EXERCISES ON THE OUTER AUTOMORPHISMS OF S6 AARON LANDESMAN 1. INTRODUCTION In this class, we investigate the outer automorphism of S6. Let’s recall some definitions, so that we can state what an outer automorphism is. Definition 1.1. Recall that the symmetric group on n elements is the group of all permutations of n elements. So elements of the symmetric group are precisely bijections f1, . , ng ! f1, . , ng with multiplication in Sn corresponding to composition of maps. Definition 1.2. An automorphism of a group G is an group isomorphism f : G ! G. Exercise 1.3. Let G be a group and g 2 G be an element. Check that the map fg : G ! G h 7! ghg−1 is a group automorphism.1 Definition 1.4. An automorphism f : G ! G is called inner if it is of the form fg (conjugation by g) as defined in Exercise 1.3. An automorphism f : G ! G is called outer if it is not inner. Of course every group has at most #G inner automorphisms (though some of these may be the same), at most one for each g 2 G. A natural question to ask is: Question 1.5. Which symmetric groups have outer automorphisms? For those that do have outer automorphisms, how many do they have? The answer is quite simple: Theorem 1.6. (1) The symmetric group S6 is the only symmetric group with an outer automorphism. 0 (2) Further, any two outer automorphisms f : S6 ! S6 and f : S6 ! S6 are related by an inner automorphism. That is, there some s 2 S6 with 0 f = fs ◦ f . 1 2 AARON LANDESMAN We’ll come back to proving this theorem later, but for most of the course, we’ll concentrate on investigating various descriptions of this mysterious outer automorphism of S6. EXERCISES ON THE OUTER AUTOMORPHISMS OF S6 3 2. MYSTIC PENTAGONS We now encounter our first incarnation of the outer automorphism of S6 in terms of mystic pentagons. Recall that a complete graph on n elements is the graph with n points and every pair of points joined by an edge. Exercise 2.1. (1) Show there are six ways to color the edges of a complete graph on 5 vertices with two colors so that the edges of one color form a 5-cycle. (See Figure 1.) Here we consider two colorings equivalent if one can move from one to the other by reversing the labels of the colors. We call these six colorings the six mystic pentagons. (2) Verify that if the edges of one color form a 5-cycle, the edges of the other color also form a 5-cycle. S{1,...,6}, sof is injective and hence an isomorphism; andi is not a usual inclusion, sof is not inner), but for the sake of simplicity we do so by way of a second description. 1 1 1 5 2 5 2 5 2 4 3 4 3 4 3 a b c 1 1 1 5 2 5 2 5 2 4 3 4 3 4 3 d e f FIGURE 1. The six mystic pentagons, with black and white (dashed) edges FIGURE 1. The six mystic pentagons 1.2. Second description of the outer automorphism: labeled triangles. We now make 6∈{1,...,6} this constructionWe label the more six mystic symmetric, pentagons not distinguishing by a, b, c, d, e, thef as element in Figure 1. For T a. Consider the 6 = 20triangles on six vertices labeled {1,...,6}. There are six ways of dividing the set3 of size n, we use ST ' Sn to denote the symmetric group acting on the trianglesset T. into two sets of 10so that (i) any two disjoint triangles have opposite colors, and (ii) every tetrahedron has two triangles of each color. (The bijection between these and the mystic pentagonsa,...,f is as follows. The triangle 6AB is colored the same as edgeAB. The triangleCDE(6 6=C,D,E ) is colored the opposite of the “complementary” edgeAB, where {A,B} = {1,...,5}−{C,D,E}.) TheS 6-action on this set is the outer automor- phism ofS 6. (Reason: (12) induces a nontrivial permutation (ad)(bc)(ef) of the mystic ∼ pentagons, so the induced mapS 6 →S {a,...,f } = S6 is injective and hence an isomorphism. But (12) does not induce a transposition on {a,...,f}, so the automorphism is not inner.) This isomorphismS {1,...,6} →S {a,...,f } is inverse to the isomorphismf of§1.1. 1.3. Another description: labeled icosahedra. Here is another description, which is pleasantlyS 6-symmetric. Up to rotations and ref ections, there are twelve ways to number the vertices of an icosahedron1through6, such that antipodal vertices have the same label. Each icosahedron gives ten triples in {1,...,6}, corresponding to the vertices around its faces. These twelve icosahedra come in six pairs, where two icosahedra are “opposite” if they have no triples in common. (It is entertaining to note that if an icosahe- dron is embedded in Q(φ)3 with vertices at (±1,±φ,0), (0,±1,±φ), and (±φ,0,±1), then conjugation in Gal(Q(φ)/ Q) sends the icosahedron to its opposite. Hereφ is the golden section.) ThenS 6 acts on these six pairs, and this is the outer automorphism. One may show this via bijections to the descriptions of§1.1 and§1.2. Each pair of mystic icosahedra corresponds to two-coloring the triangles in {1,...,6}, as in§1.2. For the bijection to§1.1, the cyclic order of the vertices around vertex6gives a mystic pentagon. (This provides a 2 4 AARON LANDESMAN Exercise 2.2. (1) Use Exercise 2.1 to describe a map i : Sf1,2,3,4,5g ! 2 Sfa,b,c,d,e,fg = S6. (2) Check your understanding of the previous part by computing i((12)), where (12) denotes the permuation sending 1 7! 2, 2 7! 1, 3 7! 3, 4 7! 4, 5 7! 5. (3) Check that this realizes S5 as a subgroup of S6 (i.e., check that the 3 kernel of i is only the identity e 2 S5). (4) Show that the inclusion i : Sf1,2,3,4,5g ! Sfa,b,c,d,e,fg is not induced by an inclusion j : f1, 2, 3, 4, 5g ! fa, b, c, d, e, fg. More precisely, show there is no inclusion of sets j so that j(s(k)) = i(s)(j(k)) for k 2 f1, 2, 3, 4, 5g , s 2 Sf1,2,3,4,5g. Exercise 2.3. Let f1, 2, 3, 4, 5, 6g label the 6 cosets of Sfa,b,c,d,e,fg/i(S5). Show the action of Sfa,b,c,d,e,fg on these 6 cosets induces a map f : Sfa,b,c,d,e,fg ! Sf1,2,3,4,5,6g. Remark 2.4. In the next class, we will relate the mystic pentagons to other problems in order to verify that the map f of Exercise 2.3 is indeed an outer automorphism. Your homework is to finish up through the previous problem. Also, check out the wobbly nobbly, a dog chew toy exhibiting the outer automorphism! If you have more time, and have not seen the bijection between conjugacy classes and partitions in Sn, we recommend you ask for Appendix C and do the exercises there. Once you have finished that attempt the following. For the following exercise, it will help to look at a nobbly wobbly. I don’t completely understand the answer to the following question myself. It is a bit open ended, but see what you can figure out. Exercise 2.5 (Loosely phrased, optional, very challenging exercise). Can you 4 figure out how the nobbly wobbly relates to the outer automorphism of S6? I don’t know a clean way to solve the following exercise directly. See if you can find one. We’ll see an indirect way to check the next exercise tomorrow. Exercise 2.6 (Optional very challenging exercise). Show directly that the map f of Exercise 2.3 is actually an outer automorphism. 5 EXERCISES ON THE OUTER AUTOMORPHISMS OF S6 5 3. LABELED TRIANGLES The action on mystical pentagons, although quite beautiful is not symmet- ric, since it preferences a certain subset of 5 of the 6 elements, corresponding to the 5 vertices of the pentagon. We now discuss a more symmetric con- struction. 6 Definition 3.1. Note that are 20 = (3) triangles on 6 vertices labeled f1, 2, 3, 4, 5, 6g A configuration of labeled triangles is a way of dividing these 20 triangles into two sets of 10 (which we think of as colored white and black, up to swapping of the colors) so that (1) two disjoint triangles are are colored different colors (2) every tetrahedron (subset of 4 vertices) has two triangles of each color Fact 3.2. There are six configurations of labeled triangles. This can be checked via a tedious combinatorial exercise, which we recommend skipping. Exercise 3.3. Construct a bijection between mystic pentagons a,b,c,d,e,f and the 6 groupings of labeled triangles as follows: (1) Given a mystic pentagon, construct a 2-coloring of the 20 triangles on 6 vertices as follows: If edge AB is colored black for A, B 2 f1, 2, 3, 4, 5g, coloring the triangle 6AB black, and the triangle CDE white (where CDE = f1, 2, 3, 4, 5g − fA, Bg). If edge AB is colored white for A, B 2 f1, 2, 3, 4, 5g, coloring the triangle 6AB white, and the triangle CDE black (where CDE = f1, 2, 3, 4, 5g − fA, Bg).