LetX be a set andG be a group. A(left) action ofG onX is a mapG×X →X given by (g, x)7→gx, where
1. ex=x for allx∈X;
2. (g1g2)x=g1(g2x) for all x∈X and all g1, g2 ∈G.
Under these considerations X is called a G-set. Notice that we are not requiringX to be related to G in any way. It is true that every group G acts on every setX by the trivial action (g, x) 7→ x; however, group actions are more interesting if the set X is somehow related to the groupG.
Example 14.1. LetG=GL2(R) and X=R2. Then Gacts on X by left multiplication.
Ifv∈R2 andI is the identity matrix, thenIv=v. IfAandB are2×2invertible matrices, then(AB)v=A(Bv) since matrix multiplication is associative.
Example 14.2. Let G = D4 be the symmetry group of a square. If X = {1,2,3,4}
is the set of vertices of the square, then we can consider D4 to consist of the following
permutations:
{(1),(13),(24),(1432),(1234),(12)(34),(14)(23),(13)(24)}.
The elements ofD4 act on X as functions. The permutation (13)(24) acts on vertex 1 by
sending it to vertex 3, on vertex 2 by sending it to vertex 4, and so on. It is easy to see that the axioms of a group action are satisfied.
In general, if X is any set and G is a subgroup of SX, the group of all permutations
acting on X, thenX is aG-set under the group action
(σ, x)7→σ(x)
forσ ∈Gand x∈X.
14.1. GROUPS ACTING ON SETS 159 Example 14.3. If we let X = G, then every group G acts on itself by the left regular representation; that is,(g, x)7→λg(x) =gx, whereλg is left multiplication:
e·x=λex=ex=x
(gh)·x=λghx=λgλhx=λg(hx) =g·(h·x).
IfH is a subgroup of G, then Gis anH-set under left multiplication by elements ofH. Example 14.4. LetGbe a group and suppose thatX=G. IfH is a subgroup ofG, then Gis anH-set under conjugation; that is, we can define an action ofH on G,
H×G→G, via
(h, g)7→hgh−1
forh∈H and g∈G. Clearly, the first axiom for a group action holds. Observing that
(h1h2, g) =h1h2g(h1h2)−1
=h1(h2gh−21)h− 1 1
= (h1,(h2, g)),
we see that the second condition is also satisfied.
Example 14.5. LetH be a subgroup ofGandLH the set of left cosets ofH. The setLH
is aG-set under the action
(g, xH)7→gxH.
Again, it is easy to see that the first axiom is true. Since (gg′)xH =g(g′xH), the second axiom is also true.
IfGacts on a setX andx, y∈X, thenxis said to beG-equivalent toy if there exists a g∈Gsuch that gx=y. We write x∼G y orx∼y if two elements areG-equivalent.
Proposition 14.6. LetX be aG-set. ThenG-equivalence is an equivalence relation onX. Proof. The relation ∼ is reflexive sinceex=x. Suppose that x∼y forx, y∈X. Then there exists a g such that gx=y. In this case g−1y = x; hence, y ∼x. To show that the relation is transitive, suppose thatx∼y and y∼z. Then there must exist group elements g and h such thatgx=y andhy =z. Soz=hy = (hg)x, and x is equivalent toz.
If X is a G-set, then each partition of X associated with G-equivalence is called an orbit ofX underG. We will denote the orbit that contains an element xof X by Ox.
Example 14.7. LetGbe the permutation group defined by
G={(1),(123),(132),(45),(123)(45),(132)(45)}
and X ={1,2,3,4,5}. Then X is a G-set. The orbits are O1 =O2 =O3 ={1,2,3} and
O4 =O5 ={4,5}.
Now suppose that G is a group acting on a set X and let g be an element of G. The fixed point setofginX, denoted byXg, is the set of allx∈Xsuch thatgx=x. We can
also study the group elementsgthat fix a givenx∈X. This set is more than a subset of G, it is a subgroup. This subgroup is called the stabilizer subgroup orisotropy subgroup of x. We will denote the stabilizer subgroup of xby Gx.
Remark 14.8. It is important to remember that Xg ⊂X andGx ⊂G.
Example 14.9. Let X = {1,2,3,4,5,6} and suppose that G is the permutation group given by the permutations
{(1),(12)(3456),(35)(46),(12)(3654)}. Then the fixed point sets of X under the action ofG are
X(1) =X,
X(35)(46)={1,2}, X(12)(3456)=X(12)(3654)=∅,
and the stabilizer subgroups are
G1 =G2 ={(1),(35)(46)},
G3 =G4 =G5 =G6={(1)}.
It is easily seen thatGx is a subgroup of G for eachx∈X.
Proposition 14.10. Let G be a group acting on a setX and x∈X. The stabilizer group of x,Gx, is a subgroup of G.
Proof. Clearly,e∈Gx since the identity fixes every element in the setX. Let g, h∈Gx.
Then gx=xand hx=x. So (gh)x=g(hx) =gx=x; hence, the product of two elements inGx is also inGx. Finally, if g∈Gx, then x =ex= (g−1g)x= (g−1)gx=g−1x. So g−1
is in Gx.
We will denote the number of elements in the fixed point set of an element g ∈ G by |Xg| and denote the number of elements in the orbit ofx ∈X by |Ox|. The next theorem
demonstrates the relationship between orbits of an elementx∈X and the left cosets ofGx
inG.
Theorem 14.11. Let Gbe a finite group and X a finite G-set. If x∈X, then |Ox|= [G:
Gx].
Proof. We know that |G|/|Gx| is the number of left cosets of Gx in G by Lagrange’s
Theorem (Theorem6.10). We will define a bijective mapϕbetween the orbit Ox of X and the set of left cosets LGx of Gx in G. Let y ∈ Ox. Then there exists a g in G such that
gx=y. Defineϕbyϕ(y) =gGx. To show thatϕis one-to-one, assume thatϕ(y1) =ϕ(y2).
Then
ϕ(y1) =g1Gx=g2Gx =ϕ(y2),
whereg1x=y1andg2x=y2. Sinceg1Gx=g2Gx, there exists ag∈Gx such thatg2 =g1g,
y2 =g2x=g1gx=g1x=y1;
consequently, the map ϕis one-to-one. Finally, we must show that the map ϕis onto. Let gGx be a left coset. Ifgx=y, thenϕ(y) =gGx.