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Lecture 3 1 Structure of the Automorphism Group

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Lecture 3 1 Structure of the Automorphism Group Hypercube problems Lecture 3 October 24, 2012 Lecturer: Petr Gregor Scribe by: Otakar Trunda Updated: November 8, 2012 1 Structure of the automorphism group In this talk we take a closer look at the structure of Γ = Aut(Qn). We already know that n 8g 2 Γ 9!π 2 Sn 9!a 2 Z2 such that g = rπta where rπ 2 Rn; ta 2 Tn: A subgroup N of a group G is normal (denoted by N C G) if it is invariant under −1 conjugation; that is, gNg = N for every g 2 G. It is easy to see that Tn C Γ for every n ≥ 1 and Rn 6 Γ for every n ≥ 2. Definition 1 Let G, H be groups. The direct product of G and H (denoted by G × H) is the group on f(g; h) j g 2 G; h 2 Hg with the operation defined by (g1; h1)(g2; h2) = (g1g2; h1h2). Note that Γ cannot be expressed as Rn ×Tn since (rπ; ta)(rρ; tb) 6= (rπrρ; tatb). However, we can use a conjugation to express the composition since (rπ; ta)(rρ; tb) = (rπ(rρrρ−1 ); ta)(rρ; tb) = (rπrρ; (rρ−1 tarρ)tb): Definition 2 Let H, N be groups and let ' : H ! Aut(N) be a homomorphism. The semidirect product (external) of H and N with respect to ' (we write H n' N) is the group on −1 f(h; n); h 2 H; n 2 Ng with the operation defined by (h1; n1)(h2; n2) = (h1h2;'(h2 )(n1)n2): If H, N are subgroups of same group and N is normal then we can use the homomorphism −1 ' defined by '(h): n 7! hnh (a conjugation by h). Then we write simply H n N. n Fact 3 Aut(Qn) = Rn n Tn ' Sn n Z2 for every n ≥ 1. To verify that Aut(Qn) can be expressed in this way, let us see the composition −1 −1 (rπ; ta)(rρ; tb) = (rπrρ;'(rρ )(ta)tb) = (rπrρ; rρ tarρtb) = (rπrρ; taρ tb) = (rπρ; taρ⊕b) −1 since rρ tarρ : u 7! (uρ−1 ⊕ a)ρ = u ⊕ aρ. The neutral element is (rid; t0) and the inverse is −1 −1 −1 −1 −1 −1 −1 −1 −1 (r ; t ) = (r ; ['(r )(t )] ) = (r ; [r t r ] ) = (r ; r t r ) = (r −1 ; t ): π a π π a π π a π π π a π π aπ−1 Aut(Qn) is called the hyperoctahedral group and it is also the group of symmetries of a cross-polytope (a dual polytope to the hypercube). Alternatively, Aut(Qn) can be written as the wreath product Aut(Qn) ' S2 o Sn, see the following definition. 3-1 Definition 4 Let A, H be groups, H acting on V . Let K be the direct product of copies Q of A indexed by elements of V (i.e. K = x2V Ax, K is called a base.) The (unrestricted) wreath product of A and H (denoted by A oV H) is A oV H = H n K. If H = Sn we take V = [n] with the natural action of H on V and write simply A o H. In the first lecture we defined the folded cube FQn and the augmented cube AQn. It is n n+3 known that Aut(FQn) ' Sn+1 n Z2 [8] and jAut(AQn)j = 2 [5]. Problem 1 What is the structure of Aut(AQn)? 2 Distance transitivity In this section we inspect how Aut(Qn) acts on ordered pairs of vertices. Definition 5 Let Γ be a group acting on V and let x; y 2 V . The orbital of (x; y) is a set Γ(x; y) = f(g(x); g(y)) j g 2 Γg: In fact, it is an orbit in Γ with the action on V × V induced by the action on V . Let n ≥ 1 be fixed, Γ = Aut(Qn) with the action on V = V (Qn). For 0 ≤ d ≤ n let Dd = f(u; v) 2 V × V j dH (u; v) = dg: n n Since every vertex u in Qn has exactly d vertices at distance d and there are 2 choices n n for u, we obtain jDdj = d 2 . Moreover, as every automorphism preserves the distances, it is clear that Γ(x; y) ⊆ Dd for every x; y 2 V where d = dH (x; y). The following lemma shows that actually the equality holds. n n Lemma 6 For every x; y 2 V it holds jΓ(x; y)j = d 2 where d = dH (x; y). Proof From the orbit-stabilizer theorem for the group Γx we have jΓx;yj:jΓxyj = jΓxj (1) where Γx;y = Γ(x;y) = Γx \ Γy. From a study of stabilizers in the last lecture n jΓ j = n! and Γ y = fz j d (x; z) = d (x; y)g; so jΓ yj = : (2) x x H H x d From (1) and (2) it follows that jΓ(x;y)j = (n − d)!n!: (3) By another use of the orbit-stabilizer theorem, jΓ(x;y)j:jΓ(x; y)j = jΓj: (4) n n n From (3), (4) and the fact that jΓj = n!2 we obtain jΓ(x; y)j = d 2 . Before we state the result let us put another definition. 3-2 Definition 7 A graph G = (V; E) is • distance-transitive if for every x; y; u; v 2 V with dG(x; y) = dG(u; v) there is g 2 Aut(G) such that g(x) = u and g(y) = v, • arc-transitive if for every x; y; u; v 2 V with xy; uv 2 E there is g 2 Aut(G) such that g(x) = u and g(y) = v (we also say that G is symmetric), • edge-transitive if for every xy; uv 2 E there is g 2 Aut(G) such that g(xy) = uv, • vertex-transitive if for every x; y 2 V there is g 2 Aut(G) such that g(x) = y. Clearly, distance-transitivity implies arc-transitivity which implies both edge-transitivity and vertex-transitivity. There are graphs that are edge-transitive but not vertex-transitive. Theorem 8 For every n ≥ 1 the hypercube Qn is distance-transitive. Proof From Lemma 6 it directly follows that Γ(x; y) = Dd for every x; y 2 V (Qn) where d = dH (x; y). A problem on hypercube automorphisms n A d-th level in Qn is the set Ld = fu 2 Z2 j juj = dg. Let n be even and A ⊆ Ln=2 be n n a basis of F2 . For a 2 Ln=2 let R(a) = fr 2 Rn j r(a) = a ⊕ 1g. Then C ⊆ Z2 is called A-compatible if for every a 2 A there is ra 2 R(a) such that ra(C) = C. Examples of trivial A-compatible sets are unions of some levels. Problem 2 (the weakest form) For some basis A in Ln=2 where n is even, find a non- trivial A-compatible set in some level Ld. Problem 3 (the strongest form) For every n ≡ 2 (mod 4), every basis A 2 Ln=2 find all (nontrivial) A-compatible sets in each level Ld. 3 Symmetry breaking In this section we inspect how to break all the symmetries of Qn. A first approach is by labeling the vertices. Let G = (V; E) be a graph. A mapping Φ : V ! f1; 2; :::; rg is called an r-labeling of G. Then (G; Φ) is called a labeled graph. Its automorphisms must preserve the labeling; that is, Aut(G; Φ) = fg 2 Aut(G) j Φ(u) = Φ(g(u)) for every u 2 V g. A labeling Φ of G is called distinguishing if Aut(G; Φ) = fidg; that is, (G; Φ) is rigid. The distinguishing number D(G) of G is the minimal number of labels in a distinguishing labeling of G. Theorem 9 (Bogstad, Cowen [1]) D(Q1) = 2, D(Q2) = D(Q3) = 3, and D(Qn) = 2 for every n ≥ 4. 3-3 Proof Clearly D(Qn) ≥ 2 since Qn is not rigid for every n ≥ 1. It is easy to see that D(Q1) = 2, D(Q2) = 3. We leave as a homework to verify D(Q3) = 3. Pi Let n ≥ 4. We define S = fsi = j=1 ej j 0 ≤ i ≤ ng and t = e1 ⊕ en. Furthermore, for u 2 V (Qn) let Φ(u) = 1 if u 2 S [ ftg, and Φ(u) = 2 otherwise. Then Φ is distinguishing 2-labeling since every automorphism of (Qn; Φ) has to map the set S to itself and cannot revert the order of vertices in S as s1t 2 E(Qn) and sn−1t 62 E(Qn). We may require, in addition, that the distinguishing labeling is proper. The distin- guishing chromatic number χD(G) of a graph G is the minimal number of labels in a distinguishing proper labeling of G; that is, adjacent vertices have distinct labels. Theorem 10 (Choi et al. [4], Kl¨ockl [7]) χD(Q1) = 2, χD(Q2) = χD(Q3) = χD(Q4) = 4, and χD(Qn) = 3 if n ≥ 5. Another approach to the symmetry breaking is by finding a set of vertices that every automorphism has to preserve point-wise. Let G = (V; E) be a graph. A set S ⊆ V is point-wise preserved by g 2 Aut(G) if g(s) = s for every s 2 S. If S is point-wise preserved only by the trivial automorphism, the we call S a symmetry breaking set.
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