Math 412. The Orbit Stabilizer Theorem

1 (c)Karen E. Smith 2018 UM math Deptlicensed under a Creative CommonsBy-NC-SA International 412. The Orbit Stabilizer TheoremFix an action of a groupGon a setX. For each pointxofX, we have two important concepts:DEFINITION: Theorbitofx Xis the subset ofXO(x) :={g x|g G} : Thestabilizerofxis the subgroup ofGStab(x) ={g G|g x=x} :If a finite groupGacts on a setX, then for everyx X, we have|G|=|O(x)| |Stab(x)|.A. LetD4be the symmetry group of the square. Consider the natural action ofD4on the setR2byrotations and reflections of the whole space.(1) Complete the following chart which records, for different points of the setR2, 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 THESTABILIZER OF EVERY POINT IS A a groupGacts on a setX.

2 Letx X.(1) Prove that the Stabilizer ofxis asubgroupofG.(2) Use the Orbit - Stabilizer Theorem to prove that the cardinality of every Orbit divides|G|.2(3) LetGbe a group of order 17 and letXbe a set with 16 elements. Explain why there is nonon-trivial action ofGonX.[The trivial action is the one in whichg x=xfor allg Gand allx X.](1) We need to show thatStab(x) ={g G|g x=x}is a subgroup ofG. It suffices to check:(a)Stab(x)is non-empty; this is easy sinceeG x=x(by definition of action), soeG Stab(x).(b) Ifa,b Stab(x), thenab Stab(x). Also pretty easy: take arbitrarya,b Stab(x).We compute(a b) x=a( (b x)by definition of action. Sinceb x=x, this becomes(a b) x=a( (b x)) =a x;and sincea x=x, we conclude that(a b) x=x. Thusa b Stab(x).(c) ifa Stab(x), thena 1 Stab(x). For this, take arbitrarya Stab(x). This meansa x=x. Applya 1to both sides to geta 1 (a x) =a 1 definition of action,a 1 (a x) = (a 1 a) x=e x= we havea 1 x=x, anda 1 Stab(x). QED.(2) This is clear:|O(x)|divides|G|since|G|=|O(x)| |Stab(x)|.)

3 (3) Since 17 is prime, the only divisors of|G|are 1 and 17. So any Orbit of anyGaction can only have in this case,Xhas only 16 points total, no Orbit can have 17 points! So all orbits have one point. Thismeansg x=xfor allgand allx. The only such action is the trivial CANONICALACTION OF THESYMMETRICGROUP. LetSnact onXn={1,2,3,4,..,n}in the natural way; that is, Snacts oni Xby i= (i).(1) In the casen= 4, write out all the elements in the Stabilizer of the element4for the canonical action ofS4on{1,2,3,4}. Prove that the Stabilizer of4is isomorphic toS3.(2) Again for the canonical action ofS4on{1,2,3,4}, what is the Stabilizer of the element2? Explain why the stabilizeris isomorphic toS3.(3) Verify the Orbit - Stabilizer Theorem for some point for the canonical action ofS4on{1,2,3,4}. Why is the verifica-tion more or less the same for any other point?(4) WhenSnacts on a setXofnobjects in the canonical way, what is the decomposition ofXinto orbits? Describethis Stabilizer of somex Xexplicitly in words.

4 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 ofS3. Note that a permuation (bijection)of{1,2,3,4}that fixes the element4is basically the same as a permutation of{1,2,3}. SoStab(4) =S3.(2)Stab(2)is the permutations of{1,3,4}(with2fixed). This is the permutation group of three objects, henceisomorphic toS3.(3) The Orbit of each point is the whole set{1,2,3,4}, so|O(x)|= 4for allx {1,2,3,4}. Likewise the stabilizerof any point is the group of permutations of the other 3. So the stabilizers are all isomorphic toS3, which hascardinality. Since4 6 = 24 =|S4|, the Orbit - Stabilizer Theorem is confirmed.(4) There is only one Orbit , since any of thenobjects can be mapped to any other (in multiple ways!) under anelement inSn. So the decomposition into orbits is trivial: there is only one Orbit . The Orbit Stabilizer theoremtells us that|Sn|=|O(x)||Stab(x)|for anyx. So|Sn|=n!

5 =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 elementx. Of course, this isthe same as a permutation ofn 1objects, all the the numbers1,2,..nexceptx, which has to go to (x) =Sn ROTATIONALSYMMETRYGROUPS OFPLATONICSOLIDS. There are exactly five convex regular solid figures 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:3 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.

6 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 edgeson verticesFor each of the three actions, does it matterwhich pointx Xyou use to compute theorbit? Why is the order of the Stabilizer thesame for eachx Xin 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 Faces6424on edges12224on vertices8324(2) For the symmetry group of the tetrahedron we have:Action# Orbit # stab|G|on Faces4312on edges6212on vertices4312 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.

7 (3) For the Octahedron, we haveAction# Orbit # stab|G|on Faces8324on edges12224on vertices6424(4) For the symmetry group of the dodecahedron, we have:Action# Orbit # stab|G|on Faces12560on edges30260on vertices20360(5) For the symmetry group of the icosahedron, we have:Action# Orbit # stab|G|on Faces20360on edges30260on vertices125604E. LINEAR ACTIONS. Consider the action ofGL2(R)onR2by matrix multiplication (where elements ofR2are 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) ForA GL2(R2)and[xy] R2,we haveA [xy]=A[xy]. It is an action becauseI2[xy]=[xy]andA (B [xy]) =A(B[xy]) = (AB)[xy],by the associativity of matrix multiplication. So the two axioms of agroup action are satisfied.(2) WhatA=[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{[1b0d]|b,d, R,d6= 0}.

8 (3) The Orbit of[10]under matrix multiplication isR2\~0. Indeed, we can get an arbitrary non-zero[xy]bymultiplying[x by d][10]=[xy]. Note that as long as[xy]is not the zero vector, we can always findb,d Rsuch that the matrix[x by d]is RIGIDMOTIONS OF THEPLANE. The groupM2of rigid motions of the plane acts naturally on the set of all triangles inthe Euclidean plane. Explain how and whyM2is a group. What is the Orbit of a fixed triangleT? 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 thatM2is a group. A fixed trianglewill be sent to a congruent triangle underM2.

9 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 isD3, 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 toD2,C2,S2orZ2. The Stabilizer of a scalene triangle is{e}.G. Prove the Orbit - Stabilizer Theorem . [Hint: Start by finding a surjective mapG O(x); show that elements ofGhavethe same image under this map if and only if they are in the same coset of the stablizer ofx.]This will be assigned on Problem Set 11.