Existence of non-Cayley Haar graphs

TL;DR

This paper proves that most finite non-abelian groups have Haar graphs that are not Cayley graphs, except for a few specific groups, resolving an open problem from 2016.

Contribution

It demonstrates the existence of non-Cayley Haar graphs for all finite non-abelian groups except a small set of exceptions, answering a previously open question.

Findings

Most finite non-abelian groups admit non-Cayley Haar graphs.

The exceptions are the dihedral groups D6, D8, D10, the quaternion group Q8, and Q8×Z2.

This resolves an open problem in the theory of Haar graphs.

Abstract

A Cayley graph of a group $H$ is a finite simple graph $\Gamma$ such that its automorphism group ${\rm Aut}(\Gamma)$ contains a subgroup isomorphic to $H$ acting regularly on $V(\Gamma)$, while a Haar graph of $H$ is a finite simple bipartite graph $\Sigma$ such that ${\rm Aut}(\Sigma)$ contains a subgroup isomorphic to $H$ acting semiregularly on $V(\Sigma)$ and the $H$-orbits are equal to the partite sets of $\Sigma$. It is well-known that every Haar graph of finite abelian groups is a Cayley graph. In this paper, we prove that every finite non-abelian group admits a non-Cayley Haar graph except the dihedral groups $D_6$, $D_8$, $D_{10}$, the quaternion group $Q_8$ and the group $Q_8\times\mathbb{Z}_2$. This answers an open problem proposed by Est\'elyi and Pisanski in 2016.

Tables

Table 1. Table 1: The vertices in Γ Γ \Gamma having distance at most 2 2 2 from 1 0 subscript 1 0 1_{0} or c 1 subscript 𝑐 1 c_{1} .

$v$	neighbors of $v$	vertices having distance 2 from $v$
$1_0$	$1_1$	${\left(a^{-1}\right)}_0$, ${\left(c^{-1}\right)}_0$, ${\left(a^{-1}bc\right)}_0$, ${\left(b^{-1}{c}^{-1}\right)}_0$
	$a_1$	$a_0$, ${\left(a{c}^{-1}\right)}_0$, ${\left(b^{-1}c\right)}_0$, ${\left(ab{c}^{-1}\right)}_0$
	$c_1$	$c_0$, ${\left(a^{-1}c\right)}_0$, ${\left(a^{-1}b{c}^2\right)}_0$, ${\left(b^{-1}\right)}_0$
	${\left(ab{c}^{-1}\right)}_1$	${\left(ab{c}^{-1}\right)}_0$, ${\left(b{c}^{-1}\right)}_0$, ${\left(ab{c}^{-2}\right)}_0$, ${\left(a^{-1}{c}^{-2}\right)}_0$
	${\left(bc\right)}_1$	${\left(bc\right)}_0$, ${\left(a^{-1}bc\right)}_0$, $b_0$, ${\left(a{c}^2\right)}_0$
$c_1$	$c_0$	${\left(ac\right)}_1$, ${\left(c^2\right)}_1$, ${\left(ab\right)}_1$, ${\left(b{c}^2\right)}_1$
	${\left(a^{-1}c\right)}_0$	${\left(a^{-1}c\right)}_1$, ${\left(a^{-1}{c}^2\right)}_1$, ${\left(b^{-1}\right)}_1$, ${\left(ab{c}^2\right)}_1$
	$1_0$	$1_1$, $a_1$, ${\left(ab{c}^{-1}\right)}_1$, ${\left(bc\right)}_1$
	${\left(a^{-1}b{c}^2\right)}_0$	${\left(a^{-1}b{c}^2\right)}_1$, ${\left(b{c}^2\right)}_1$, ${\left(a^{-1}b{c}^3\right)}_1$, ${\left(a^{-1}{c}^3\right)}_1$
	${\left(b^{-1}\right)}_0$	${\left(b^{-1}\right)}_1$, ${\left(a^{-1}b\right)}_1$, ${\left(b^{-1}c\right)}_1$, ${\left(a{c}^{-1}\right)}_1$

$i$	$H_i$	$S$
1	$⟨a,b,c|a^8,b^2,c^2,[a,c],[b,c],a^b=a^{-1}⟩\cong D_8\times {\mathbb{Z}}_2$	$\{1,a,b,c,ab,abc\}$
2	$⟨a,b,c|a^4,b^2,c^2,[a,b],[a,c],[b,c]=a^2⟩$	$\{1,a,b,ab,ac,abc\}$
3	$⟨a,b\mid a^8,b^2=a^4,a^b=a^{-1}⟩$	$\{1,a,b,a^5,ab,a^{5}b\}$
4	$⟨a,b\mid a^8,b^2,a^b=a^3⟩$	$\{1,a,b,ab\}$
5	$⟨a,b,c,d\mid a^4,b^4,c^2,d^2,a^2=b^2,a^b=a^{-1},[a,c],[a,d],$	$\{1,a,b,b^{-1},ab,ac,$
	$[b,c],[b,d],[c,d]⟩\cong Q{}_8\times \mathbb{Z}{}_2\times \mathbb{Z}{}_2$	$bd,abd\}$
6	$⟨a,b,c\mid a^4,b^4,c^3,a^2=b^2,a^b=a^{-1},a^c=b^{\pm 1},b^c=a^{\pm 1}b⟩$	$\{1,a,bc,abc\}$
7	$⟨a,b,c\mid a^2,b^2,c^3,[a,b],a^c=b,b^c=ab⟩\cong A_4$	$\{1,a,c,abc\}$
8	$⟨a,g\mid a^5,g^4,a^g=a^2⟩\cong F_{20}$	$\{1,a,g\}$
9	$⟨a,c,b\mid a^p,c^p,b^2,[a,c],a^b=a^{-1},c^b=c^{-1}⟩\cong {\mathbb{Z}}_p^2⋊{\mathbb{Z}}_2$, $p=3,5$	$\{1,a,c,b,ab,cb\}$

Equations

$$R(g):h_i\mapsto (hg{)}_i,\forall i\in \{0,1\},\forall h,g\in H.$$

$${\delta }_{\alpha ,x,y}:h_0\mapsto (x{h}^{\alpha }{)}_1,h_1\mapsto (y{h}^{\alpha }{)}_0,$$

$${\sigma }_{\alpha ,g}:h_0\mapsto (h^{\alpha }{)}_0,h_1\mapsto (g{h}^{\alpha }{)}_1,$$

$$\mathrm{I}$$

$$\mathrm{F}$$

$${\cal BC}=\big{\{}H\;\text{is a finite group}:\hbox{\rm H}(H,S)\;\text{is a Cayley graph for any }S\subseteq H\big{\}}.$$

$$H=D_{2n}\times {\mathbb{Z}}_p=⟨a,b,c|a^n=b^2=c^p=[a,c]=[b,c]=1,a^b=a^{-1}⟩,$$

$$S^{\beta }=\{1^{\beta },a^{\beta },b^{\beta },c^{\beta },(abc{)}^{\beta }\}=y^{-1}\{1,a^{-1},b,c^{-1},ab{c}^{-1}\}.$$

$$H=Q_8\times {\mathbb{Z}}_p=⟨a,b,c|a^4=b^4=c^p=[a,c]=[b,c]=1,a^2=b^2,a^b=a^{-1}⟩,$$

$$\Gamma ((b^{-1}{)}_1)=\{(b^{-1}{)}_0,(ab{)}_0,(b^{-1}{c}^{-1}{)}_0,(a^{-1}c{)}_0,(b^2{c}^{-1}{)}_0\},$$

$$\Gamma ((b{c}^2{)}_1)=\{(b{c}^2{)}_0,(a^{-1}b{c}^2{)}_0,(bc{)}_0,(a{c}^3{)}_0,c_0\}.$$

$$C_1=(1_0,1_1,(a^{-1}bc{)}_0,(bc{)}_1),C_2=(1_0,a_1,(ab{c}^{-1}{)}_0,(ab{c}^{-1}{)}_1),$$

$$C_3=(c_1,c_0,(b{c}^2{)}_1,(a^{-1}b{c}^2{)}_0),C_4=(c_1,(b^{-1}{)}_0,(b^{-1}{)}_1,(a^{-1}c{)}_0).$$

$$S^{\beta }=\{1^{\beta },a^{\beta },c^{\beta },(ab{c}^{-1}{)}^{\beta },(bc{)}^{\beta }\}=y^{-1}\{1,a^{-1},c^{-1},a^{-1}bc,b^{-1}{c}^{-1}\}.$$

$$\begin{array}[]{ll}&H_{1}=\langle a,b,c~{}|~{}a^{8}=b^{2}=c^{2}=[a,c]=[b,c]=1,a^{b}=a^{-1}\rangle~{}(\cong D_{8}\times\mathbb{Z}_{2});\\ &H_{2}=\langle a,b,c~{}|~{}a^{4}=b^{2}=c^{2}=[a,b]=[a,c]=1,[b,c]=a^{2}\rangle;\\ &H_{3}=\langle a,b\mid a^{8}=1,b^{2}=a^{4},a^{b}=a^{-1}\rangle;\\ &H_{4}=\langle a,b\mid a^{8}=b^{2}=1,a^{b}=a^{3}\rangle;\ \ \ H_{5}=D_{16};\ \ \ H_{6}=Q_{8}\times\mathbb{Z}_{2}.\\ \end{array}$$

$$\begin{array}[]{lll}H_{1}(p)&=&\langle a,b\mid a^{p^{2}}=b^{2}=1,b^{-1}ab=a^{-1}\rangle;\\ H_{2}(p)&=&\langle a,b,c\mid a^{p}=b^{p}=c^{2}=[a,b]=1,c^{-1}ac=a^{-1},c^{-1}bc=b^{-1}\rangle;\\ H_{3}(p)&=&\langle a,b,c\mid a^{p}=b^{p}=c^{2}=1,[a,b]=[a,c]=1,c^{-1}bc=b^{-1}\rangle.\end{array}$$

$$H\cong {\mathbb{Z}}_p⋊D_8=⟨a,b,g\mid b^p=a^4=g^2=1,a^g=a^{-1},b^a=b,b^g=b^{-1}⟩\cong D_{8p}.$$

Full text

Keywords: Haar graph, Cayley graph, vertex-transitive graph.

MSC 2010: 05E18 (primary), 20B25 (secondary).

1 Introduction

All groups in this paper are finite and all graphs are finite and undirected. Let $H$ be a group, and let $R$, $L$ and $S$ be three subsets of $H$ such that $R^{-1}=R$, $L^{-1}=L$, and $R\cup L$ does not contain the identity element $1$ of $H$. The Cayley graph of $H$ relative to the subset $R,$ denoted by $\hbox{\rm Cay}(H,R),$ is the graph having vertex set $H,$ and edge set $\{\{h,xh\}:x\in R,h\in H\}$, and the bi-Cayley graph of $H$ relative to the triple $(R,L,S)$, denoted by $\hbox{\rm BiCay}(H,R,L,S)$, is the graph having vertex set the union of the right part $H_{0}=\{h_{0}:h\in H\}$ and the left part $H_{1}=\{h_{1}:h\in H\}$, and edge set being the union of the following three sets

• •

\big{\{}\{h_{0},(xh)_{0}\}:x\in R,\,h\in H\big{\}} (right edges),

• •

\big{\{}\{h_{1},(xh)_{1}\}:x\in L,\,h\in H\big{\}} (left edges),

• •

\big{\{}\{h_{0},(xh)_{1}\}:x\in S,\,h\in H\big{\}} (spokes).

In the special case when $R=L=\emptyset$, the bi-Cayley graph $\hbox{\rm BiCay}(H,\emptyset,\emptyset,S)$ is called a Haar graph of $H$ relative to the set $S$, denoted by $\hbox{\rm H}(H,S)$. A Haar graph $H(H,S)$ of a finite group $H$ was first defined as a voltage graph of a dipole with no loops and $|S|$ parallel edges (see [15]), and the name Haar graph comes from the fact that, when $H$ is an abelian group the Schur norm of the corresponding adjacency matrix can be easily evaluated via the so called Haar integral on $H$ (see [14]).

Symmetries of Cayley graphs have always been an active topic among algebraic combinatorics, and lately, the symmetries of bi-Cayley graphs received considerable attention. For various results and constructions in connection with bi-Cayley graphs and their automorphisms, we refer the reader to [1, 2, 6, 9, 20, 23, 28, 29] and all the references therein. In particular, Estélyi and Pisanski [9] initiated the investigation for the relationship between Cayley graphs and Haar graphs. A Cayley graph is a Haar graph exactly when it is bipartite, but no simple condition is known for a Haar graph to be a Cayley graph. An elementary argument shows that every Haar graph of abelian groups is a Cayley graph (this also follows from Proposition 2.1). On the other hand, Lu et al. [22] constructed cubic semi-symmetric graphs, that is, edge- but not vertex-transitive graphs, as Haar graphs of alternating groups. Clearly, as these graphs are not vertex-transitive, they are examples of Haar graphs which are not Cayley graphs. It is natural to ask which non-abelian groups admit a Haar graph that is not a Cayley graph, or putting it another way, we have the following problem, which was first posed by Estélyi and Pisanski [9, Problem 1].

Problem 1.1**.**

([9])* Determine the finite non-abelian groups $H$ for which all Haar graphs $\hbox{\rm H}(H,S)$ are Cayley graphs.*

We denote by $\mathbb{Z}_{n}$ the cyclic group of order $n$, by $D_{2n}$ the dihedral group of order $2n,$ and by $Q_{8}$ the quaternion group. Estélyi and Pisanski [9, Theorem 8] solved Problem 1.1 for dihedral groups.

Proposition 1.2**.**

([9])* Each Haar graph of the dihedral group $D_{2n}$ is a Cayley graph if and only if $n=2,3,4,5$.*

A group $H$ is called inner abelian if $H$ is non-abelian, and all proper subgroups of $H$ are abelian. Recently, Feng et al. [10, Theorem 1.2] solved Problem 1.1 for the class of inner abelian groups.

Proposition 1.3**.**

([10])* Each Haar graph of an inner abelian group $H$ is a Cayley graph if and only if $H\cong D_{6}$, $D_{8}$, $D_{10}$ or $Q_{8}$.*

In this paper we solve Problem 1.1 completely.

Theorem 1.4**.**

Let $H$ be a non-abelian group with the property that every Haar graph of $H$ is a Cayley graph. Then $H$ is isomorphic to $D_{6}$, $D_{8}$, $D_{10}$, $Q_{8}$ or $Q_{8}\times\mathbb{Z}_{2}$.

The main idea of the proof of Theorem 1.4 is to construct non-Cayley Haar graphs. It is worth mentioning that all non-Cayley Haar graphs of non-abelian groups, constructed in [9, 10] and this paper, are not vertex-transitive. It seems difficulty to construct vertex-transitive non-Cayley Haar graphs. Estélyi and Pisanski [9] raised a question whether there exists a vertex-transitive non-Cayley Haar graph. Later, infinitely many vertex-transitive non-Cayley Haar graphs were constructed by Conder et al. [6] and Feng et al. [12], and this prompts us to consider the following problem.

Problem 1.5**.**

Determine the finite non-abelian groups $H$ for which all vertex-transitive Haar graphs $\hbox{\rm H}(H,S)$ are Cayley graphs.

Note that Problem 1.5 is closely related to the so called non-Cayley numbers. A positive integer $n$ is called a Cayley number if every vertex-transitive graph of order $n$ is a Cayley graph, and otherwise it is a non-Cayley number. In 1983, Marušič [24] posed the problem of determining Cayley numbers, and this question has generated a fair amount of interests. For some works about Cayley numbers and vertex-transitive non-Cayley graphs, one may refer to [7, 21, 30].

By a graphical regular representation (GRR for short) for a group $H$ we mean a Cayley graph $\Gamma$ of $H$ such that $\hbox{\rm Aut}(\Gamma)\cong H$. When studying a Cayley graph $\Gamma$ of a finite group $H$, a very important question is to determine whether $H$ is in fact the full automorphism group of $\Gamma$. For this reason, GRRs have been widely studied. The most natural question is classifying finite groups admitting a GRR, and the solution was derived in several papers (see, for instance, [4, 11, 17, 18, 19, 25, 26, 27]). A bi-Cayley graph $\Sigma$ of a group $H$ is called a bi-graphical regular representation (bi-GRR for short) if $\hbox{\rm Aut}(\Sigma)\cong H$. The problem of classifying finite groups admitting a bi-GRR was posed by Zhou [31] (also see [16]), and it was solved by Du et al. [8] recently. Motivated by GRR and bi-GRR, a GHRR of a group $H$ is a Haar graph $\Gamma$ of $H$ with $\hbox{\rm Aut}(\Gamma)\cong H$. Since every Haar graph of abelian groups is a Cayley graph, abelian groups have no GHRR. However, many non-abelian groups have GHRRs, for example, see [9, 10] and Section 3 of this paper. Moreover, Theorem 1.4 implies that the non-abelian groups $D_{6}$, $D_{8}$, $D_{10}$, $Q_{8}$ and $Q_{8}\times\mathbb{Z}_{2}$ have no GHRRs, and to the best of our knowledge, they are the only known non-abelian groups that have no GHRRs. In the end of this section, we would like to pose the following problem.

Problem 1.6**.**

Determine the finite non-abelian groups that have no GHRRs.

The rest of the paper is organized as follows. In the next section we collect all concepts and results that will be used later. In Section 3, we introduce some Haar graphs that are not vertex-transitive, and prove Theorem 1.4 in Section 4.

2 Preliminaries

For a graph $\Gamma$, we denote by $V(\Gamma),$ $E(\Gamma)$ and $\hbox{\rm Aut}(\Gamma)$ the vertex set, the edge set and the group of all automorphisms of $\Gamma$. Given a vertex $v\in V(\Gamma),$ we denote by $\Gamma(v)$ the set of vertices adjacent to $v$. For a subgroup $G$ of $\hbox{\rm Aut}(\Gamma),$ denote by $G_{v}$ the stabilizer of the vertex $v$ in $G$, that is, the subgroup of $G$ fixing $v$. We say that $G$ is semiregular on $V(\Gamma)$ if $G_{v}=1$ for every $v\in V(\Gamma)$, and regular if $G$ is transitive and semiregular.

Let $\Gamma=\hbox{\rm H}(H,S)$ be a Haar graph of a group $H$ with identity element $1$. By [28, Lemma 3.1(2)], up to graph isomorphism, we may always assume that $1\in S$. The graph $\Gamma$ is then connected exactly when $H=\langle S\rangle$. For $g\in H$, the right translation $R(g)$ is the permutation of $H$ defined by $R(g):h\mapsto hg$ for $h\in H,$ and the left translation $L(g)$ is the permutation of $H$ defined by $L(g):h\mapsto g^{-1}h$ for $h\in H$. Set $R(H)=\{R(h)\,:\,h\in H\}$. Recall that $V(\Gamma)=H_{0}\cup H_{1}$. It is easy to see that $R(H)$ can be regarded as a group of automorphisms of $\hbox{\rm H}(H,S)$ acting on $V(\Gamma)$ by the rule

[TABLE]

Furthermore, $R(H)$ acts semiregularly on $V(\Gamma)$ with two orbits $H_{0}$ and $H_{1}$.

For an automorphism $\alpha\in\hbox{\rm Aut}(H)$ and $x,y,g\in H$, define two permutations on $V(\Gamma)=H_{0}\cup H_{1}$ as follows

[TABLE]

Set

[TABLE]

By [28, Lemma 3.3], ${\rm F}\leq\hbox{\rm Aut}(\Gamma)_{1_{0}}$, and if $\Gamma$ is connected, then ${\rm F}$ acts on the set $\Gamma(1_{0})$ consisting of all neighbours of $1_{0}$ faithfully. By [28, Theorem 1.1 and Lemma 3.2], we have the following proposition.

Proposition 2.1**.**

Let $\Gamma=\hbox{\rm H}(H,S)$ be a connected Haar graph, and let $A=\hbox{\rm Aut}(\Gamma)$.

1. (i)

If ${\rm I}=\emptyset,$ then the normalizer $N_{A}(R(H))=R(H)\rtimes{\rm F}$. 2. (ii)

If ${\rm I}\neq\emptyset$, then $N_{A}(R(H))=R(H)\langle{\rm F},\delta_{\alpha,x,y}\rangle$ for some $\delta_{\alpha,x,y}\in{\rm I}$.

Moreover, $\langle R(H),\delta_{\alpha,x,y}\rangle$ acts transitively on $V(\Gamma)$ for any $\delta_{\alpha,x,y}\in{\rm I}$.

Throughout the paper we follow the notation defined in [10]:

[TABLE]

The following proposition was given by [10, Lemma 3.1].

Proposition 2.2**.**

The class ${\cal BC}$ is closed under taking subgroups.

In view of [10, Theorem 1.3 and Corollary 4.6], we have the following proposition.

Proposition 2.3**.**

Let $H$ be a group belonging to the class ${\cal BC}$. Then the following hold.

1. (i)

The group $H$ is solvable. 2. (ii)

Each Sylow $p$-subgroup of $H$ with a prime $p\geq 3$ is abelian. 3. (iii)

If $H$ is non-abelian, then $H$ has a subgroup isomorphic to $D_{6}$, $D_{8}$, $D_{10}$ or $Q_{8}$.

The following proposition is well-known, and one may see [3, (1.12)].

Proposition 2.4**.**

Let $P$ be a finite abelian $p$-group. Then $P=\mathbb{Z}_{p^{e_{1}}}\times\mathbb{Z}_{p^{e_{2}}}\times\cdots\times\mathbb{Z}_{p^{e_{n}}}$, where $1\leq e_{1}\leq e_{2}\leq\cdots\leq e_{n}$. Moreover, the integers $n$ and $e_{i}$ with $1\leq i\leq n$ are uniquely determined by $P$.

3 Haar graphs that are not vertex-transitive

In this section, we introduce some Haar graphs that are not vertex-transitive, which will be used in the proof of Theorem 1.4. First we describe two infinite families of Haar graphs that are not vertex-transitive.

Lemma 3.1**.**

Let $n$ be an integer with $n\geq 3$, and let $p$ be an odd prime. Let

[TABLE]

and $S=\{1,a,b,c,abc\}$. Then $\hbox{\rm Aut}(\hbox{\rm H}(H,S))=R(H)$ and $H\not\in{\cal BC}$.

Proof.

Let $\Gamma=\hbox{\rm H}(H,S)$ and let $A=\hbox{\rm Aut}(\Gamma)$. Note that $R(H)\leq A$ has exactly two orbits on $V(\Gamma)$. Then $A$ is vertex-transitive or has two orbits, that is, $H_{0}$ and $H_{1}$. For the former, $A_{1_{0}}$ and $A_{1_{1}}$ are conjugate in $A$, and for the latter, the Frattini argument implies that $A=R(H)A_{1_{0}}=R(H)A_{1_{1}}$. In the both cases, $|A_{1_{0}}|=|A_{1_{1}}|$, and hence $|A_{1_{0}}|=|A_{h_{0}}|=|A_{k_{1}}|$ for any $h,k\in H$. To finish the proof, it suffices to show that $A_{1_{0}}=1$ and $\Gamma$ is not vertex-transitive.

We depicted the subgraph of $\Gamma$ induced by the vertices at distance at most $2$ from $1_{0}$ in Figure 1.

Consider the $4$-cycles of $\Gamma$ passing through the vertex $1_{0}$. For each $h\in H$, denote by $\Gamma(h_{0})$ and $\Gamma(h_{1})$ the neighborhoods of $h_{0}$ and $h_{1}$ in $\Gamma$ respectively, that is, $\Gamma(h_{0})=\{(sh)_{1}~{}|~{}s\in S\}$ and $\Gamma(h_{1})=\{(s^{-1}h)_{0}~{}|~{}s\in S\}$. By Figure 1, the numbers of $4$-cycles passing through the edges $\{1_{0},1_{1}\}$ and $\{1_{0},b_{1}\}$ are $1$ and $4$, respectively, while there are exactly three $4$-cycles passing through the edge $\{1_{0},u_{1}\}$ for each $u_{1}=a_{1},c_{1}$ or $(abc)_{1}$. This implies that $A_{1_{0}}$ fixes $1_{1}$ and $b_{1}$, and $\{a_{1},c_{1},(abc)_{1}\}$ setwise. It follows that $A_{1_{0}}\leq A_{1_{1}}$ and $A_{1_{0}}\leq A_{b_{1}}$, and since there is a unique $4$-cycle passing through $1_{0}$ and $1_{1}$, we have $A_{1_{0}}\leq A_{b_{0}}$. Since $|A_{1_{0}}|=|A_{h_{0}}|=|A_{k_{1}}|$ for any $h,k\in H$, we have $A_{1_{0}}=A_{1_{1}}=A_{b_{1}}=A_{b_{0}}$.

By Figure 1, there are $4$-cycles passing through $(a_{1},1_{0},b_{1})$ but no $4$-cycles passing through $(c_{1},1_{0},b_{1})$ or $((abc)_{1},1_{0},b_{1})$, and since $A_{1_{0}}$ fixes $b_{1}$ and $\{a_{1},c_{1},(abc)_{1}\}$ setwise, $A_{1_{0}}$ fixes $a_{1}$, and $\{c_{1},(abc)_{1}\}$ setwise. Thus, $A_{1_{0}}$ fixes $\Gamma(a_{1})$ setwise, and since there exist $4$-cycles passing through $1_{0}$, $a_{1}$ and a vertex in $\Gamma(a_{1})$ except $a_{0}$, we have $A_{1_{0}}\leq A_{a_{0}}$. It follows that $A_{1_{0}}=A_{a_{0}}=A_{a_{1}}$.

Now we claim that $A_{1_{0}}$ fixes $c_{1}$ and $(abc)_{1}$. Note that $A_{1_{0}}$ fixes $\{c_{1},(abc)_{1}\}$ setwise. Suppose that $\alpha\in A_{1_{0}}$ interchanges $c_{1}$ and $(abc)_{1}$. By Figure 1, there exist $4$-cycles passing through $1_{0}$, $c_{1}$ (resp. $(abc)_{1}$) and a vertex in $\Gamma(c_{1})$ (resp. $\Gamma((abc)_{1})$) except $c_{0}$ (resp. $(abc)_{0}$), and hence $\alpha$ interchanges $c_{0}$ and $(abc)_{0}$. Since $A_{1_{0}}$ fixes $a_{0}$, we have $(\Gamma(a_{0})\cap\Gamma(c_{0}))^{\alpha}=\Gamma(a_{0})\cap\Gamma((abc)_{0})$. Clearly, $(ac)_{1}\in\Gamma(a_{0})\cap\Gamma(c_{0})$. Then $|\Gamma(a_{0})\cap\Gamma((abc)_{0})|\not=0$, and hence there exist $s,t\in S$ such that $sa=tabc$, that is, $t^{-1}s=a^{2}bc\in S^{-1}S$. This is impossible as $S=\{1,a,b,c,abc\}$. Thus $A_{1_{0}}$ fixes $c_{1}$ and $(abc)_{1}$, and hence $c_{0}$ and $(abc)_{0}$. It follows that $A_{1_{0}}=A_{c_{0}}=A_{c_{1}}$.

Now we have that $A_{1_{0}}=A_{x_{0}}$ for each $x\in T:=\{a,b,c\}$. For any $y\in T$, we have $A_{1_{0}}^{R(y)}=A_{x_{0}}^{R(y)}$, that is, $A_{y_{0}}=A_{(xy)_{0}}$. It follows that $A_{1_{0}}=A_{(xy)_{0}}$, and an easy inductive argument implies that $A_{1_{0}}=A_{(x_{1}x_{2}\cdots x_{n})_{0}}$ for any $x_{1},\cdots,x_{n}\in T$. Since $\langle T\rangle=H$, $A_{1_{0}}$ fixes $H_{0}$ pointwise. Since $A_{1_{0}}=A_{1_{1}}$, we have $A_{h_{0}}=A_{h_{1}}$ for any $h\in H$, and hence $A_{1_{0}}$ fixes $H_{1}$ pointwise. Thus, $A_{1_{0}}=1$.

To finish the proof, we are left with showing that $A$ is not vertex-transitive. Suppose to the contrary that $A$ is vertex-transitive. Since $A_{1_{0}}=1$, we have $|A|=|V(\Gamma)|=2|R(H)|$ and hence $R(H)\unlhd A$. By Proposition 2.1, there exists $\delta_{\beta,x,y}\in A$ for some $\beta\in\hbox{\rm Aut}(H)$ and $x,y\in H$ such that $S^{\beta}=y^{-1}S^{-1}x$. Since $R(H)$ acts transitively on $H_{1},$ we may further assume that $1_{0}^{\delta_{\beta,x,y}}=1_{1}$. By Eq. (1), $1_{0}^{\delta_{\beta,x,y}}=(x1^{\beta})_{1}=1_{1}$, forcing $x=1$. Thus $S^{\beta}=y^{-1}S^{-1}$, that is,

[TABLE]

Since $1\in S$, we have $1\in S^{\beta}$ and so $y^{-1}=1,a,b^{-1},c$ or $abc$.

Note that $H=D_{2n}\times\mathbb{Z}_{p}=\langle a,b,c~{}|~{}a^{n}=b^{2}=c^{p}=[a,c]=[b,c]=1,a^{b}=a^{-1}\rangle$. If $n$ is odd then the center $Z(H)=\mathbb{Z}_{p}$, and if $n=2m$ is even then $Z(H)=\langle a^{m}\rangle\times\mathbb{Z}_{p}\cong\mathbb{Z}_{2}\times\mathbb{Z}_{p}$, where $\mathbb{Z}_{p}$ is characteristic in $\langle a^{m}\rangle\times\mathbb{Z}_{p}$. It follows that $\mathbb{Z}_{p}=\langle c\rangle$ is characteristic in $H$, and since $\beta\in\hbox{\rm Aut}(H)$, we have $c^{\beta}\in\langle c\rangle$.

If $y^{-1}=a,b^{-1},c$ or $abc$, we have from Eq. (3) that $S^{\beta}=\{a,1,ab,ac^{-1},a^{2}bc^{-1}\}$, $\{b,ba^{-1}$, $1$, $bc^{-1}$, $a^{-1}c^{-1}\},\{c,ca^{-1},cb,1,ab\}$ or $\{abc,a^{2}bc,ac,ab,1\}$, respectively. This is impossible because $c^{\beta}\in\langle c\rangle$. Thus, $y=1$ and $S^{\beta}=\{1,a^{-1},b,c^{-1},abc^{-1}\}$. This implies $c^{\beta}=c^{-1}$ because $c^{\beta}\in\langle c\rangle$. Since all involutions of $H$ generate the dihedral subgroup $\langle a,b\rangle$, $\langle a,b\rangle$ is characteristic in $H$, and since $\langle a,b\rangle$ is dihedral, $\langle a\rangle$ is characteristic in $H$. Thus, $a^{\beta}\in\langle a\rangle$ and $b^{\beta}\in\langle a,b\rangle$, and since $S^{\beta}=\{1,a^{-1},b,c^{-1},abc^{-1}\}$, we have $a^{\beta}=a^{-1}$, $b^{\beta}=b$ and $(abc)^{\beta}=abc^{-1}$. However, $abc^{-1}=(abc)^{\beta}=a^{\beta}b^{\beta}c^{\beta}=a^{-1}bc^{-1}$, that is, $a^{2}=a$, contrary the hypothesis $n\geq 3$. This completes the proof. ∎

Lemma 3.2**.**

Let $p$ be an odd prime, and let

[TABLE]

and $S=\{1,a,c,abc^{-1},bc\}$. Then $\hbox{\rm Aut}(\hbox{\rm H}(H,S))=R(H)$ and $H\not\in{\cal BC}$.

Proof.

Let $\Gamma=\hbox{\rm H}(H,S)$ and let $A=\hbox{\rm Aut}(\Gamma)$. The lemma holds for $p=3$ and $5$ by Magma [5], and we assume that $p\geq 7$ in the rest of the proof. Since $A$ is transitive or has the two orbits $H_{0}$ and $H_{1}$ as same as $R(H)$, we have $|A_{h_{0}}|=|A_{k_{1}}|$ for any $h,k\in H$.

For each $h\in H$, the neighborhood of the vertices $h_{0}$ and $h_{1}$ in $\Gamma$ are $\{(sh)_{1}~{}|~{}s\in S\}$ and $\{(s^{-1}h)_{0}~{}|~{}s\in S\}$, respectively. From this, it is easy to list the vertices in $\Gamma$ having distance at most $2$ from $1_{0}$ or $c_{1}$ in Table 1.

Furthermore, we have the following equations:

[TABLE]

By Table 1, there are exactly two $4$-cycles $C_{1}$ and $C_{2}$ passing through $1_{0}$:

[TABLE]

and there are exactly two $4$-cycles $C_{3}$ and $C_{4}$ passing through $c_{1}$:

[TABLE]

We depicted, using Table 1, Eqs. (4) and (5), an induced subgraph of $\Gamma$ in Figure 2.

There exist $4$-cycles passing through $1_{0}$ and any given vertex in $\Gamma(1_{0})$ except $c_{1}$. Then $A_{1_{0}}\leq A_{c_{1}}$ and hence $A_{1_{0}}$ fixes $\{C_{1},C_{2}\}$ and $\{C_{3},C_{4}\}$ setwise. Furthermore, $A_{1_{0}}$ fixes $\{(b^{-1})_{1},(bc^{2})_{1}\}$ setwise because these two vertices are antipodal to $c_{1}$ in $C_{3}$ and $C_{4}$ respectively, and since $|A_{h_{0}}|=|A_{k_{1}}|$ for any $h,k\in H$, we have $A_{1_{0}}=A_{c_{1}}$.

We first prove that $A_{1_{0}}$ fixes the $4$-cycle $C_{1}$ setwise. Recall that $A_{1_{0}}$ fixes $\{C_{1},C_{2}\}$ setwise. Suppose to the contrary that $\alpha\in A_{1_{0}}$ interchanges $C_{1}$ and $C_{2}$. Then $\{1_{1},(bc)_{1}\}^{\alpha}=\{a_{1},(abc^{-1})_{1}\}$, and since $A_{1_{0}}$ fixes $\{(b^{-1})_{1},(bc^{2})_{1}\}$ setwise, we have $\{[\Gamma(1_{1})\cup\Gamma((bc)_{1})]\cap[\Gamma((b^{-1})_{1})\cup\Gamma((bc^{2})_{1})]\}^{\alpha}=[\Gamma(a_{1})\cup\Gamma((abc^{-1})_{1})]\cap[\Gamma((b^{-1})_{1})\cup\Gamma((bc^{2})_{1})]$, which is impossible because $[\Gamma(a_{1})\cup\Gamma((abc^{-1})_{1})]\cap[\Gamma((b^{-1})_{1})\cup\Gamma((bc^{2})_{1})]=\emptyset$ and $(bc)_{0}\in\Gamma((bc)_{1})\cap\Gamma((bc^{2})_{1})$ by Table 1, Eqs. (4) and (5). Thus, $A_{1_{0}}$ fixes $C_{1}$ setwise.

Now we prove that $A_{1_{0}}$ fixe $C_{1}$ pointwise. Since $A_{1_{0}}$ fixes $C_{1}$ setwise, it fixes $C_{2}$ setwise, implying $A_{1_{0}}$ fixes $(abc^{-1})_{0}$. Suppose to the contrary that $\beta\in A_{1_{0}}$ interchanges $1_{1}$ and $(bc)_{1}$. By Table 1, $\Gamma((bc)_{1})\cap[\Gamma((b^{-1})_{1})\cup\Gamma((bc^{2})_{1})]=\{(bc)_{0}\}$ and $\Gamma(1_{1})\cap[\Gamma((b^{-1})_{1})\cup\Gamma((bc^{2})_{1})]=\{(b^{-1}c^{-1})_{0}\}$. Since $A_{1_{0}}$ fixes $\{(b^{-1})_{1},(bc^{2})_{1}\}$ setwise, $\beta$ interchanges $(bc)_{0}$ and $(b^{-1}c^{-1})_{0}$, implying $\Gamma((bc)_{0})^{\beta}=\Gamma((b^{-1}c^{-1})_{0})$. Since $A_{1_{0}}$ fixes $(abc^{-1})_{0}$, we have $[\Gamma((abc^{-1})_{0})\cap\Gamma((bc)_{0})]^{\beta}=\Gamma((abc^{-1})_{0})\cap\Gamma((b^{-1}c^{-1})_{0})$. It is easy to see that $(b^{-1}c^{-1})_{1}\in\Gamma((abc^{-1})_{0})\cap\Gamma((b^{-1}c^{-1})_{0})$. Then $\Gamma((abc^{-1})_{0})\cap\Gamma((bc)_{0})\not=\emptyset$, and there is $s,t\in S$ such that $sabc^{-1}=tbc$, that is, $s^{-1}t=ac^{-2}\in S^{-1}S$, which is impossible because $S^{-1}S=\{1,a,c,abc^{-1},bc,a^{-1},a^{-1}c,bc^{-1},a^{-1}bc,c^{-1},ac^{-1},abc^{-2},b,b^{-1}c,a^{-1}bc^{2}$, $ac^{2}$, $b^{-1}c^{-1}$, $b^{-1}$, $a^{-1}c^{-2}\}$. Thus, $A_{1_{0}}$ fixes $C_{1}$ pointwise, and hence $A_{1_{0}}=A_{1_{1}}=A_{(bc)_{1}}=A_{(a^{-1}bc)_{0}}$.

Since $A_{1_{0}}=A_{c_{1}}$, we have $A_{1_{0}}^{R(c^{-1})}=A_{c_{1}}^{R(c^{-1})}$, and so $A_{c^{-1}_{0}}=A_{1_{1}}=A_{1_{0}}$. Similarly, we have $A_{1_{0}}=A_{(b^{-1}c^{-1})_{0}}$ because $A_{1_{0}}=A_{(bc)_{1}}$. It follows that $A_{1_{0}}=A_{x_{0}}$ for any $x\in T:=\{c^{-1},b^{-1}c^{-1},a^{-1}bc\}$, and an easy inductive argument implies that $A_{1_{0}}=A_{x_{0}}$ for any $x\in\langle T\rangle=H$. Thus, $A_{1_{0}}$ fixes $H_{0}$ pointwise. Also, since $A_{1_{0}}=A_{1_{1}}$ implies $A_{h_{0}}=A_{h_{1}}$ for any $h\in H$, it follows that $A_{1_{0}}$ fixes $H_{1}$ pointwise too. Thus, $A_{1_{0}}=1$.

To finish the proof, we are left with showing that $A$ is not vertex-transitive. Suppose to the contrary that $A$ is vertex-transitive. Since $A_{1_{0}}=1$, we have $|A|=|V(\Gamma)|=2|R(H)|$ and hence $R(H)\unlhd A$. By Proposition 2.1, there exists $\delta_{\beta,x,y}\in A$ for some $\beta\in\hbox{\rm Aut}(H)$ and $x,y\in H$ such that $S^{\beta}=y^{-1}S^{-1}x$. By the transitivity of $R(H)$ on $H_{1}$, we may assume $1_{0}^{\delta_{\beta,x,y}}=1_{1}$, anb by Eq. (1), $1_{0}^{\delta_{\beta,x,y}}=(x1^{\beta})_{1}=1_{1}$, forcing $x=1$. Recall that $S=\{1,a,c,abc^{-1},bc\}$, and $S^{\beta}=y^{-1}S^{-1}$, that is,

[TABLE]

Since $1\in S$, we have $1\in S^{\beta}$ and so $y^{-1}=1,a,c,abc^{-1}$ or $bc$.

Note that $H=Q_{8}\times\mathbb{Z}_{p}=\langle a,b,c~{}|~{}a^{4}=b^{4}=c^{p}=[a,c]=[b,c]=1,a^{2}=b^{2},a^{b}=a^{-1}\rangle$. Then $Q_{8}$ and $\mathbb{Z}_{p}$ are characteristic in $H$, and hence $c^{\beta}\in\langle c\rangle$ and $\langle a,b\rangle^{\beta}=\langle a,b\rangle$.

Let $y^{-1}=a,abc^{-1}$ or $bc$. By Eq (6), $S^{\beta}=\{a,1,ac^{-1},bc,ab^{-1}c^{-1}\}$, $\{abc^{-1}$, $b^{-1}c^{-1}$, $abc^{-2}$, $1$, $ac^{-2}\}$ or $\{bc,ba^{-1}c,b,a^{-1}c^{2},1\}$, which is impossible because $c^{\beta}\in\langle c\rangle$.

Let $y=1$. Then $S^{\beta}=\{1,a^{-1},c^{-1},a^{-1}bc,b^{-1}c^{-1}\}$ by Eq. (6). Since $a^{\beta}\in\langle a,b\rangle$ and $c^{\beta}\in\langle c\rangle$, we have $a^{\beta}=a^{-1}$, $c^{\beta}=c^{-1}$ and $\{(abc^{-1})^{\beta},(bc)^{\beta}\}=\{a^{-1}bc,b^{-1}c^{-1}\}$, and since $(bc)^{\beta}=b^{\beta}c^{-1}\in\langle a,b\rangle c^{-1}$, we have $(bc)^{\beta}=b^{-1}c^{-1}$. Hence $b^{\beta}=b^{-1}$ and $(abc^{-1})^{\beta}=a^{-1}bc$. However, $a^{-1}bc=(abc^{-1})^{\beta}=a^{\beta}b^{\beta}(c^{-1})^{\beta}=a^{-1}b^{-1}c$, forcing $b^{2}=1$, a contradiction.

Let $y^{-1}=c$. Then $S^{\beta}=\{c,a^{-1}c,1,a^{-1}bc^{2},b^{-1}\}$, yielding that $a^{\beta}=b^{-1}$ and $c^{\beta}=c$. Since $(abc^{-1})^{\beta}=(ab)^{\beta}c^{-1}\in S^{\beta}$, we have $(ab)^{\beta}c^{-1}=a^{-1}c$ or $a^{-1}bc^{2}$, forcing $c^{2}=1$ or $c^{3}=1$, contradicting $p\geq 7$. This completes the proof. ∎

To end this section, we describe some Haar graphs of small orders that are not vertex-transitive, and this can be checked easily by the computer software Magma [5].

Lemma 3.3**.**

Let $G=H_{i}$ and $S$ be given in the following table for each $1\leq i\leq 9$:

Then $\hbox{\rm Aut}(\hbox{\rm H}(G,S))$ is not vertex-transitive and $G\not\in{\cal BC}$.

4 Proof of Theorem 1.4

In this section, we aim to prove Theorem 1.4. First, we need two lemmas.

Lemma 4.1**.**

A non-abelian $2$-group belongs to the class ${\cal BC}$ if and only if it is isomorphic to $D_{8}$, $Q_{8}$ or $Q_{8}\times\mathbb{Z}_{2}$.

Proof.

By Proposition 1.3, $D_{8}\in{\cal BC}$ and $Q_{8}\in{\cal BC}$. For $Q_{8}\times\mathbb{Z}_{2}$, let $\Gamma=\hbox{\rm H}(Q_{8}\times\mathbb{Z}_{2},S)$ be a Haar graph with $1\in S$. If $\Gamma$ is not connected, then $\langle S\rangle<Q_{8}\times\mathbb{Z}_{2}$ (see [9, Lemma 1 (i)]), and either $\langle S\rangle$ is abelian or $\langle S\rangle\cong Q_{8}$. This implies that the Haar graph $\hbox{\rm H}(\langle S\rangle,S)$ is a Cayley graph, and since $\Gamma$ is a union of components with each isomorphic to $\hbox{\rm H}(\langle S\rangle,S)$, $\Gamma$ is a Cayley graph. If $\Gamma$ is connected, a computation by Magma [5] shows that all connected Haar graphs of $Q_{8}\times\mathbb{Z}_{2}$ are Cayley graphs. Thus, $Q_{8}\times\mathbb{Z}_{2}\in{\cal BC}$.

Let $H$ be a non-abelian $2$-group and $H\in{\cal BC}$. To prove the necessity, it suffices to show that $H\cong D_{8},Q_{8}$ or $Q_{8}\times\mathbb{Z}_{2}$.

Case 1: $|H|\leq 8$.

Since $H$ is a non-abelian $2$-group, we have $H\cong D_{8}$ or $Q_{8}$.

Case 2: $|H|=16$.

Note that all non-abelian groups of order $16$ can be found in [13] (this can also be obtained by the computer software Magma [5]). By Proposition 2.3 (iii), $H$ has a subgroup isomorphic to $D_{8}$ or $Q_{8}$, and hence $H\cong H_{i}$ for some $1\leq i\leq 6$:

[TABLE]

By Proposition 1.2, $H_{5}\not\in{\cal BC}$, and by Lemma 3.3, $H_{i}\not\in{\cal BC}$ for each $1\leq i\leq 4$. It follows that $H\cong H_{6}=Q_{8}\times\mathbb{Z}_{2}$.

Case 3: $|H|\geq 32$.

Since $H\in{\cal BC}$, Proposition 2.2 implies that each subgroup of $H$ belongs to ${\cal BC}$. If each subgroup of $H$ of order $32$ is abelian, then $H$ has an inner abelian subgroup of order at least $64$, which is impossible by Proposition 1.3. Thus, $H$ has a non-abelian subgroup of order $32$, say $L$. Then $L\in{\cal BC}$. Similarly, $L$ has a non-abelian subgroup of order $16$, and by the proof of Case 2, each non-abelian subgroup of $L$ of order $16$ is isomorphic to $Q_{8}\times\mathbb{Z}_{2}$. By checking the non-abelian groups of order $32$ listed in [13], we have $L\cong Q_{8}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2}$, and by Lemma 3.3, $L\notin{\cal BC}$, a contradiction. ∎

Lemma 4.2**.**

Let $p$ be an odd prime, and let $H$ be a non-abelian $\{2,p\}$-group with $p\mid|H|$. Then $H\in\mathcal{BC}$ if and only if $H\cong D_{6}$ or $D_{10}$.

Proof.

By Proposition 1.2, $D_{6}\in{\cal BC}$ and $D_{10}\in{\cal BC}$. Let $H$ be a non-abelian $\{2,p\}$-group with $p\mid|H|$. To prove the necessity, suppose to the contrary that $H$ is a minimal counterexample, that is, $H\in{\cal BC}$ has the smallest order such that $H\not\cong D_{6}$ or $D_{10}$.

Denote by $P$ and $P_{2}$ a Sylow $p$-subgroup and a Sylow $2$-subgroup of $H$, respectively. Then $H=PP_{2}$, and by Proposition 2.2, $P\in{\cal BC}$ and $P_{2}\in{\cal BC}$. By Proposition 2.3 (ii), $P$ is abelian, and by Lemma 4.1, either $P_{2}$ is abelian or $P_{2}\cong D_{8}$, $Q_{8}$, or $Q_{8}\times\mathbb{Z}_{2}$. Now we consider the two cases depending whether $P_{2}$ is normal in $H$.

Case 1: $P_{2}\unlhd H$.

Suppose that $P_{2}$ is abelian. It follows from Proposition 2.3 (iii) that $H$ has a subgroup $D_{2p}$ with $p=3$ or $5$. Since $P_{2}$ is the unique Sylow $2$-subgroup of $H$, all involutions of $D_{2p}$ are contained in $P_{2}$, and since $D_{2p}$ can be generated by its two involutions, we have $D_{2p}\leq P_{2}$, which is impossible. Hence $P_{2}$ is non-abelain. By Proposition 2.2, $P_{2}\in{\cal BC}$, and by Lemma 4.1, $P_{2}\cong D_{8}$, $Q_{8}$ or $Q_{8}\times\mathbb{Z}_{2}$. It follows $H=P_{2}\rtimes P$, and by the minimality of $H$, $|P|=p$. Thus, $H=P_{2}\rtimes P\cong D_{8}\rtimes\mathbb{Z}_{p}$, $Q_{8}\rtimes\mathbb{Z}_{p}$ or $(Q_{8}\times\mathbb{Z}_{2})\rtimes\mathbb{Z}_{p}$.

Consider the centralizer $C_{H}(P_{2})$ of $P_{2}$ in $H$. If $P\leq C_{H}(P_{2})$, then $H=P_{2}\times P\cong D_{8}\times\mathbb{Z}_{p}$, $Q_{8}\times\mathbb{Z}_{p}$, or $Q_{8}\times\mathbb{Z}_{2}\times\mathbb{Z}_{p}$, and by Lemmas 3.1 and 3.2, $D_{8}\times\mathbb{Z}_{p}\not\in{\cal BC}$ and $Q_{8}\times\mathbb{Z}_{p}\not\in{\cal BC}$. Also we have $Q_{8}\times\mathbb{Z}_{2}\times\mathbb{Z}_{p}\not\in{\cal BC}$ because otherwise $Q_{8}\times\mathbb{Z}_{2}\times\mathbb{Z}_{p}\in{\cal BC}$ implies $Q_{8}\times\mathbb{Z}_{p}\in{\cal BC}$ by Proposition 2.2. Thus, $H\not\in{\cal BC}$, a contradiction. Hence $P\nleq C_{H}(P_{2})$, and since $|P|=p$, we have $P\cap C_{H}(P_{2})=1$ and so $C_{H}(P_{2})\leq P_{2}$. Note that $\hbox{\rm Aut}(D_{8})\cong D_{8}$, $\hbox{\rm Aut}(Q_{8})\cong S_{4}$, and $\hbox{\rm Aut}(Q_{8}\times\mathbb{Z}_{2})\cong\mathbb{Z}_{2}^{3}\rtimes S_{4}$. Since $N_{H}(P_{2})/C_{H}(P_{2})=H/C_{H}(P_{2})\lesssim\hbox{\rm Aut}(P_{2})$, we have $P_{2}\cong Q_{8}$ or $Q_{8}\times\mathbb{Z}_{2}$, and $P\cong\mathbb{Z}_{3}$, which implies that a generator of $P$ induces (by conjugacy) an automorphism of $P_{2}$ of order $3$. For $P_{2}\cong Q_{8}$, let $P_{2}=\langle a,b\mid a^{4}=b^{4}=1,a^{2}=b^{2},a^{b}=a^{-1}\rangle$, and let $\alpha$ be the automorphism of $P_{2}$ of order $3$ induced by the map $a\mapsto b$ and $b\mapsto ab$. Since all the automorphisms of $P_{2}$ of order $3$ are conjugate in $\hbox{\rm Aut}(P_{2})$, we have $H\cong P_{2}\rtimes\langle\alpha\rangle=\langle a,b,\alpha\mid a^{4}=b^{4}=\alpha^{3}=1,a^{2}=b^{2},a^{b}=a^{-1},a^{\alpha}=b,b^{\alpha}=ab\rangle$. By Lemma 3.3, $H\cong H_{6}$ and $H\notin{\cal BC}$, a contradiction. Similarly, for $P_{2}\cong Q_{8}\times\mathbb{Z}_{2}$, let $P_{2}=\langle a,b\mid a^{4}=b^{4}=1,a^{2}=b^{2},a^{b}=a^{-1}\rangle\times\langle c\rangle$ with $\langle c\rangle\cong\mathbb{Z}_{2}$, and let $\alpha$ be the automorphism of $P_{2}$ of order $3$ induced by the map $a\mapsto b$, $b\mapsto ab$ and $c\mapsto c$. It follows that $H\cong(\langle a,b\rangle\times\langle c\rangle)\rtimes\langle\alpha\rangle\cong H_{6}\times\mathbb{Z}_{2}$ and hence $H\notin{\cal BC}$ as $H_{6}\not\in{\cal BC}$, a contradiction.

Case 2: $P_{2}\ntrianglelefteq H$.

Since $P$ is abelian, we have $P\leq C_{H}(P)\leq N_{H}(P)$. If $C_{H}(P)=N_{H}(P)$, then the Burnside’s $p$-nilpotency criterion implies that $P$ has a normal complement $N$, that is, $N$ is a normal Sylow $2$-subgroup of $H$, which is impossible by Case 1. Thus, we may assume that $P\leq C_{H}(P)<N_{H}(P)$, and hence there exists a $2$-element $g$ such that $g\in N_{H}(P)$ and $g\not\in C_{H}(P)$. In particular, $P\rtimes\langle g\rangle$ is a non-abelian subgroup of $H$ and $p\ |\ |P\rtimes\langle g\rangle|$. By the minimality of $H$, either $H=P\rtimes\langle g\rangle$, or $H>P\rtimes\langle g\rangle\cong D_{6}$ or $D_{10}$.

Subcase 2.1: $H=P\rtimes\langle g\rangle$.

In this case, $P_{2}=\langle g\rangle$ and $P\unlhd H$. By Proposition 1.3, $H$ contains a proper subgroup $K$ such that $K\cong D_{6}$ or $D_{10}$. Then $p=3$ or $5$. Let $K=\langle a,b\ |\ a^{p}=b^{2}=1,a^{b}=a^{-1}\rangle$. We may assume $b\in P_{2}$, and since $P_{2}=\langle g\rangle$, $b$ is the unique involution in $P_{2}$. Furthermore, $PK\leq H$ is non-abelian. Let $|P|=p^{s}$ for some $s\geq 1$. Then $|PK|=2p^{s}$.

If $s=1$ then $P\cong\mathbb{Z}_{p}$ and hence $P\leq K$. Since $K\cong D_{6}$ or $D_{10}$, we have $K<H$, and since $K/P<H/P\cong P_{2}$, $H$ contains a subgroup of order $4p$ with $K$ as a subgroup of index $2$. By the minimality of $H$, we have $P_{2}=\langle g\rangle\cong\mathbb{Z}_{4}$. It follows $b=g^{2}$, and since $a^{b}=a^{-1}$, we have $p=5$ and $a^{g}=a^{2}$ or $a^{3}$. This implies that $H\cong F_{20}$ and by Lemma 3.3, $H\not\in{\cal BC}$, a contradiction. Thus, $s\geq 2$.

Suppose $s\geq 3$. Note that $P\unlhd H$ and $K=\langle a,b\rangle\cong D_{2p}<H$ with $o(a)=p$. By the minimality of $H$, we have $H=PK$. Then $|H|=2p^{s}$ and $P_{2}=\langle g\rangle=\langle b\rangle\cong\mathbb{Z}_{2}$. If $P$ is cyclic, it is easy to see that $H$ is dihedral, which is impossible by Proposition 1.2. Thus, $P$ is not cyclic, and since $P$ is abelian, Proposition 2.4 implies that $P$ has an element $c$ of order $p$ with $\langle c\rangle\cap\langle a\rangle=1$. If $c^{b}\not\in\langle c\rangle$ then $\langle c,c^{b},b\rangle$ is a non-abelian subgroup of order $2p^{2}$, and by the minimality of $H$, we have $H=\langle c,c^{b},b\rangle$, which is impossible because $|H|=2p^{s}$ with $s\geq 3$. Similarly, if $c^{b}\in\langle c\rangle$ then $\langle a,c,b\rangle$ is a non-abelian subgroup of order $2p^{2}$, which is also impossible.

Thus, $s=2$ and $|H|=2p^{2}$. From the elementary group theory we know that up to isomorphism there are three non-abelian groups of order $2p^{2}$ defined as:

[TABLE]

Thus, $H\cong H_{1}(p),H_{2}(p)$ or $H_{3}(p)$. Note that $H_{1}(p)\cong D_{2p^{2}}$ and $H_{3}(p)\cong D_{2p}\times\mathbb{Z}_{p}$. By Proposition 1.2 and Lemma 3.1, $H_{1}(p)\not\in{\cal BC}$ and $H_{3}(p)\not\in{\cal BC}$. Recall that $p=3$ or $5$. By Lemma 3.3, $H_{2}(p)\not\in{\cal BC}$. It follows that $H\not\in{\cal BC}$, a contradiction.

Subcase 2.2: $H>P\rtimes\langle g\rangle\cong D_{6}$ or $D_{10}$.

Clearly, $\langle g\rangle\cong\mathbb{Z}_{2}$, $p=3$ or $5$, and $P\cong\mathbb{Z}_{3}$ or $\mathbb{Z}_{5}$. Let $P=\langle a\rangle\cong\mathbb{Z}_{p}$. Without any loss of generality, we may assume that $g\in P_{2}$. First we prove two claims.

Claim 1: $P_{2}\cong D_{8}$, $Q_{8}$, or $Q_{8}\times\mathbb{Z}_{2}$.

Recall that either $P_{2}$ is abelian, or $P_{2}\cong D_{8},Q_{8}$ or $Q_{8}\times\mathbb{Z}_{2}$.

Suppose that $P_{2}$ is abelian. Then $P_{2}\leq C_{H}(P_{2})\leq N_{H}(P_{2})$. Since$|H:P_{2}|=p$ and $P_{2}\ntrianglelefteq H$, we have $P_{2}=C_{H}(P_{2})=N_{H}(P_{2})$. By the Burnside’s $p$-nilpotency criterion, $P$ is the normal complement of $P_{2}$ in $H$, that is, $P\unlhd H$ and $H=P\rtimes P_{2}$. Note that $\mathbb{Z}_{2}\cong\langle g\rangle\cong(P\rtimes\langle g\rangle)/P<H/P\cong P_{2}$. Then $H/P$ contains a subgroup of order $4$, and hence $H$ has a non-abelian subgroup $L$ with $P\rtimes\langle g\rangle$ as a subgroup of index $2$. By the minimality of $H$, $H=L$ and so $|P_{2}|=4$. In particular, $P_{2}\cong\mathbb{Z}_{4}$ or $\mathbb{Z}_{2}^{2}$.

Let $P_{2}=\langle b\rangle\cong\mathbb{Z}_{4}$. Then $H=P\rtimes P_{2}=\langle a,b,g\mid a^{p}=b^{4}=1,g=b^{2},a^{b}=a^{i},a^{g}=a^{-1}\rangle$ with $i\in\mathbb{Z}_{p}^{*}$. Since $a^{-1}=a^{g}=a^{b^{2}}=a^{i^{2}}$, we have $i^{2}=-1$ in $\mathbb{Z}_{p}$, and since $p=3$ or $5$, we have $i=\pm 2$ and $p=5$. Hence $H$ is isomorphic to the Frobenius group $F_{20}$ of order 20 and so $H\notin{\cal BC}$ by Lemma 3.3, a contradiction.

Let $P_{2}=\langle b,g\rangle\cong\mathbb{Z}_{2}^{2}$. Considering the action of $P_{2}$ on $P$, we have $H=P\rtimes P_{2}\cong\langle a,b,g\mid a^{p}=b^{2}=g^{2}=[b,g]=1,a^{b}=a,a^{g}=a^{-1}\rangle\cong D_{4p}$, and by Proposition 1.2, $H\notin{\cal BC}$, a contradiction.

It follows that $P_{2}\cong D_{8}$, $Q_{8}$, or $Q_{8}\times\mathbb{Z}_{2}$, as claimed.

Claim 2: $P\unlhd H$.

Let $N$ be a minimal normal subgroup of $H$. Since $P\cong\mathbb{Z}_{p}$, we have $N=P$ or $N=\mathbb{Z}_{2}^{\ell}$ for some $\ell\geq 1$. If $N=P$ then $P\unlhd H$, as claimed. If $N=\mathbb{Z}_{2}^{\ell}$, then $(P\rtimes\langle g\rangle)N>P\rtimes\langle g\rangle$ because $P\rtimes\langle g\rangle$ ($\cong D_{6}$ or $D_{10}$) has non-normal Sylow $2$-subgroups. By the minimality of $H$, we have $H=N(P\rtimes\langle g\rangle)$. Since $P_{2}\ntrianglelefteq H$, we have $N<P_{2}$ and hence $NP<H$. Clearly, $NP\not\cong D_{6}$ or $D_{10}$, and the minimality of $H$ implies that $NP$ is abelian. It follows $NP=N\times P$ and $H=(N\times P)\rtimes\langle g\rangle$. Thus, $P\unlhd H$, as claimed.

By Claim 2, $PK\leq H$ for any subgroup $K\leq H$, and by Claim 1, $P_{2}\cong D_{8}$, $Q_{8}$, or $Q_{8}\times\mathbb{Z}_{2}$. If $P_{2}\cong Q_{8}\times\mathbb{Z}_{2}$ then $H$ has a proper subgroup isomorphic to $P\rtimes Q_{8}$, and the minimality of $H$ implies that either $P\rtimes Q_{8}$ is abelian, or $P\rtimes Q_{8}\cong D_{6}$ or $D_{10}$, both of which are impossible. If $P_{2}\cong Q_{8}$, then $P\rtimes\langle a\rangle$ is a proper subgroup of $H$ for any element $a$ of order $4$ in $P_{2}$, and by the minimality of $H$, $P\rtimes\langle a\rangle$ is abelian because $P\rtimes\langle a\rangle\ncong D_{6}$ or $D_{10}$. This implies that $[P,P_{2}]=1$ as $P_{2}\cong Q_{8}$, and hence $H=P\times P_{2}$, which is impossible because $P\rtimes\langle g\rangle\cong D_{6}$ or $D_{10}$. If $P_{2}\cong D_{8}$, let $P_{2}=\langle a,g\mid a^{4}=g^{2}=1,a^{g}=a^{-1}\rangle$ and $P=\langle b\rangle$, and a similar argument as above implies $[P,\langle a\rangle]=1$. Since $P\rtimes\langle g\rangle\cong D_{6}$ or $D_{10}$, we have $b^{g}=b^{-1}$. It follows that

[TABLE]

By Proposition 1.2, $H\notin{\cal BC}$, a contradiction.

∎

Now we are ready to prove Theorem 1.4.

Proof of Theorem 1.4:.

By Lemmas 4.1 and 4.2, $D_{8}$, $Q_{8}$, $Q_{8}\times\mathbb{Z}_{2}$, $D_{6}$ and $D_{10}$ belong to ${\cal BC}$. To prove the necessity, let $H$ be a non-abelian group with $H\in{\cal BC}$.

By Proposition 2.3 (i), $H$ is solvable, and by Proposition 2.3 (iii), $2\mid|H|$. Let $L$ and $K$ be a Sylow $2$-subgroup and a Hall $2^{\prime}$-subgroup of $H$, respectively. By Proposition 2.3, $K\in{\cal BC}$ and $L\in{\cal BC}$. Furthermore, $H=KL$, and by Proposition 2.3 (iii), $K$ is abelian. Let $p_{1},\ldots,p_{k}$ be all distinct odd prime divisors of $|H|$, and let $P_{i}$ be a Sylow $p_{i}$-subgroup of $H$ contained in $K$ for $1\leq i\leq k$. Then $K=P_{1}\times P_{2}\times\cdots\times P_{k}$. If $k=0$ then $H=L$ is a $2$-group, and by Lemma 4.1, $H\cong D_{8}$, $Q_{8}$ or $Q_{8}\times\mathbb{Z}_{2}$. If $k=1$ then $H$ is a $\{2,p_{1}\}$-group, and by Lemma 4.2, then $H\cong D_{6}$ or $D_{10}$. In what follows, we assume $k\geq 2$.

If each Hall $\{2,p_{i}\}$-subgroup of $H$ is abelian for each $1\leq i\leq k$, then $L$ is abelian and $H=K\times L$, forcing that $H$ is abelian, a contradiction. Hence $H$ has a non-abelian Hall $\{2,p_{\ell}\}$-subgroup for some prime $p_{\ell}$, say $M$. It follows from Proposition 2.3 that $M\in{\cal BC}$ and from Lemma 4.2 that $M\cong D_{6}$ or $D_{10}$, yielding that $L\cong\mathbb{Z}_{2}$. Hence $K\unlhd H$ and $H=K\rtimes P_{2}=(P_{1}\times\cdots\times P_{k})\rtimes\mathbb{Z}_{2}$. It follows that $P_{i}\unlhd H$ for each $1\leq i\leq k$, and hence $P_{i}L\leq H$. Furthermore, we may assume $M=P_{\ell}L$. Again by Lemma 4.2, for each $1\leq i\leq k$ we have either $P_{i}L=P_{i}\times L$ (abelian), or $P_{i}L\cong D_{6}$ or $D_{10}$.

Suppose $P_{j}L=P_{j}\times L$ for some $1\leq j\leq k$. Recall that $M=LP_{\ell}$ is a Hall $\{2,p_{\ell}\}$-subgroup of $H$, and $M\cong D_{6}$ or $D_{10}$. Clearly, $p_{\ell}\not=p_{j}$, and $MP_{j}=LP_{\ell}P_{j}=M\times P_{j}$. Then $H$ contains a subgroup isomorphic to $D_{6}\times\mathbb{Z}_{p_{j}}$ or $D_{10}\times\mathbb{Z}_{p_{j}}$, which is impossible by Lemma 3.1. Note that if $p_{i}\not=3,5$, then $P_{i}L=P_{i}\times L$ because $P_{i}L\ncong D_{6}$ or $D_{10}$. This implies that $k=2$ as $k\geq 2$, and $\{p_{1},p_{2}\}=\{3,5\}$. Furthermore, $\{P_{1}L,P_{2}L\}=\{D_{6},D_{10}\}$ and hence $H=P_{1}P_{2}L\cong D_{30}$ because a group of order $15$ must be cyclic, which is impossible by Proposition 1.2. This completes the proof. ∎