HAMILTON CYCLES IN CAYLEY GRAPHS

Hamiltonicity of (2, s, 3)- University of Primorska July, 2011

Hamiltonicity of (2, s, 3)- Lovász, 1969 Does every connected vertex-transitive graph have a Hamilton path?

Hamiltonicity of (2, s, 3)- Hamiltonicity of vertex - transitive graphs All known VTG have Hamilton path. Only 4 CVTG (having at least three vertices) not having a Hamilton cycle are known to exist. None of them is a Cayley graph. Folklore conjecture Every connected Cayley graph contains a Hamilton cycle.

Hamiltonicity of (2, s, 3)- The current situation Hamilton cycles (paths) are known to exist in these cases: VTG of order p, 2p, 3p, 4p, 5p, 6p, 2p 2, p k (for k 4) (Alspach, Chen, Du, Marušič, Parsons, Šparl, KK, etc.); CG of p-groups (Witte); VTG having groups with a cyclic commutator subgroup of order p k (Durenberger, Gavlas, Keating, Marušič, Morris, Morris-Witte, etc.). CG Cay(G, {a, b, a b }), where a is an involution (Pak, Radoičić). Cubic CG Cay(G, S), where S = {a, b, c} and a 2 = b 2 = c 2 = 1 and ab = ba (Cherkassoff, Sjerve). Cubic CG Cay(G, S), where S = {a, x, x 1 } and a 2 = 1, x s = 1 and (ax) 3 = 1 (Glover, Marušič). CG on groups whose orders have small prime factorization. This shows, among other, that every CG on a group of order n < 120, n / {48, 72, 80, 96, 108, 112}, has a Hamilton cycle (KK, Marušič, Morris, Morris, Šparl, 2010). and in some other cases.

Hamiltonicity of (2, s, 3)- Definition Given a group G and a subset S of G \ {1}, S = S 1, the Cayley graph Cay(G, S) has vertex set G and edges of the form {g, gs} for all g G and s S. The group G acts on itself by the left multiplication. This action may be viewed as the action of G on its Cayley graph. Sabidussi characterized as follows. A graph is a Cayley graph of a group G if and only if its automorphism group contains a regular subgroup isomorphic to G.

Hamiltonicity of (2, s, 3)- Cubic If Cay(G, S) is a cubic Cayley graph then S = 3, and either S = {a, b, c a 2 = b 2 = c 2 = 1}, or S = {a, x, x 1 a 2 = x s = 1} where s 3.

Hamiltonicity of (2, s, 3)- (2, s, t)- Definition Let G = a, x a 2 = x s = (ax) t = 1,... be a group. Then the Cayley graph X = Cay(G, S) of G with respect to the generating set S = {a, x, x 1 } is called (2, s, t)-cayley graph. Examples: - Z 6 = a, x a 2 = x 6 = (ax) 3 = 1,... - A 4 = a, x a 2 = x 3 = (ax) 3 = 1 - A 5 = a, x a 2 = x 5 = (ax) 3 = 1 - S 4 = a, x a 2 = x 4 = (ax) 3 = 1 - Most of simple groups - Hurwitz groups

Hamiltonicity of (2, s, 3)- Hamiltonicity of (2, s, 3)- Hamiltonicity of (2, s, 3)- Glover, Marušič, J. Eur. Math. Soc., 2007 Let s 3 be an integer and let X = Cay(G, S) be a (2, s, 3)-Cayley graph of a group G. Then X has a Hamilton cycle when G is congruent to 2 modulo 4, and a cycle of length G 2, and also a Hamilton path, when G is congruent to 0 modulo 4. Glover, KK, Marušič, J. Algebraic Combin., 2009 Let s 0 (mod 4) 4 be an integer. Then a (2, s, 3)-Cayley graph X has a Hamilton cycle. Glover, KK, Malnič, Marušič, 2011 Let s be an odd integer. Then a (2, s, 3)-Cayley graph X has a Hamilton cycle.

Henry H. Glover Hamiltonicity of (2, s, 3)- Hamiltonicity of (2, s, 3)-

Hamiltonicity of (2, s, 3)- Hamiltonicity of (2, s, 3)- Hamilton trees on surfaces approach the beginning H. H. Glover and T. Y. Yang, A Hamilton cycle in the Cayley graph of the (2, p, 3) presentation of PSL 2 (p), Discrete Math. 160 (1996), 149 163.

Hamiltonicity of (2, s, 3)- Hamiltonicity of (2, s, 3)- A Hamilton cycle given as the boundary of a Hamilton tree in the Cayley graph for the (5, 3, 2)-generation of PSL 2 (5).

Cayley map Hamiltonicity of (2, s, 3)- Hamiltonicity of (2, s, 3)- (2, s, 3)-Cayley graph Cay(G, {a, x, x 1 }) has a canonical Cayley map given by the embedding in the closed orientable surface of genus 1 + (s 6) G /12s, whose set of faces consists of G /s disjoint s-gons and G /3 hexagons. This map is given by taking, at each vertex, the same rotation of the edges labelled by x, a, x 1, and results in one s-gon and two hexagons adjacent at each vertex.

Hamiltonicity of (2, s, 3)- Hamiltonicity of (2, s, 3)-

Hamiltonicity of (2, s, 3)- Hamiltonicity of (2, s, 3)- Example: (2, 6, 3)-Cayley graph on Z 6 Z 6 = a, x a 2 = x 6 = (ax) 3 = 1,... where a = 3 and x = 1.

Soccer ball Hamiltonicity of (2, s, 3)- Hamiltonicity of (2, s, 3)- (2, 5, 3)-Cayley graph on A 5 = a, x a 2 = x 5 = (ax) 3 = 1.

Hamiltonicity of (2, s, 3)- Hamilton trees on surfaces approach Hamiltonicity of (2, s, 3)-

Hamiltonicity of (2, s, 3)- Hamiltonicity of (2, s, 3)- A necessary condition for the existence of a Hamilton cycle in a (2, s, 3)-Cayley graph given as the boundary of the Hamiltonian tree HEAD Equation (Glover, Yanq, 1996) If there is a Hamilton cycle given as the boundary of a tree of α s-gons and β hexagons in the Cayley map of the (2, s, 3)-generation of G, then (s 2)α + 4β = G 2.

Hamiltonicity of (2, s, 3)- Hamiltonicity of (2, s, 3)- How to find a Hamiltonian tree of faces?

Hexagon graph Hamiltonicity of (2, s, 3)- Hamiltonicity of (2, s, 3)- With the Cayley map of a (2, s, 3)-Cayley graph X on a group G we associate the so-called hexagon graph Hex(X ) having all the hexagons in X as vertices, where two hexagons are adjacent in Hex(X ) whenever they share an edge in X.

Hamiltonicity of (2, s, 3)- Properties of the hexagon graph Hamiltonicity of (2, s, 3)- Let X be a (2, s, 3)-Cayley graph on a group G = a, x a 2, x s, (ax) 3,.... Then the hexagon graph has the following properties Hex(X ): - Hex(X ) is of order G /3. - Hex(X ) is isomorphic to the orbital graph of the left action of G on the set of left cosets H of the subgroup H = ax, arising from the suborbit {ah, axah, axaxah} = {ah, x 1 H, ax 2 H} of length 3. More precisely, the graph has vertex set H, with adjacency defined as follows: a coset yh, y G, is adjacent to the three cosets yah, yx 1 H and yax 2 H.

Hamiltonicity of (2, s, 3)- Example: G 2(mod 4) Hamiltonicity of (2, s, 3)- G = Z 6 with a (2, 6, 3)-presentation a, x a 2 = x 6 = (ax) 3 = 1, where a = 3 and x = 1.

Hamiltonicity of (2, s, 3)- Example: G 2(mod 4) Hamiltonicity of (2, s, 3)- G = S 3 Z 3 with a (2, 6, 3)-presentation a, x a 2 = x 6 = (ax) 3 = 1,..., where a = ((12), 0) and x = ((13), 1).

Hamiltonicity of (2, s, 3)- Example: G 0(mod 4) Hamiltonicity of (2, s, 3)- G = A 4 with a (2, 3, 3)-presentation a, x a 2 = x 3 = (ax) 3 = 1, where a = (1 2)(3 4) and x = (1 2 3).

Hamiltonicity of (2, s, 3)- Example: G 0(mod 4) Hamiltonicity of (2, s, 3)- G = S 4 with a (2, 4, 3)-presentation a, x a 2 = x 4 = (ax) 3 = 1, where a = (12) and x = (1234).

Hamiltonicity of (2, s, 3)- Example: G 0(mod 4) Hamiltonicity of (2, s, 3)- (2, 8, 3)-Cayley graph on G = Q 3 S 3

Hamiltonicity of (2, s, 3)- Cyclically stable subsets Hamiltonicity of (2, s, 3)-

Hamiltonicity of (2, s, 3)- Payan, Sakarovitch, 1975 Hamiltonicity of (2, s, 3)- Payan, Sakarovitch, 1975 Let X be a cyclically 4-edge-connected cubic graph of order n, and let S be a maximum cyclically stable subset of V (X ). Then S = (3n 2)/2 and more precisely, the following hold. If n 2 (mod 4) then S = (3n 2)/4, and X [S] is a tree and V (X ) \ S is an independent set of vertices; If n 0 (mod 4) then S = (3n 4)/4, and either X [S] is a tree and V (X ) \ S induces a graph with a single edge, or X [S] has two components and V (X ) \ S is an independent set of vertices.

Hamiltonicity of (2, s, 3)- Cyclically k-edge-connected graphs Hamiltonicity of (2, s, 3)- Cycle-separating subset A subset F E(X ) of edges of X is said to be cycle-separating if X F is disconnected and at least two of its components contain cycles. Cyclically k-edge-connected graphs A graph X is cyclically k-edge-connected, if no set of fewer than k edges is cycle-separating in X.

Is it c.4.c? Hamiltonicity of (2, s, 3)- Hamiltonicity of (2, s, 3)-

Is it c.4.c? Hamiltonicity of (2, s, 3)- Hamiltonicity of (2, s, 3)- No.

Is it c.4.c? Hamiltonicity of (2, s, 3)- Hamiltonicity of (2, s, 3)-

Is it c.4.c? Hamiltonicity of (2, s, 3)- Hamiltonicity of (2, s, 3)- Yes. It is in fact c.5.c.

Hamiltonicity of (2, s, 3)- Hamiltonicity of (2, s, 3)- of cubic arc-transitive graphs Nedela, Škoviera, 1995 of a vertex-transitive graph equals its girth. Proposition Let Y be a cubic arc-transitive graph. Then: (i) The girth g(y ) of Y is at least 6; or (ii) Y is one of the following graphs: the theta graph θ 2, K 4, K 3,3, Q 3, GP(5, 2) or GP(10, 2). Proof: exercise.

Hamiltonicity of (2, s, 3)- Hamiltonicity of (2, s, 3)- Hamiltonicity of a (2, s, 3)-Cayley graph X on a group G By previous examples we can conclude: If G 2 (mod 4) then Payan-Sakarovitch result implies the existence of a Hamilton cycle in X. If G 0 (mod 4) then Payan-Sakarovitch result implies the existence of a Hamilton cycle in X.

Hamiltonicity of (2, s, 3)- Hamiltonicity of (2, s, 3)- (2, s, 3)- of order 0 (mod 4) For a (2, s, 3)-Cayley graph of order 0 (mod 4) three cases can occur: s 0 (mod 4). s 2 (mod 4). s odd.

Hamiltonicity of (2, s, 3)- Hamiltonicity of (2, s, 3)- (2, s, 3)- of order 0 (mod 4) For a (2, s, 3)-Cayley graph of order 0 (mod 4) three cases can occur: s 0 (mod 4). s 2 (mod 4). s odd. HEAD Equation (Glover, Yanq, 1996) If there is a Hamilton cycle given as the boundary of a tree of α s-gons and β hexagons in the Cayley map of the (2, s, 3)-generation of G, then (s 2)α + 4β = G 2.

Hamiltonicity of (2, s, 3)- Hamiltonicity of (2, s, 3)- Hamiltonicity of (2, s, 3)-, s 0 (mod 4) Glover, KK, Marušič, 2009 Let s 0 (mod 4) 4 be an integer. Then a (2, s, 3)-Cayley graph X has a Hamilton cycle. Glover-Marušič method. Classification of cubic ATG of girth 6. Results on cubic ATG admitting a 1-regular subgroup.

Hamiltonicity of (2, s, 3)- Modification process in (2, 4k, 3) case Hamiltonicity of (2, s, 3)- We consider the graph of hexagons HexX. The graph of hexagons HexX is modified so as to get a graph ModX with 2 (mod 4) vertices and cyclically 4-edge-connected, so to be able to get by [Payan, Sakarovitch, 1975] a tree whose complement is an independent set of vertices giving rise to a Hamilton cycle in X.

Hamiltonicity of (2, s, 3)- Modification process in (2, 4k, 3) case Hamiltonicity of (2, s, 3)- The local structure

Hamiltonicity of (2, s, 3)- Modification process in (2, 4k, 3) case Hamiltonicity of (2, s, 3)- Graph of hexagons Hex(X )

Hamiltonicity of (2, s, 3)- Modification process in (2, 4k, 3) case Hamiltonicity of (2, s, 3)- Graph of hexagons Hex(X ) Hex(X ) is of order G /3.

Hamiltonicity of (2, s, 3)- Modification process in (2, 4k, 3) case Hamiltonicity of (2, s, 3)-

Hamiltonicity of (2, s, 3)- Modification process in (2, 4k, 3) case Hamiltonicity of (2, s, 3)- Modified graph ModX is of order V (HexX ) (3s 6). Since V (HexX ) 0(mod 4) we have that ModX is of order 2(mod 4).

Hamiltonicity of (2, s, 3)- Modification process in (2, 4k, 3) case Hamiltonicity of (2, s, 3)- HexX is cubic symmetric graph. Lemma If Hex(X ) is cyclically c-edge-connected for c 5 then it is Θ 2, K 4, K 3,3, Q 3 or GP(10, 2). If Hex(X ) cyclically 6-edge-connected then it is an Ik n (r)-path or a suitable cover of the cube or some Ik n(r)-path. If Hex(X ) is cyclically 7-edge-connected then s = 7. In all other cases Hex(X ) is cyclically 8-edge-connected.

Hamiltonicity of (2, s, 3)- Modification process in (2, 4k, 3) case Hamiltonicity of (2, s, 3)- Using this lemma we show ModX is cyclically 4-edge-connected. If cyclical edge-connectivity c of HexX is at most 5 then HexX is Θ 2 or Q 3 ; If c = 6 then it is first proved that HexX is some Ik n (r)-path and then shown that the modification process does not reduce c by more than 2. The case c = 7 is impossible; If c is at least 8, then it is shown that the modification process does not reduce c by more than 4. Applying [Payan, Sakarovitch] to ModX gives a Hamilton cycle in X.

Example Hamiltonicity of (2, s, 3)- Hamiltonicity of (2, s, 3)-

Hamiltonicity of (2, s, 3)- Hamiltonicity of (2, s, 3)- Hamiltonicity of (2, s, 3)-, s odd x is corefree in G = a, x a 2 = x s = (ax) 3 = 1,... : A method similar to the method used in s 0 (mod 4) case gives us a Hamilton cycle as a boundary of a Hamilton tree of faces consisting of hexagons and two s-gons. x is not corefree in G = a, x a 2 = x s = (ax) 3 = 1,... : Factor Group Lemma. Factor Group Lemma Let G be a group with a generating set S such that S 1 = S and 1 S, and let N be a cyclic normal subgroup of G of index n. If [s 1 N,..., s n N] is a Hamilton cycle in Cay(G/N, {sn, s S}) and the product s 1 s 2 s n generates N, where s 1,..., s n S, then Cay(G, S) has a Hamilton cycle.

Hamiltonicity of (2, s, 3)- Hamiltonicity of (2, s, 3)- Hamiltonicity of (2, s, 3)-, s odd

Hamiltonicity of (2, s, 3)- Hamiltonicity of (2, s, 3)- Hamiltonicity of (2, s, 3)-, s 2 (mod 4) For s 2 (mod 4) and G 0 (mod 4) the HEAD equation has no solution, and thus in order to prove the existence of Hamilton cycles in this remaining case a new approach needs to be developed.

Hamiltonicity of (2, s, 3)- Hamiltonicity of (2, s, 3)- Thanks!