[PPT] - On non-normal 4-valent arc transitive dihedrants Aleksander Malni c PowerPoint Presentation

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On non-normal 4-valent arc transitive dihedrants

Aleksander Malniˇ c University of Ljubljana

Joint work with Istv´ an Kov´ acs and Boˇ stjan Kuzman Banff, Canada November, 2008

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Dihedrants and Bicirculants

An n-dihedrant is a Cayley graph of a dihedral group Dn. An n-bicirculant is a regular Zn-cover of a dipole. n-dihedrant ⇒ n-bicirculant It is often convenient if we consider dihedrants as bicirculants.

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Dihedrants and Bicirculants

An n-dihedrant is a Cayley graph of a dihedral group Dn. An n-bicirculant is a regular Zn-cover of a dipole. n-dihedrant ⇒ n-bicirculant It is often convenient if we consider dihedrants as bicirculants. BCn1, BCn2, BCn3, BCn4 Bicirculants of valency 4 fall into 4 classes,

wrt. the number of perfect matchings between the two orbits of Zn.

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4 valent edge transitive bicirculants

BCn1, BCn3 No such graphs. Kov´ acs, Kuzman, M., Wilson, 2008

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4 valent edge transitive bicirculants

BCn1, BCn3 No such graphs. Kov´ acs, Kuzman, M., Wilson, 2008 BCn2 Rose window graphs. Kov´ acs, Kutnar, Maruˇ siˇ c, 2008 Some of these are dihedrans. Generalized Rose window graphs. Still open. Conjecture: empty.

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4 valent edge transitive bicirculants

BCn1, BCn3 No such graphs. Kov´ acs, Kuzman, M., Wilson, 2008 BCn2 Rose window graphs. Kov´ acs, Kutnar, Maruˇ siˇ c, 2008 Some of these are dihedrans. Generalized Rose window graphs. Still open. Conjecture: empty.

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4 valent edge transitive bicirculants

BCn1, BCn3 No such graphs. Kov´ acs, Kuzman, M., Wilson, 2008 BCn2 Rose window graphs. Kov´ acs, Kutnar, Maruˇ siˇ c, 2008 Some of these are dihedrans. Generalized Rose window graphs. Still open. Conjecture: empty. BCn4 X ∈ BCn4 is necessarily a dihedrant. Normal, Dn ⊳ Aut(X). Kov´ acs, Kuzman, M., Wilson, 2008 Non-normal. Kov´ acs, Kuzman, M., 2008 This talk.

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Non-normal arc-transitive BCn4

I. The lexicographic product Cn[2K1].

n ≥ 4 even, S = {b, ba, ba

n 2 , ba n 2 +1}

(in picture, n = 16).

II. The graph K5,5 − 5K2.

n = 5, S = {b, ba, ba2, ba3}.

III. The non-incidence graph of

PG(2, 2). n = 7, S = {b, ba, ba2, ba4}.

IV. The incidence graph of PG(2, 3).

n = 13, S = {b, ba, ba3, ba9}.

V. A 2-cover of the graph III.

n = 14, S = {b, ba, ba4, ba6}.

VI. A 3-cover of the graph II.

n = 15, S = {b, ba, ba3, ba7} Table 1: Non-normal 4-valent arc-transitive dihedrants satisfying the bipartition condition.

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References

Arc transitive dihedrants val 4, 1-regular

C. Q. Wang, M. Xu, Non-normal one-regular and 4-valent Cayley

graphs of dihedral groups D2n, European J. Combin. 27 (2006), 750–766.

C. Q. Wang, Z. Y. Zhou, One-regularity of 4-valent and normal

Cayley graphs of dihedral groups D2n, Acta Math. Sinica (Chin. ser.) 49 (2006), 669–678.

Y. H. Kwak, Y. M. Oh, One-regular normal Cayley graphs on

dihedral groups of valency 4 or 6 with cyclic vertex stabilizer, Acta

Math. Sinica (Engl. ser.) 22 (2006), 1305–1320.

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References

Arc transitive dihedrants val 4, 1-regular

C. Q. Wang, M. Xu, Non-normal one-regular and 4-valent Cayley

graphs of dihedral groups D2n, European J. Combin. 27 (2006), 750–766.

C. Q. Wang, Z. Y. Zhou, One-regularity of 4-valent and normal

Cayley graphs of dihedral groups D2n, Acta Math. Sinica (Chin. ser.) 49 (2006), 669–678.

Y. H. Kwak, Y. M. Oh, One-regular normal Cayley graphs on

dihedral groups of valency 4 or 6 with cyclic vertex stabilizer, Acta

Math. Sinica (Engl. ser.) 22 (2006), 1305–1320.

Lemma X ∈ BCn4 non-normal ⇒ X not 1-regular.

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References

Arc transitive dihedrants val 4, 1-regular

C. Q. Wang, M. Xu, Non-normal one-regular and 4-valent Cayley

graphs of dihedral groups D2n, European J. Combin. 27 (2006), 750–766.

C. Q. Wang, Z. Y. Zhou, One-regularity of 4-valent and normal

Cayley graphs of dihedral groups D2n, Acta Math. Sinica (Chin. ser.) 49 (2006), 669–678.

Y. H. Kwak, Y. M. Oh, One-regular normal Cayley graphs on

dihedral groups of valency 4 or 6 with cyclic vertex stabilizer, Acta

Math. Sinica (Engl. ser.) 22 (2006), 1305–1320.

Lemma X ∈ BCn4 non-normal ⇒ X not 1-regular. 2-arc transitive dihedrants

S. F. Du, M., D. Maruˇ

siˇ c, Classification of 2-arc transitive dihedrants, J.Combin. Theory B, in print.

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Non-normal arc-transitive BCn4

I. The lexicographic product Cn[2K1].

n ≥ 4 even, S = {b, ba, ba

n 2 , ba n 2 +1}

(in picture, n = 16).

II. The graph K5,5 − 5K2.

n = 5, S = {b, ba, ba2, ba3}.

III. The non-incidence graph of

PG(2, 2). n = 7, S = {b, ba, ba2, ba4}.

IV. The incidence graph of PG(2, 3).

n = 13, S = {b, ba, ba3, ba9}.

V. A 2-cover of the graph III.

n = 14, S = {b, ba, ba4, ba6}.

VI. A 3-cover of the graph II.

n = 15, S = {b, ba, ba3, ba7} Table 1: Non-normal 4-valent arc-transitive dihedrants satisfying the bipartition condition.

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Non-normal arc transitive BCn4, Reduction to circulants

With X ∈ BCn4 we associate a certain circulant Y (step two blue graph in figure below). Graphs X are classified by finding all possible graphs Y .

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Non-normal arc transitive BCn4, Reduction to circulants

With X ∈ BCn4 we associate a certain circulant Y (step two blue graph in figure below). Graphs X are classified by finding all possible graphs Y .

Y. G. Baik, Y. Q. Feng, H. S. Sim, M. Y. Xu, On the normality of

Cayley graphs of abelian groups, Algebra Colloq. 5 (1998), 227–234.

I. Kov´

acs, Classifying Arc-Transitive Circulants, J. Algebraic

Combin. 20 (2004), 353–358.
C. H. Li, Permutation groups with a cylic regular subgroup and

arc-transitive circulants, J. Algebraic Combin. 21 (2005), 131-136.

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Non-normal arc transitive BCn4, Reduction to circulants

With X ∈ BCn4 we associate a certain circulant Y (step two blue graph in figure below). Graphs X are classified by finding all possible graphs Y .

Y. G. Baik, Y. Q. Feng, H. S. Sim, M. Y. Xu, On the normality of

Cayley graphs of abelian groups, Algebra Colloq. 5 (1998), 227–234.

I. Kov´

acs, Classifying Arc-Transitive Circulants, J. Algebraic

Combin. 20 (2004), 353–358.
C. H. Li, Permutation groups with a cylic regular subgroup and

arc-transitive circulants, J. Algebraic Combin. 21 (2005), 131-136. In order this to work we need to trasfer symmetry properties between X and Y

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Reduction to circulants, Why it works?

Lemma X ∈ BCn4 non-normal ⇔ X non-normal Zn-cover of dip4.

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Reduction to circulants, Why it works?

Lemma X ∈ BCn4 non-normal ⇔ X non-normal Zn-cover of dip4. Lemma X ∈ BCn4 non-normal ⇒ Y non-normal circulant.

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Reduction to circulants, Why it works?

Lemma X ∈ BCn4 non-normal ⇔ X non-normal Zn-cover of dip4. Lemma X ∈ BCn4 non-normal ⇒ Y non-normal circulant.

Kerφ = Ker¯

G = 1. Then Z ⊳ Aut(Y ) ⇒ Z ⊳ ¯ G ⇒ Z ⊳ G Kerφ = Ker¯ G = 1. Then X = Cn[2K1], n ≥ 4 even, Y = Cn/2[K2].

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Why it works? The structure of Y

X = Cay(Dn, S), S = {b, bax, bay, baz} Y = Cay(Zn, T), T = {a±x, a±y, a±z, a±(x−y), a±(y−z), a±(x−z)} might not be arc transitive. However: it is an edge-disjoint union of arc transitive circulants (of which at least one of them is connected).

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Why it works? The structure of Y

X = Cay(Dn, S), S = {b, bax, bay, baz} Y = Cay(Zn, T), T = {a±x, a±y, a±z, a±(x−y), a±(y−z), a±(x−z)} might not be arc transitive. However: it is an edge-disjoint union of arc transitive circulants (of which at least one of them is connected). Lemma Either Y connected arc transitive, non normal, and T = {a±x, a±y, a±z, a±(x−y), a±(y−z), a±(x−z)}

r

Y = Y1 +Y2, where Y2 is connected, arc transitive, non-normal, and T1 = {a±x, a±(y−z)}, T2 = {a±y, a±z, a±(x−y), a±(x−z)}

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The structure of Y , uniformity index of X

For e ∈ Y , let r(e) be the number of 3-cycles in X ∪ X 2 containing e.

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The structure of Y , uniformity index of X

For e ∈ Y , let r(e) be the number of 3-cycles in X ∪ X 2 containing e. Lemma If Y is arc transitive, then r(e) = 12/|T| for each e ∈ E(Y ). If Y = Y1 + Y2 is an edge disjoint union of two arc transitive graphs, then r(e) =

4/|T1|,

for each e ∈ E(Y1) 8/|T2|, for each e ∈ E(Y2).

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The structure of Y , uniformity index of X

For e ∈ Y , let r(e) be the number of 3-cycles in X ∪ X 2 containing e. Lemma If Y is arc transitive, then r(e) = 12/|T| for each e ∈ E(Y ). If Y = Y1 + Y2 is an edge disjoint union of two arc transitive graphs, then r(e) =

4/|T1|,

for each e ∈ E(Y1) 8/|T2|, for each e ∈ E(Y2). X is k-uniform if r(e) = k for all e ∈ E(Y ). The paramter k is the uniformity index. If Y = Y1 + Y2, then Y1 is k1-uniform and Y2 is k2-uniform. Possibly, k1 = k2 = k.

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X is non-uniform

Then Y = Y1 + Y2. Since |T1|k1 = 4, |T2|k2 = 8 ⇒ k1, k2 ∈ {1, 2, 4}.

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X is non-uniform

Then Y = Y1 + Y2. Since |T1|k1 = 4, |T2|k2 = 8 ⇒ k1, k2 ∈ {1, 2, 4}. Lemma If X is non-uniform, then X = Cn[2K1], n ≥ 4 even.

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X is non-uniform

Then Y = Y1 + Y2. Since |T1|k1 = 4, |T2|k2 = 8 ⇒ k1, k2 ∈ {1, 2, 4}. Lemma If X is non-uniform, then X = Cn[2K1], n ≥ 4 even. k1 = 4. Then elements in T1 = {a±x, a±(y−z)} coincide. Hence X = Cn[2K1].

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X is non-uniform

Then Y = Y1 + Y2. Since |T1|k1 = 4, |T2|k2 = 8 ⇒ k1, k2 ∈ {1, 2, 4}. Lemma If X is non-uniform, then X = Cn[2K1], n ≥ 4 even. k1 = 4. Then elements in T1 = {a±x, a±(y−z)} coincide. Hence X = Cn[2K1]. k1 = 2. No graphs.

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X is non-uniform

Then Y = Y1 + Y2. Since |T1|k1 = 4, |T2|k2 = 8 ⇒ k1, k2 ∈ {1, 2, 4}. Lemma If X is non-uniform, then X = Cn[2K1], n ≥ 4 even. k1 = 4. Then elements in T1 = {a±x, a±(y−z)} coincide. Hence X = Cn[2K1]. k1 = 2. No graphs. k1 = 1. Then T1 = {a±(y+z), a±(y−z)} and T2 = {a±y, a±z}. Now Y2 is a connected 4-valent arc transitive circulant, and non-normal. By a result of Baik, Feng, Sim, Xu, the graph Y2 is one of K5, K5,5 − 5K2, and Cn/2[2K1] for n ≥ 6 even. None of these can appear as Y2.

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X is non-uniform

Then Y = Y1 + Y2. Since |T1|k1 = 4, |T2|k2 = 8 ⇒ k1, k2 ∈ {1, 2, 4}. Lemma If X is non-uniform, then X = Cn[2K1], n ≥ 4 even. k1 = 4. Then elements in T1 = {a±x, a±(y−z)} coincide. Hence X = Cn[2K1]. k1 = 2. No graphs. k1 = 1. Then T1 = {a±(y+z), a±(y−z)} and T2 = {a±y, a±z}. Now Y2 is a connected 4-valent arc transitive circulant, and non-normal. By a result of Baik, Feng, Sim, Xu, the graph Y2 is one of K5, K5,5 − 5K2, and Cn/2[2K1] for n ≥ 6 even. None of these can appear as Y2.

Y. G. Baik, Y. Q. Feng, H. S. Sim, M. Y. Xu, On the normality of Cayley

graphs of abelian groups, Algebra Colloq. 5 (1998), 227–234.

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X is k-uniform, k > 1

Lemma If X is k-uniform, k > 1, then k ≤ 4, and k = 4 and X = K4,4. k = 3 and X = K5,5 − 5K2. k = 2 and X is the non-incidence graph of PG(2, 2).

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X is k-uniform, k > 1

Lemma If X is k-uniform, k > 1, then k ≤ 4, and k = 4 and X = K4,4. k = 3 and X = K5,5 − 5K2. k = 2 and X is the non-incidence graph of PG(2, 2). These graphs are constructed using elementary combinatorial arguments.

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X is 1-uniform

We use quotienting by the action of some cyclic subgroup ¯ J ≤ ¯ Z such that

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X is 1-uniform

We use quotienting by the action of some cyclic subgroup ¯ J ≤ ¯ Z such that

rbits of ¯

J are blocks of imprimitivity for G = Aut(X).

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X is 1-uniform

We use quotienting by the action of some cyclic subgroup ¯ J ≤ ¯ Z such that

rbits of ¯

J are blocks of imprimitivity for G = Aut(X). As a quotient we obtain either a cycle, or a smaller 4-valent graph.

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X is 1-uniform

We use quotienting by the action of some cyclic subgroup ¯ J ≤ ¯ Z such that

rbits of ¯

J are blocks of imprimitivity for G = Aut(X). As a quotient we obtain either a cycle, or a smaller 4-valent graph. In the latter case, X → X/¯ J is a regular cyclic covering projection. Moreover,

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X is 1-uniform

We use quotienting by the action of some cyclic subgroup ¯ J ≤ ¯ Z such that

rbits of ¯

J are blocks of imprimitivity for G = Aut(X). As a quotient we obtain either a cycle, or a smaller 4-valent graph. In the latter case, X → X/¯ J is a regular cyclic covering projection. Moreover, X/¯ J is an arc transitive dihedrant, and non-normal.

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X is 1-uniform

We use quotienting by the action of some cyclic subgroup ¯ J ≤ ¯ Z such that

rbits of ¯

J are blocks of imprimitivity for G = Aut(X). As a quotient we obtain either a cycle, or a smaller 4-valent graph. In the latter case, X → X/¯ J is a regular cyclic covering projection. Moreover, X/¯ J is an arc transitive dihedrant, and non-normal. The uniformity index for the small dihedrant is > 1.

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X is 1-uniform

We use quotienting by the action of some cyclic subgroup ¯ J ≤ ¯ Z such that

rbits of ¯

J are blocks of imprimitivity for G = Aut(X). As a quotient we obtain either a cycle, or a smaller 4-valent graph. In the latter case, X → X/¯ J is a regular cyclic covering projection. Moreover, X/¯ J is an arc transitive dihedrant, and non-normal. The uniformity index for the small dihedrant is > 1. So X is a regular cyclic cover of Cm[2K1], K4,4, K5,5 − 5K2, non-inc. PG(2, 2).

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X is 1-uniform

We use quotienting by the action of some cyclic subgroup ¯ J ≤ ¯ Z such that

rbits of ¯

J are blocks of imprimitivity for G = Aut(X). As a quotient we obtain either a cycle, or a smaller 4-valent graph. In the latter case, X → X/¯ J is a regular cyclic covering projection. Moreover, X/¯ J is an arc transitive dihedrant, and non-normal. The uniformity index for the small dihedrant is > 1. So X is a regular cyclic cover of Cm[2K1], K4,4, K5,5 − 5K2, non-inc. PG(2, 2). How to find ¯ J?

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X is 1-uniform

We use quotienting by the action of some cyclic subgroup ¯ J ≤ ¯ Z such that

rbits of ¯

J are blocks of imprimitivity for G = Aut(X). As a quotient we obtain either a cycle, or a smaller 4-valent graph. In the latter case, X → X/¯ J is a regular cyclic covering projection. Moreover, X/¯ J is an arc transitive dihedrant, and non-normal. The uniformity index for the small dihedrant is > 1. So X is a regular cyclic cover of Cm[2K1], K4,4, K5,5 − 5K2, non-inc. PG(2, 2). How to find ¯ J? It is enough to find J < Z < Aut(Y ) such that the orbits of J are blocks of imprimitivity for Aut(Y ).

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X is 1-uniform

We use quotienting by the action of some cyclic subgroup ¯ J ≤ ¯ Z such that

rbits of ¯

J are blocks of imprimitivity for G = Aut(X). As a quotient we obtain either a cycle, or a smaller 4-valent graph. In the latter case, X → X/¯ J is a regular cyclic covering projection. Moreover, X/¯ J is an arc transitive dihedrant, and non-normal. The uniformity index for the small dihedrant is > 1. So X is a regular cyclic cover of Cm[2K1], K4,4, K5,5 − 5K2, non-inc. PG(2, 2). How to find ¯ J? It is enough to find J < Z < Aut(Y ) such that the orbits of J are blocks of imprimitivity for Aut(Y ). Since X = Cn[2K1], we have a monomorphism of ¯ G → Aut(Y ), and we can transfer the action of J back to an action of ¯ J on X.

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Finding J - using a result of Kov´ acs

Y or (Y2) is a connected, arc transitive, non-normal circulant. Hence the connection set T (or T2) has restricted structure, by a result of Kov´ acs.

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Finding J - using a result of Kov´ acs

Y or (Y2) is a connected, arc transitive, non-normal circulant. Hence the connection set T (or T2) has restricted structure, by a result of Kov´ acs.

I. Kov´

acs, Classifying Arc-Transitive Circulants, J. Algebraic Combin. 20 (2004), 353–358.

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Finding J - using a result of Kov´ acs

Y or (Y2) is a connected, arc transitive, non-normal circulant. Hence the connection set T (or T2) has restricted structure, by a result of Kov´ acs.

I. Kov´

acs, Classifying Arc-Transitive Circulants, J. Algebraic Combin. 20 (2004), 353–358.

C. H. Li, Permutation groups with a cylic regular subgroup and

arc-transitive circulants, J. Algebraic Combin. 21 (2005), 131-136.

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Structure of non-normal arc transitive circulants, by Kov´ acs

(A) There exists 1 < E < Z such that T (or T2) is a union of E-cosets,

(B) There exists a coprime decomposition Z = E × F, |E| > 3, such that

T (or T2) is of the form E ♯T ′, for some T ′ < F.

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Structure of non-normal arc transitive circulants, by Kov´ acs

(A) There exists 1 < E < Z such that T (or T2) is a union of E-cosets,

(B) There exists a coprime decomposition Z = E × F, |E| > 3, such that

T (or T2) is of the form E ♯T ′, for some T ′ < F.

In (A) and (B), cosets of E and F are blocks of imprimitivity for Aut(Y ).

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1 < E < Z, take J = E

X → X/¯ E not a cover

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1 < E < Z, take J = E

X → X/¯ E not a cover T ∩ E = ∅. So Y = Y1 + Y2, and T2 ∩ E = ∅. Then T1 ∩ E = ∅ and moreover, T1 ⊂ E. |T2| = 8 and |E| ≥ 5 forces |E| = 8, T2 is one coset. Also, E is of index 2. So n = 16. Hence T1 ⊂ a±{2,4,6,8} and T2 = a±{1,3,5,7}. So x ∈ ±{2, 4, 6}, y = 1, z ∈ {3, 5, 7}. By MAGMA, none of these graphs is arc transitive.

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1 < E < Z, take J = E

X → X/¯ E not a cover T ∩ E = ∅. So Y = Y1 + Y2, and T2 ∩ E = ∅. Then T1 ∩ E = ∅ and moreover, T1 ⊂ E. |T2| = 8 and |E| ≥ 5 forces |E| = 8, T2 is one coset. Also, E is of index 2. So n = 16. Hence T1 ⊂ a±{2,4,6,8} and T2 = a±{1,3,5,7}. So x ∈ ±{2, 4, 6}, y = 1, z ∈ {3, 5, 7}. By MAGMA, none of these graphs is arc transitive. X → X/¯ E is a cover

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1 < E < Z, take J = E

X → X/¯ E not a cover T ∩ E = ∅. So Y = Y1 + Y2, and T2 ∩ E = ∅. Then T1 ∩ E = ∅ and moreover, T1 ⊂ E. |T2| = 8 and |E| ≥ 5 forces |E| = 8, T2 is one coset. Also, E is of index 2. So n = 16. Hence T1 ⊂ a±{2,4,6,8} and T2 = a±{1,3,5,7}. So x ∈ ±{2, 4, 6}, y = 1, z ∈ {3, 5, 7}. By MAGMA, none of these graphs is arc transitive. X → X/¯ E is a cover The quotient graph must be k-uniform, k > 1. Then k = |E|. By direct case analysis, also using MAGMA, we have: k = 2. A 2-cover of the graph III. We obtain graph V. k = 3. A 3-cover of the graph II. We obtain graph VI. k = 4. A 4-cover of K4,4. No graphs.

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Z = E × F, take J = E

X → X/¯ E not a cover

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Z = E × F, take J = E

X → X/¯ E not a cover T2 ∩ E = ∅. Then T2 = E ♯ = Z♯

n, so n = 13. We obtain graph IV.

T1 ⊂ E, T1 = T2. More involved. No other graphs.

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Z = E × F, take J = E

X → X/¯ E not a cover T2 ∩ E = ∅. Then T2 = E ♯ = Z♯

n, so n = 13. We obtain graph IV.

T1 ⊂ E, T1 = T2. More involved. No other graphs. X → X/¯ E is a cover. Consider X → X/¯ F

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Z = E × F, take J = E

X → X/¯ E not a cover T2 ∩ E = ∅. Then T2 = E ♯ = Z♯

n, so n = 13. We obtain graph IV.

T1 ⊂ E, T1 = T2. More involved. No other graphs. X → X/¯ E is a cover. Consider X → X/¯ F Then ¯ E ⊳ G. So X → X/¯ F is not a cover. Otherwise ¯ F ⊳ G, and so ¯ Z = ¯ E ¯ F ⊳ G, a contradiction. Thus, T1 ⊂ F. It follows that X/¯ E is non-uniform, with parameters (1, |E| − 1), a contradiction.

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Non-normal arc-transitive BCn4

I. The lexicographic product Cn[2K1].

n ≥ 4 even, S = {b, ba, ba

n 2 , ba n 2 +1}

(in picture, n = 16).

II. The graph K5,5 − 5K2.

n = 5, S = {b, ba, ba2, ba3}.

III. The non-incidence graph of

PG(2, 2). n = 7, S = {b, ba, ba2, ba4}.

IV. The incidence graph of PG(2, 3).

n = 13, S = {b, ba, ba3, ba9}.

V. A 2-cover of the graph III.

n = 14, S = {b, ba, ba4, ba6}.

VI. A 3-cover of the graph II.

n = 15, S = {b, ba, ba3, ba7} Table 1: Non-normal 4-valent arc-transitive dihedrants satisfying the bipartition condition.

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Thank you!

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