(c)Karen E. Smith 2018 UM Math Deptlicensed under a Creative CommonsBy-NC-SA 4.0 International License.

Math 412. The Orbit Stabilizer Theorem

Fix an action of a group G on a set X . For each point x of X , we have two important concepts:

DEFINITION: The orbit of x ∈ X is the subset of X

O(x) := {g · x | g ∈ G} ⊂ X.

DEFINITION: The stabilizer of x is the subgroup of G

Stab(x) = {g ∈ G | g · x = x} ⊂ G.

THEOREM: If a finite group G acts on a set X , then for every x ∈ X , we have

|G| = |O(x)| × |Stab(x)|.

A. Let D4 be the symmetry group of the square. Consider the natural action of D4 on the set R2 byrotations and reflections of the whole space.

(1) Complete the following chart which records, for different points of the set R2, the orbit, stabilizer,and cardinalities of each. Use the notation {e, r, r2, r3, x, y, d, a}.

(2) Now verify the orbit stabilizer theorem for each of the five points in your chart. Easy: the numberof elements in the orbit times the number of elements in the stabilizer is the same, always 8, foreach point.

B. THE STABILIZER OF EVERY POINT IS A SUBGROUP. Assume a group G acts on a set X . Letx ∈ X .

(1) Prove that the stabilizer of x is a subgroup of G.(2) Use the Orbit-Stabilizer theorem to prove that the cardinality of every orbit divides |G|.

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(3) Let G be a group of order 17 and let X be a set with 16 elements. Explain why there is nonon-trivial action of G on X . [The trivial action is the one in which g · x = x for all g ∈ G and all x ∈ X .]

(1) We need to show that Stab(x) = {g ∈ G | g · x = x} is a subgroup of G. It suffices to check:(a) Stab(x) is non-empty; this is easy since eG · x = x (by definition of action), so eG ∈ Stab(x).(b) If a, b ∈ Stab(x), then ab ∈ Stab(x). Also pretty easy: take arbitrary a, b ∈ Stab(x). We compute

(a◦ b) ·x = a(·(b ·x) by definition of action. Since b ·x = x, this becomes (a◦ b) ·x = a(·(b ·x)) = a ·x;and since a · x = x, we conclude that (a ◦ b) · x = x. Thus a ◦ b ∈ Stab(x).

(c) if a ∈ Stab(x), then a−1 ∈ Stab(x). For this, take arbitrary a ∈ Stab(x). This means a · x = x. Applya−1 to both sides to get a−1 ·(a·x) = a−1 ·x.By definition of action, a−1 ·(a·x) = (a−1◦a)·x = e·x = x.So we have a−1 · x = x, and a−1 ∈ Stab(x). QED.

(2) This is clear: |O(x)| divides |G| since |G| = |O(x)| × |Stab(x)|.(3) Since 17 is prime, the only divisors of |G| are 1 and 17. So any orbit of any G action can only have size 1 or 17.

Since in this case, X has only 16 points total, no orbit can have 17 points! So all orbits have one point. Thismeans g · x = x for all g and all x. The only such action is the trivial action.

C. CANONICAL ACTION OF THE SYMMETRIC GROUP. Let Sn act on Xn = {1, 2, 3, 4, . . . , n} in the natural way; that is,σ ∈ Sn acts on i ∈ X by σ · i = σ(i).

(1) In the case n = 4, write out all the elements in the stabilizer of the element 4 for the canonical action of S4 on{1, 2, 3, 4}. Prove that the stabilizer of 4 is isomorphic to S3.

(2) Again for the canonical action of S4 on {1, 2, 3, 4}, what is the stabilizer of the element 2? Explain why the stabilizeris isomorphic to S3.

(3) Verify the orbit-stabilizer theorem for some point for the canonical action of S4 on {1, 2, 3, 4}. Why is the verifica-tion more or less the same for any other point?

(4) When Sn acts on a set X of n objects in the canonical way, what is the decomposition of X into orbits? Describethis stabilizer of some x ∈ X explicitly in words. To what well-known group is it isomorphic? Verify the Orbit-Stabilizer theorem in this case.

(1) e, (12), (13), (23), (123), (132). These are the elements of S3. Note that a permuation (bijection)of {1, 2, 3, 4}that fixes the element 4 is basically the same as a permutation of {1, 2, 3}. So Stab(4) ∼= S3.

(2) Stab(2) is the permutations of {1, 3, 4} (with 2 fixed). This is the permutation group of three objects, henceisomorphic to S3.

(3) The orbit of each point is the whole set {1, 2, 3, 4}, so |O(x)| = 4 for all x ∈ {1, 2, 3, 4}. Likewise the stabilizerof any point is the group of permutations of the other 3. So the stabilizers are all isomorphic to S3, which hascardinality. Since 4× 6 = 24 = |S4|, the orbit-stabilizer theorem is confirmed.

(4) There is only one orbit, since any of the n objects can be mapped to any other (in multiple ways!) under anelement in Sn. So the decomposition into orbits is trivial: there is only one orbit. The orbit stabilizer theoremtells us that |Sn| = |O(x)||Stab(x)| for any x. So |Sn| = n! = n |Stab(x)|. Thus |Stab(x)| = n!/n = (n−1)!.The elements in the stabilizer are the bijections of {1, 2, . . . , n} that fix the one element x. Of course, this isthe same as a permutation of n − 1 objects, all the the numbers 1, 2, . . . n except x, which has to go to itself.So Stab(x) ∼= Sn−1.

D. ROTATIONAL SYMMETRY GROUPS OF PLATONIC SOLIDS. There are exactly five convex regular solid figures in R3.Each is constructed by congruent regular polygonal faces with the same number of faces meeting at each vertex. The chartbelow describes each of these platonic solids:

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Each platonic solid has a rotational symmetry groupwhich acts naturally on the solid. In particular, eachsymmetry group also acts on the set of vertices, theset of edges and the set of faces of the correspondingsolid. By analyzing these three actions, we can betterunderstand the symmetry group of each solid.

(1) For the cube, complete the chart below describing the size of the orbits and stabilizers for each of the three differentactions (on faces, vertices, edges) of the symmetry group of the cube. What is the order of the symmetry group ofthe cube?

(2) For the tetahedron, again complete the chart below for each of the three different actions (on faces, vertices, edges)of the tetrahedral group (= the symmetry group of the tetrahedon).

(3) Now complete the chart below for the octahedral group acting on the faces, vertices, and edges of the regularoctahedron. What is the order of this symmetry group (the octahedral group)?

(4) Repeat again for the dodecahedron. What is the order of its symmetry group?(5) Finally, use the same method to compute the order of the Icosahedral group (the symmetry group of the icosahedron).

Action # orbit # stab |G|on Faceson edges

on vertices

For each of the three actions, does it matterwhich point x ∈ X you use to compute theorbit? Why is the order of the stabilizer thesame for each x ∈ X in each of the threeactions? Is this true in general for a groupacting on a set? What is special in this case?

(1) For the symmetry group of the cube, we have:

Action # orbit # stab |G|on Faces 6 4 24on edges 12 2 24

on vertices 8 3 24

(2) For the symmetry group of the tetrahedron we have:

Action # orbit # stab |G|on Faces 4 3 12on edges 6 2 12

on vertices 4 3 12

Note that here,

it is a bit tricky to find the stabilizer of an edge, but since we know there are 2 elements in the stabilizer fromthe Orbit-Stabilizer theorem, we can look.

(3) For the Octahedron, we have

Action # orbit # stab |G|on Faces 8 3 24on edges 12 2 24

on vertices 6 4 24

(4) For the symmetry group of the dodecahedron, we have:

Action # orbit # stab |G|on Faces 12 5 60on edges 30 2 60

on vertices 20 3 60

(5) For the symmetry group of the icosahedron, we have:

Action # orbit # stab |G|on Faces 20 3 60on edges 30 2 60

on vertices 12 5 60

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E. LINEAR ACTIONS. Consider the action of GL2(R) on R2 by matrix multiplication (where elements of R2 are written ascolumns).

(1) Describe the action in mathematical symbols and prove it is really an action.

(2) What is the stabilizer of the point[10

]?

(3) What is the orbit of the point[10

]?

(1) For A ∈ GL2(R2) and[xy

]∈ R2, we have A ·

[xy

]= A

[xy

]. It is an action because I2

[xy

]=

[xy

]and

A · (B ·[xy

]) = A(B

[xy

]) = (AB)

[xy

], by the associativity of matrix multiplication. So the two axioms of a

group action are satisfied.

(2) What A =

[a bc d

]stabilize

[10

]? Since

[a bc d

] [10

]=

[ac

], a necessary and sufficient condition is that[

ac

]=

[10

]. So the stabilizer of

[10

]is the subgroup of matrices {

[1 b0 d

]| b, d,∈ R, d 6= 0}.

(3) The orbit of[10

]under matrix multiplication is R2 \ ~0. Indeed, we can get an arbitrary non-zero

[xy

]by

multiplying[x by d

] [10

]=

[xy

]. Note that as long as

[xy

]is not the zero vector, we can always find b, d ∈ R

such that the matrix[x by d

]is invertible.

F. RIGID MOTIONS OF THE PLANE. The group M2 of rigid motions of the plane acts naturally on the set of all triangles inthe Euclidean plane. Explain how and why M2 is a group. What is the orbit of a fixed triangle T ? What is the stabilizer of anequilateral triangle? The stabilizer of an isosceles (but non-equilateral) triangle? A scalene triangle?

Each rigid motion is a distance (and angle preserving) transformation of the plane, and the composition of suchtransformations is another; composition of functions is always associate (proved in class, also appendix), and the identitymap is an identity. The inverse of such a transformation is always another as well, so that M2 is a group. A fixed trianglewill be sent to a congruent triangle underM2. The orbit of a triangle is the set of all congruent triangles. The stabilizer ofan equilateral triangle consists of all the rigid motions which fix the triangle (as a set). This is D3, basically by definitionofD3. The stabilizer of an isoceles triangle is much smaller: we only have reflections across one axis, so it is a 2-elementgroup, isomorphic to D2, C2, S2 or Z2. The stabilizer of a scalene triangle is {e}.

G. Prove the Orbit-Stabilizer Theorem. [Hint: Start by finding a surjective map G → O(x); show that elements of G havethe same image under this map if and only if they are in the same coset of the stablizer of x.]

This will be assigned on Problem Set 11.