8.2
Orbits and Stabilizers
If the groupGacts on the setX, theorbitof a pointx∈Xis the set
Ox:={αx|α∈G} ⊂X .
Thestabilizer ofxis the subgroup
Gx :={α ∈G|αx =x}. Example 8.3. If we take X ={1,2,3,4} and G=V′ ={(1),(12),(34),(12)(34)} ⊂S4, then O1 =O2 ={1,2}, O3 =O4 ={3,4} and V1′ =V2′ =⟨(34)⟩, V3′ =V4′ =⟨(12)⟩. The group V ={(1),(12)(34),(13)(24),(14)(23)}
also acts on this set. However
O1 =O2 =O3 =O4 =X, V1 =V2 =V3 =V4 ={(1)}.
In chapter 3 we looked at the decomposition of a permutationα ∈ Sninto
a product of cycles. The cycles we found correspond to the orbits of the permu- tation group⟨α⟩acting on the set{1,2, . . . , n}. The orbit of1is
O1 ={1, α(1), α2(1), . . . , αr1−1(1)},
which gives us the first cycle. We then pick the smallest numberi2 which does
not occur inO1 and compute its orbit:
Oi2 ={i2, α(i2), α
2
This gives us the second cycle, and so on.
In general if a groupGacting on a setXhas only one orbit, we say thatGacts transitivelyonX. IfGis a permutation group of degreenwhich acts transitively on{1,2, . . . , n}we say thatGis transitive.
Remark8.4. In general,Ox = Oy, for x, y ∈ X, if and only if there exists an
α∈Gsuch thatαx =y. Why is this so? Suppose thatαx =y. Ifz =βy ∈Oy,
thenβαx=βy =zso thatz ∈Ox. Thus
Oy ⊂Ox .
By writingα−1y=x, we can reverse the roles ofxandyand see that
Ox ⊂Oy .
ThereforeOx =Oy. Conversely, ifOx =Oy, theny∈ Ox and therefore there
existsα ∈ Gsuch that αx =y. If there is only one orbit, then for allxandy, we haveOx = Oy or equivalently, for all xandythere exists α ∈ Gsuch that
αx=y.
Many of the permutation groups that we have seen act transitively on the set {1,2, . . . , n}. However the stabilizers may not be trivial. For example, taking
G = Sn, we see that any permutation which fixes 1 can permute {2, . . . , n}
quite arbitrarily. So(Sn)1is the group of all permutations of{2, . . . , n}, which
is isomorphic toSn−1.
The group of proper symmetries of a Platonic solid acts transitively on the set of vertices, the set of edges, and the set of faces. What do the stabilizers look like? Let's look at the regular tetrahedron. First pick a vertex. The rotations whose axis passes through the vertex and the centre of the opposite face leave it fixed. No other rotations of order3do so. A rotation of order2whose axis passes through the centres of a pair of opposite edges, does not fix any vertex. So the stabilizer of a vertex is the cyclic group of order3consisting of the rotations about the line through the vertex and the centre of the opposite side. The only rotation which fixes a given edge is the half-turn about the line joining its midpoint and that of
8.2. ORBITS AND STABILIZERS 117
the opposite edge. So its stabilizer is the cyclic group of order2generated by this half-turn. And the stabilizer of a face is just the stabilizer of the opposite vertex. In example (i) aboveGacts transitively on itself. Why? Takeν, ξ ∈ G. Let
α=ξν−1. Then
αν = (ξν−1)ν=ξ.
The stabilizer of any elementξis trivial, becauseαξ=ξimplies thatα= 1. The action of G on itself by conjugation (example (ii)) is more interesting: first some terminology. The orbit of an elementξ ∈ G under conjugation is called itsconjugacy class and will be denoted byCξ. So
Cξ :={αξα−1 |α ∈G}.
An element αξα−1 is called a conjugate of ξ. The stabilizer of ξ is called its
centralizer, denoted byZξ(G)or justZξ:
Zξ :={α∈G|αξα−1 =ξ}={α |αξ=ξα}.
Thus the centralizer ofξis the set of all elements which commute with ξ. The centre ofG, writtenZ(G), is
Z(G) ={α|αβ =βα, for allβ ∈G}
So
Z(G) =∩ξZξ(G).
IfGis abelian then of courseZ(G) =G. Let's compute the conjugacy classes ofSn.
Lemma 8.5. Supposeν = (i1, . . . , ir)∈Snis anr-cycle. Thenανα−1 is ther-cycle
( α(i1), . . . , α(ir) ) . Proof. We have α(ij) α−1 −−−→ ij ν −−−→ ij+1 α −−−→ α(ij+1),
where as usual,ir+1 :=i1. (Remember that we are reading from right to left.) If i̸∈ {α(i1), . . . , α(ir)}, thenα−1(i)̸∈ {i1, . . . , ir}, so that i −−−→α−1 α−1(i) −−−→ν α−1(i) −−−→α i . Thus ανα−1 =(α(i1), . . . , α(ir) ) .
It also follows from this lemma that if we are given a secondr-cycleξ, there exists anα∈Gsuch thatξ =ανα−1. Namely, ifξ= (j
1, . . . , jr), then define
α(i1) = j1, . . . , α(ir) = jr,
and extend α to the rest of {1,2, . . . , n} in any way you like as long as α is bijective. Then by the lemma, ξ = ανα−1. Thus the set of all r-cycles is a
conjugacy class.
Now suppose thatν =ν1· · ·νswhereν1, . . . , νsare disjoint cycles of length
r1, . . . , rs respectively. We say thatν is ofcycle type {r1, . . . , rs}. Then for any
α∈G, we have
ξ:=ανα−1 =αν1α−1· · ·ανsα−1.
Thereforeξtoo is of cycle type{r1, . . . , rs}. On the other hand, given any two
elementsν andξ of the same cycle type, we can refine the argument just given to see that there exists an αsuch that ξ = ανα−1. This proves the following theorem:
Theorem 8.6. A conjugacy class inSnconsists of all permutations of a given cycle type.
For example, the possible cycle types inS5 are
{1} {2} {3} {2,2} {4} {2,3} {5}.
And to each of these there corresponds a conjugacy class inS5. Notice that a