_(k=1)^m)[G:(Subscript[G, (Subscript[g, k])])]

Theorem  If G is a finite group whose order is divisible by a prime p, then G contains an element of order p.

Lemma 
  (i)   Let G act on X.  If x∈X and σ∈G, then (Subscript[G, σ x])= σ (Subscript[G, x]) (σ^(-1)).
  (ii)  If finite group G acts on finite set X and if x and y lie in the same orbit, then  (Abs[(Subscript[G, y])])= (Abs[(Subscript[G, x])])

Burnside's Lemma.  Let G acto on a finite set X.  If N is the number of orbits, then
	N = (1/(Abs[G]))(∑_(τ∈G))F(τ) 
where F(τ) is the number of x∈X fixed by τ.