Finite groups whose intersection power graphs are toroidal and projective-planar

Huani Li; Xuanlong Ma; Ruiqin Fu

doi:10.1515/math-2021-0071

Article Open Access

Finite groups whose intersection power graphs are toroidal and projective-planar

Huani Li , Xuanlong Ma and Ruiqin Fu

Published/Copyright: August 20, 2021

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$Open Mathematics$

From the journal Open Mathematics Volume 19 Issue 1

Abstract

The intersection power graph of a finite group G is the graph whose vertex set is G , and two distinct vertices x and y are adjacent if either one of x and y is the identity element of G , or ⟨ x ⟩ ∩ ⟨ y ⟩ is non-trivial. In this paper, we completely classify all finite groups whose intersection power graphs are toroidal and projective-planar.

Keywords: intersection power graph; finite group; genus

MSC 2010: 05C25; 05C10

1 Introduction

A simple graph is an undirected graph without loops and multiple edges. A graph is called finite if its vertex set is finite. All graphs considered in this paper are finite and simple. Let Γ be a graph. Denote by V ( Γ ) and E ( Γ ) the vertex set and the edge set of Γ , respectively. An embedding of a graph into a surface is a drawing of the graph on the surface in such a way that its edges may intersect only at their endpoints. A graph is called planar if it can be embedded in the plane. A non-planar graph can be embedded in some surface obtained from the sphere by attaching some handles or crosscaps. We denote by S k a sphere with k ≥ 0 handles and by N k a sphere with k crosscaps. Note that S 0 = N 0 is the sphere, while S 1 and N 1 are a torus and a projective plane, respectively. The smallest non-negative integer k such that a graph Γ can be embedded on S k is called the orientable genus or genus of Γ and is denoted by γ ( Γ ) . The non-orientable genus of Γ , denoted by γ ¯ ( Γ ) , is instead the smallest integer k such that Γ can be embedded on N k . A graph Γ is called toroidal if γ ( Γ ) = 1 and Γ is called projective-planar if γ ¯ ( Γ ) = 1 . Note that Γ is planar if and only if γ ( Γ ) = 0 , if and only if, γ ¯ ( Γ ) = 0 . We follow the book [1] for undefined notation and terminology.

Graphs associated with groups and other algebraic structures have been actively investigated in the literature, because they have valuable applications [2] and are related to automata theory [3,4]. For example, the Cayley graph of a group, which has a long history. The undirected power graph of a group G , denoted P ( G ) , is a simple graph whose vertex set is G and two distinct vertices are adjacent if one is a power of the other. Kelarev and Quinn [5] first introduced the concept of power graph of a group, as a directed graph. The concept of the undirected power graph of a group was introduced first by Chakrabarty et al. [6]. In the past two decades, the study of power graphs of groups has been growing. See, for example, [7,8,9, 10,11,12, 13,14,15] and the survey paper [16] with many results and open questions on power graphs.

In 2018, Bera [17] defined the intersection power graph P I ( G ) of a group G , where the vertex set of P I ( G ) is G , and distinct vertices x and y are adjacent if either one of x and y is the identity element of G or ∣ ⟨ x ⟩ ∩ ⟨ y ⟩ ∣ > 1 . In [17], the author studied some properties of P I ( G ) and determined the full automorphism group of the intersection power graph of a cyclic group. A book is a collection of half-planes having the same line as their boundary. A planar embedding of a graph into a book is called the book embedding. The smallest possible number of half-planes for any book embedding of a graph is called the book thickness of the graph. In [18], the authors determined the finite groups whose intersection power graphs have book thickness at most two.

All groups considered in this paper are finite. Let G be a group. The set of orders of elements of G is denoted by π e ( G ) . Z n denotes the cyclic group of order n . Also, we use Ψ to denote the set of all groups G such that the following two conditions hold:

{ 4 } ⊆ π e ( G ) ⊆ { 1 , 2 , 3 , 4 } ;
G has exactly two cyclic subgroups of order 4 intersecting at a subgroup of order 2.

Note that Z 2 × Z 4 ∈ Ψ . Also, we identify the relevant groups using their unique identifiers in the SmallGroups library [19], which is distributed with GAP [20]. The mth group of order n in the SmallGroups library is identified as SmallGroup ( n , m ) .

One can easily see that, for any group G , P ( G ) is a spanning subgraph of P I ( G ) . In [21], the authors classified all finite groups whose power graphs have (non)orientable genus one. In [22], the authors classified all abelian groups whose intersection power graphs have (non)orientable genus one. In this paper, we completely classify all groups whose intersection power graphs have (non)orientable genus one. Our main results are the following theorems, where D 2 n denotes the dihedral group of order 2 n .

Theorem 1.1

Let G be a group. Then P I ( G ) is toroidal if and only if

G ∈ Ψ ∪ { Z 5 , Z 6 , Z 7 , D 10 , D 12 , D 14 , SmallGroup ( 20 , 3 ) , SmallGroup ( 21 , 1 ) } .

Theorem 1.2

Let G be a group. Then P I ( G ) is projective-planar if and only if

G ∈ Ψ ∪ { Z 5 , Z 6 , D 10 , D 12 , S m a l l G r o u p ( 20 , 3 ) } .

2 Preliminaries

In this section, we briefly recall some notations, terminologies, and basic results and prove a lemma, which we need in the sequel.

G always denotes a group and its identity element is e . Let g ∈ G . The order of g , denoted o ( g ) , is the cardinality of the cyclic subgroup ⟨ g ⟩ generated by g . The exponent of G , denoted by exp ( G ) , is defined as the least common multiple of the orders of all elements of the group. Let a , b ∈ G . Then [ a , b ] = a − 1 b − 1 a b is called the commutator of a and b . Note that [ a , b ] = e if and only if a and b commute.

Recall that D 2 n is the dihedral group of order 2 n , and a presentation of D 2 n is given by

D 2 n = ⟨ a , b : a n = b 2 = e , b a b = a − 1 ⟩ .

Observe that D 2 n is abelian if and only if n ∈ { 1 , 2 } . Moreover, we have

(1) D 2 n = ⟨ a ⟩ ∪ { b , a b , a 2 b , … , a n − 1 b } , o ( a i b ) = 2 for any 1 ≤ i ≤ n .

Note that Z 2 × D 8 ∈ Ψ .

For n ≥ 2 , Johnson [23, pp. 44–45] defined the generalized quaternion group, which is denoted by Q 4 n and has the presentation

Q 4 n = ⟨ x , y : x n = y 2 , x 2 n = e , y − 1 x y = x − 1 ⟩ .

It is known that Q 4 n has order 4 n and its unique involution is x n . Moreover, o ( x i y ) = 4 for any 1 ≤ i ≤ 2 n , and Q 4 n = ⟨ x ⟩ ∪ { x i y : 1 ≤ i ≤ 2 n } . If n = 2 , then Q 8 is the quaternion group of order 8. Note that Q 8 ∉ Ψ .

Recall now the following elementary results.

Lemma 2.1

[24, Theorem 5.4.10(ii)] Let p be a prime. Then a p -group having a unique subgroup of order p is either cyclic or generalized quaternion.

Lemma 2.2

[25, Lemma 3.1] There is no group that has precisely two cyclic subgroups of order 6.

A graph is called complete if every two distinct vertices of this graph are adjacent. The complete graph of order n is denoted by K n . A complete bipartite graph Γ is a graph whose vertex set can be partitioned into two non-empty partite sets V 1 and V 2 such that for every two distinct vertices u ∈ V i and v ∈ V j where i , j ∈ { 1 , 2 } , u and v are adjacent if and only if i ≠ j . If ∣ V 1 ∣ = m and ∣ V 2 ∣ = n , then the complete bipartite graph Γ is denoted by K m , n .

Theorem 2.3

[1] Let n ≥ 3 and m ≥ 2 be two integers. Then

γ ( K n ) = 1 12 ( n − 3 ) ( n − 4 ) .
For n ≠ 7 , γ ¯ ( K n ) = 1 6 ( n − 3 ) ( n − 4 ) . Moreover, γ ¯ ( K 7 ) = 3 .
γ ( K m , n ) = 1 4 ( m − 2 ) ( n − 2 ) .
γ ¯ ( K m , n ) = 1 2 ( m − 2 ) ( n − 2 ) .

The following corollary is an immediate consequence of Theorem 2.3.

Corollary 2.4

γ ( K n ) = 1 if and only if n ∈ { 5 , 6 , 7 } .
γ ¯ ( K n ) = 1 if and only if n ∈ { 5 , 6 } .
γ ( K 4 , 6 ) = 2 and γ ¯ ( K 4 , 6 ) = 4 .

A block of a graph Γ is a subgraph B of Γ maximal with respect to the property that removing any single vertex of B does not disconnect B . It is easy to see that the intersection of two distinct blocks has at most one vertex (see (1) of [26]). Given a graph Γ , as stated in [26], there exists a unique finite collection ℬ of blocks B of Γ such that

Γ = ⋃ B ∈ ℬ B .

The collection ℬ is called the block decomposition of Γ . For more information on the block decomposition of a graph, the readers are referred to [26] and [27].

We use the following example to illustrate the block decomposition of a graph.

Example 2.5

Let G = Z 5 × Z 5 . Then P I ( G ) is a union of six complete graphs of order 5 that share the identity element of G , as displayed in Figure 1. Thus, the block decomposition of P I ( G ) is { B 1 , B 2 , … , B 6 } , where B i ≅ K 5 for each 1 ≤ i ≤ 6 .

$Figure 1 P I ( Z 5 × Z 5 ) {{\mathcal{P}}}_{I}\left({{\mathbb{Z}}}_{5}\times {{\mathbb{Z}}}_{5}) .$

Figure 1

P I ( Z 5 × Z 5 ) .

The following result tells us how to compute the (non)orientable genus of a graph using its blocks.

Theorem 2.6

[26, Theorem 1], [28, Corollary 3] Let Γ be a connected graph with block decomposition { B 1 , … , B n } . Then

γ ( Γ ) = ∑ i = 1 n γ ( B i ) .
If γ ¯ ( B i ) = 2 γ ( B i ) + 1 for each i , then
γ ¯ ( Γ ) = 1 − n + ∑ i = 1 n γ ¯ ( B i ) .
Otherwise,
γ ¯ ( Γ ) = 2 n − ∑ i = 1 n μ ( B i ) ,
where μ ( B i ) = max { 2 − 2 γ ( B i ) , 2 − γ ¯ ( B i ) } .

The following result is an easy observation and will be frequently used in the sequel sometimes without explicit reference.

Observation 2.7

Let Γ be a graph. Then Γ is planar if and only if for any subgraph Ω of Γ , Ω is planar.
If Ω is a subgraph of a graph Γ , then γ ( Ω ) ≤ γ ( Γ ) and γ ¯ ( Ω ) ≤ γ ¯ ( Γ ) .
If H is a subgroup of G , then P I ( H ) is an induced subgraph of P I ( G ) . In particular, γ ( P I ( H ) ) ≤ γ ( P I ( G ) ) and γ ¯ ( P I ( H ) ) ≤ γ ¯ ( P I ( G ) ) .
Every generator of Z n is adjacent to every other vertex in P I ( Z n ) . Moreover, Z n has φ ( n ) generators, where φ is Euler’s totient function.

Lemma 2.8

[17, Theorem 3.1] P I ( G ) is complete if and only if G is either a cyclic p -group or a generalized quaternion 2-group.

Lemma 2.9

[12, Theorem 2] P ( G ) is planar if and only if π e ( G ) ⊆ { 1 , 2 , 3 , 4 } .

Lemma 2.10

P I ( G ) is planar if and only if G satisfies the following two conditions:

π e ( G ) ⊆ { 1 , 2 , 3 , 4 } ;
If G has two distinct cyclic subgroups ⟨ x ⟩ and ⟨ y ⟩ of order 4, then ∣ ⟨ x ⟩ ∩ ⟨ y ⟩ ∣ = 1 .

Proof

If G satisfies (I) and (II), then it is easy to see that P I ( G ) = P ( G ) , and so P I ( G ) is planar by Lemma 2.9. Conversely, suppose that P I ( G ) is planar. Then P ( G ) also is planar. It follows from Lemma 2.9 that (I) holds. Now suppose, by contradiction, that G has two distinct cyclic subgroups of order 4, say, ⟨ x ⟩ and ⟨ y ⟩ , such that ∣ ⟨ x ⟩ ∩ ⟨ y ⟩ ∣ = 2 . Then the subgraph induced by ⟨ x ⟩ ∪ ⟨ y ⟩ is isomorphic to K 6 , against the fact that K 6 is non-planar.□

Lemma 2.10 implies that γ ( P I ( Z n ) ) = 0 if and only if n ∈ { 1 , 2 , 3 , 4 } , if and only if γ ¯ ( P I ( Z n ) ) = 0 . In the following, we classify all cyclic groups whose intersection power graphs have (non)orientable genus one.

Lemma 2.11

γ ( P I ( Z n ) ) = 1 if and only if n ∈ { 5 , 6 , 7 } .
γ ¯ ( P I ( Z n ) ) = 1 if and only if n ∈ { 5 , 6 } .

Proof

(i) Clearly, we have that P I ( Z 5 ) ≅ K 5 and P I ( Z 7 ) ≅ K 7 by Lemma 2.8. Thus, γ ( P I ( Z 5 ) ) = γ ( P I ( Z 7 ) ) = 1 by Corollary 2.4. Moreover, it is easy to see that P I ( Z 6 ) = P ( Z 6 ) , as displayed in Figure 2, where Z 6 = ⟨ g ⟩ . Thus, it follows from [21, Theorem 3.2] that γ ( P ( Z 6 ) ) = 1 , and so γ ( P I ( Z 6 ) ) = 1 .

Conversely, suppose that γ ( P I ( Z n ) ) = 1 . It follows from Lemma 2.10 that n ≥ 5 . Suppose, by contradiction that n ≥ 10 . Then φ ( n ) ≥ 4 , and so Z n has at least four generators. Since every generator of Z n is adjacent to every other vertex in P I ( Z n ) , we have that P I ( Z n ) has a subgraph isomorphic to K 4 , 6 . Since γ ( K 4 , 6 ) = 2 by Corollary 2.4, we also have γ ( P I ( Z n ) ) ≥ 2 , a contradiction. We conclude that n ≤ 9 . Moreover, since γ ( P I ( Z 8 ) ) = 2 and γ ( P I ( Z 9 ) ) = 3 by Lemma 2.8 and Theorem 2.3, we deduce n ≤ 7 , as desired.

(ii) By Lemma 2.8 and Corollary 2.4, we have γ ¯ ( P I ( Z 5 ) ) = γ ¯ ( K 5 ) = 1 . Moreover, since γ ¯ ( K 5 ) ≤ γ ¯ ( P I ( Z 6 ) ) ≤ γ ¯ ( K 6 ) and γ ¯ ( K 6 ) = 1 , we have γ ¯ ( P I ( Z 6 ) ) = 1 . Conversely, suppose that γ ¯ ( P I ( Z n ) ) = 1 . It follows from Lemma 2.10 that n ≥ 5 . Similar to the proof of (i), if n ≥ 10 , then P I ( Z n ) has a subgraph isomorphic to K 4 , 6 , which is impossible as γ ¯ ( K 4 , 6 ) = 4 by Corollary 2.4. Hence, n ≤ 9 . Moreover, since γ ¯ ( P I ( Z 7 ) ) = 3 , γ ¯ ( P I ( Z 8 ) ) = 4 and γ ¯ ( P I ( Z 9 ) ) = 5 by Lemma 2.8 and Theorem 2.3, we deduce n ≤ 6 , as desired.□

$Figure 2 P I ( Z 6 ) {{\mathcal{P}}}_{I}\left({{\mathbb{Z}}}_{6}) .$

Figure 2

P I ( Z 6 ) .

Lemma 2.12

Let G ∈ Ψ . Then γ ( P I ( G ) ) = 1 and γ ¯ ( P I ( G ) ) = 1 .

Proof

By definition, G has exactly one pair of cyclic subgroups ⟨ x ⟩ and ⟨ y ⟩ of order 4 such that ∣ ⟨ x ⟩ ∩ ⟨ y ⟩ ∣ = 2 . Then the subgraph induced by ⟨ x ⟩ ∪ ⟨ y ⟩ is isomorphic to K 6 . Moreover, by the definition of an intersection power graph, we have that P I ( G ) is a union of some complete graphs of order at most 4 and a complete graph of order 6 that share the identity element of G , as displayed in Figure 3. It follows that P I ( G ) has a block decomposition ℬ such that every block B ∈ ℬ is isomorphic to K 2 , K 3 , K 4 , or K 6 . Since P I ( G ) has precisely one block isomorphic to K 6 , we have that γ ( P I ( G ) ) = 1 and γ ¯ ( P I ( G ) ) = 1 by Theorem 2.6.□

$Figure 3 P I ( G ) {{\mathcal{P}}}_{I}\left(G) with G ∈ Ψ G\in \Psi .$

Figure 3

P I ( G ) with G ∈ Ψ .

3 Preliminary results

In this section, we prove some lemmas needed to give the proofs of our main theorems.

Lemma 3.1

Let G be a group with π e ( G ) ⊆ { 1 , 2 , 3 , 4 } . The following statements are equivalent:

γ ( P I ( G ) ) = 1 ;
γ ¯ ( P I ( G ) ) = 1 ;
G ∈ Ψ .

Proof

We first prove that (a) and (c) are equivalent. It follows from Lemma 2.12 that (c) implies (a). Suppose now that γ ( P I ( G ) ) = 1 . If G has no cyclic subgroup of order 4 or has precisely one cyclic subgroup of order 4, then Lemma 2.10 implies that P I ( G ) is planar, a contradiction. Thus, { 4 } ⊆ π e ( G ) ⊆ { 1 , 2 , 3 , 4 } and there exist at least two cyclic subgroups ⟨ x ⟩ and ⟨ y ⟩ of order 4 such that ∣ ⟨ x ⟩ ∩ ⟨ y ⟩ ∣ = 2 , we want to show that they are unique. If there exists z ∈ G ⧹ ( ⟨ x ⟩ ∪ ⟨ y ⟩ ) such that o ( z ) = 4 and ∣ ⟨ x ⟩ ∩ ⟨ z ⟩ ∣ = 2 , then the subgraph induced by ⟨ x ⟩ ∪ ⟨ y ⟩ ∪ ⟨ z ⟩ is isomorphic to K 8 , which has genus 2 by Theorem 2.3, a contradiction. We conclude that if there exists an element z ∈ G ⧹ ( ⟨ x ⟩ ∪ ⟨ y ⟩ ) of order 4, then ∣ ⟨ z ⟩ ∩ ⟨ x ⟩ ∣ = ∣ ⟨ z ⟩ ∩ ⟨ y ⟩ ∣ = 1 . Suppose, by contradiction, that there exist two distinct cyclic subgroups ⟨ x ′ ⟩ and ⟨ y ′ ⟩ of order 4 such that ∣ ⟨ x ′ ⟩ ∩ ⟨ y ′ ⟩ ∣ = 2 and x ′ , y ′ ∈ G ⧹ ( ⟨ x ⟩ ∪ ⟨ y ⟩ ) . By the previous argument we know that ∣ ⟨ x ′ ⟩ ∩ ⟨ x ⟩ ∣ = ∣ ⟨ x ′ ⟩ ∩ ⟨ y ⟩ ∣ = 1 . Thus, the subgraph Ω induced by ⟨ x ⟩ ∪ ⟨ y ⟩ ∪ ⟨ x ′ ⟩ ∪ ⟨ y ′ ⟩ is a union of two complete graphs of order 6 that share the identity vertex, as displayed in Figure 4. Thus, Ω has two blocks isomorphic to K 6 , and so γ ( Ω ) = 2 by Theorems 2.3 and 2.6. It follows that γ ( P I ( G ) ) ≥ γ ( Ω ) = 2 , contrary to γ ( P I ( G ) ) = 1 . We conclude that G has exactly one pair of cyclic subgroups of order 4 such that their intersection has size 2, and so G ∈ Ψ .

Note next that γ ¯ ( K 8 ) = 4 and γ ¯ ( Ω ) = 2 by Theorems 2.3 and 2.6. Then similar to the above proof, we get that γ ¯ ( P I ( G ) ) = 1 if and only if G ∈ Ψ .□

$Figure 4 The subgraph Ω \Omega .$

Figure 4

The subgraph Ω .

Lemma 3.2

Let G be a group with { 5 } ⊆ π e ( G ) ⊆ { 1 , 2 , 3 , 4 , 5 } . Then the following facts are equivalent:

γ ( P I ( G ) ) = 1 ;
G is isomorphic to one of the groups
(2) Z 5 , D 10 , SmallGroup ( 20 , 3 ) ≅ ⟨ g , w : g 5 = w 4 = e , w g w − 1 = g 2 ⟩ ;
γ ¯ ( P I ( G ) ) = 1 .

Proof

We first prove that if G is isomorphic to one group in (2), then γ ( P I ( G ) ) = 1 . By Lemma 2.11, we have γ ( P I ( Z 5 ) ) = 1 . Moreover, by (1), it is easy to see that P I ( D 10 ) has exactly one block isomorphic to K 5 , and every other block is isomorphic to K 2 . It follows that γ ( P I ( D 10 ) ) = 1 by Theorems 2.3 and 2.6. Now let G = SmallGroup ( 20 , 3 ) . Using GAP [20], we deduce that G has a unique subgroup of order 5 and has five distinct cyclic subgroups of order 4 such that the intersection of each two of them has size 1. It follows that P I ( G ) is a union of a complete graph of order 5 and four complete graphs of order 4 that share the identity vertex. Consequently, every block of P I ( G ) is isomorphic to K 4 or K 5 . Since P I ( G ) has precisely one block isomorphic to K 5 , we infer that γ ( P I ( G ) ) = 1 by Theorems 2.3 and 2.6, as desired.

We next prove that if γ ( P I ( G ) ) = 1 , then G is isomorphic to one group in (2). Suppose, by contradiction, that G has two distinct subgroups of order 5. Then the subgraph Δ induced by the two distinct subgroups of order 5 is a union of two complete graphs of order 5 that share the identity vertex. Thus, Δ has two distinct blocks isomorphic to K 5 , and so γ ( Δ ) = 2 by Theorems 2.3 and 2.6. This implies a contradiction as γ ( P I ( G ) ) ≥ γ ( Δ ) = 2 . As a consequence, G has precisely one subgroup ⟨ g ⟩ of order 5, and so ⟨ g ⟩ is normal in G . If G has an element x of order 3, then ⟨ g , x ⟩ = ⟨ g ⟩ ⟨ x ⟩ has order 15 and hence is cyclic, against the fact that G has no element of order 15. Thus, π e ( G ) ⊆ { 1 , 2 , 4 , 5 } . If π e ( G ) = { 1 , 5 } , since G has precisely one subgroup ⟨ g ⟩ of order 5, we have G = ⟨ g ⟩ ≅ Z 5 by Lemma 2.1. Therefore, in the following we may assume that G has some involutions.

Case 1

G has no elements of order 4.

Then π e ( G ) = { 1 , 2 , 5 } . Let a be an involution of G . Since every group of order 10 is isomorphic to D 10 or Z 10 , we have ⟨ a , g ⟩ ≅ D 10 by Lemma 2.11. Suppose, by contradiction, that G ≠ ⟨ a , g ⟩ . Then there exists an involution b ∈ G ⧹ ⟨ a , g ⟩ . Using again Lemma 2.11 we have ⟨ b , g ⟩ ≅ D 10 . If [ a , b ] = e , since a g a = b g b = g − 1 , we have ( g a b ) 2 = g b ( a g a ) b = g ( b g − 1 b ) = g 2 and so ( g a b ) 5 = g 4 g a b = a b , which implies o ( ( g a b ) 2 ) = 5 and o ( ( g a b ) 5 ) = 2 , it follows that o ( g a b ) = 10 , a contradiction. We conclude that [ a , b ] ≠ e . Recall that two distinct involutions generate a dihedral group. Since π e ( G ) = { 1 , 2 , 5 } , we deduce that ⟨ a , b ⟩ ≅ D 10 . Consequently, g ∈ ⟨ a , b ⟩ , and so ⟨ a , g ⟩ ⊆ ⟨ a , b ⟩ . Since ⟨ a , g ⟩ ≅ D 10 , we then have ⟨ a , b ⟩ = ⟨ a , g ⟩ , which implies b ∈ ⟨ a , g ⟩ , a contradiction. We conclude that G = ⟨ a , g ⟩ ≅ D 10 , as desired.

Case 2

G has an element w of order 4.

Then π e ( G ) = { 1 , 2 , 4 , 5 } . Let a = w 2 . By Lemma 2.11, we necessarily have ⟨ a , g ⟩ ≅ D 10 . We first claim that every involution of G belongs to ⟨ a , g ⟩ . Suppose, by contradiction, that G has an involution b ∈ G ⧹ ⟨ a , g ⟩ . Then, as in the proof of Case 1, we deduce that [ a , b ] ≠ e . It follows that ⟨ a , b ⟩ ≅ D 8 or D 10 . If ⟨ a , b ⟩ ≅ D 8 , since o ( a b ) = 4 and a g a = b g b = g − 1 , we have

( g a b ) 4 = ( g a b ) 2 ( g a b g a b ) = ( g a b ) 2 ( g a ( b g b ) b a b ) = ( g a b ) 2 ( g ( a g − 1 a ) a b a b ) = ( g a b ) 2 g 2 ( a b ) 2 = g 2 ( a b ) 2 g 2 ( a b ) 2 = g 2 ( a b a ) ( b g 2 b ) ( b a b a b ) = g 2 ( a b ) ( a g − 2 a ) ( a b a b a b ) = g 2 ( a ) ( b g 2 b ) ( b a b a b a b ) = g 2 ( a g − 2 a ) ( a b a b a b a b ) = g 4 ( a b a b a b a b ) = g 4

and so ( g a b ) 5 = a b , which implies o ( ( g a b ) 4 ) = 5 and o ( ( g a b ) 5 ) = 4 , it follows that o ( g a b ) = 20 , a contradiction. Hence, ⟨ a , b ⟩ ≅ D 10 . Therefore, we have b ∈ ⟨ a , b ⟩ = ⟨ a , g ⟩ , a contradiction.

Let P be a Sylow 2-subgroup of G with w ∈ P . Since G has a unique subgroup of order 5 by Lemma 2.1, we have that ⟨ g ⟩ is the unique Sylow 5-subgroup of G . Since π e ( G ) = { 1 , 2 , 4 , 5 } , we have G = P ⟨ g ⟩ . Note that G has precisely five involutions a , g a , g 2 a , g 3 a , g 4 a by the above claim. If P has an involution u with u ≠ a , then u = g i a for some 1 ≤ i ≤ 4 , and so g i a a − 1 = g i ∈ P , a contradiction as o ( g i ) = 5 . We conclude that P has a unique involution. Note that π e ( G ) = { 1 , 2 , 4 , 5 } . By Lemma 2.1, we know that P is isomorphic to either Z 4 or Q 8 . If P ≅ Q 8 , then the subgraph induced by P would be isomorphic to K 8 by Lemma 2.8, which is impossible as γ ( K 8 ) = 2 . Therefore, P ≅ Z 4 , and so G ≅ Z 4 ⋉ Z 5 . By GAP [20], there are five groups of order 20 up to isomorphism, and every group H of order 20 has an element of order 10 if H ≇ SmallGroup ( 20 , 3 ) . Thus, G ≅ SmallGroup ( 20 , 3 ) ≅ ⟨ g , w : g 5 = w 4 = e , w g w − 1 = g 2 ⟩ , as desired. Thus, (a) and (b) are equivalent.

Note next that γ ¯ ( K 8 ) = 4 and γ ¯ ( Δ ) = 2 by Theorems 2.3 and 2.6. Then similar to the above proof, we get that (b) and (c) are equivalent.□

Lemma 3.3

Let G be a group with { 7 } ⊆ π e ( G ) ⊆ { 1 , 2 , 3 , 4 , 7 } . Then γ ( P I ( G ) ) = 1 if and only if G is isomorphic to one of the following groups:

Z 7 , D 14 , SmallGroup ( 21 , 1 ) ≅ ⟨ g , w : g 7 = w 3 = e , w − 1 g w = g 4 ⟩ .

Proof

By Lemma 2.11, γ ( P I ( Z 7 ) ) = 1 . Moreover, by (1), it is easy to see that P I ( D 14 ) has exactly one block isomorphic to K 7 , and every other block is isomorphic to K 2 . As a consequence, γ ( P I ( D 14 ) ) = 1 by Theorem 2.6. Now let G = SmallGroup ( 21 , 1 ) . By GAP [20], we check that G has a unique subgroup of order 7, and has seven distinct subgroups of order 3. Thus, every block of P I ( G ) is isomorphic to K 7 or K 3 . Note that P I ( G ) has precisely one block isomorphic to K 7 . It follows that γ ( P I ( G ) ) = 1 by Theorem 2.6, as desired.

Conversely, suppose that γ ( P I ( G ) ) = 1 . Suppose, by contradiction, that G has two distinct subgroups of order 7. Then the subgraph Ω induced by the two distinct subgroups of order 7 is a union of two complete graphs of order 7 that share the identity vertex. Thus, Ω has two distinct blocks isomorphic to K 7 , and so γ ( P I ( G ) ) ≥ γ ( Ω ) = 2 by Theorems 2.3 and 2.6, a contradiction. As a consequence, G has precisely one subgroup ⟨ g ⟩ of order 7, and so ⟨ g ⟩ is normal in G . If there exists an element x ∈ G such that o ( x ) = 4 , then the subgroup ⟨ x , g ⟩ has order 28, which is impossible since every group of order 28 has an element of order 14 by GAP [20]. Consequently, we have π e ( G ) ⊆ { 1 , 2 , 3 , 7 } .

Assume that π e ( G ) = { 1 , 7 } . Since G has precisely one subgroup ⟨ g ⟩ of order 7, by Lemma 2.1, we have that G = ⟨ g ⟩ ≅ Z 7 , as desired. In the following, we assume that { 1 , 7 } ⊂ π e ( G ) .

Case 1

G has an involution a .

Then ⟨ g , a ⟩ has order 14. Since G has no elements of order 14, we have

⟨ g , a ⟩ = ⟨ g , a : g 7 = a 2 = e , a g a = g − 1 ⟩ ≅ D 14 .

Suppose, by contradiction, that there exists an involution b ∈ G such that b ∉ ⟨ g , a ⟩ . As before ⟨ g , b ⟩ ≅ D 14 and b g b = g − 1 . If [ a , b ] = e , then g a b g a b = g a ( b g b ) a = g a g − 1 a = g 2 , and so ( g a b ) 7 = a b , which implies o ( g a b ) = 14 , which is impossible. We conclude that [ a , b ] ≠ e . It follows that ⟨ a , b ⟩ ≅ D 6 or D 14 by π e ( G ) ⊆ { 1 , 2 , 3 , 7 } . If ⟨ a , b ⟩ ≅ D 14 , then g , a ∈ ⟨ a , b ⟩ , and so ⟨ a , b ⟩ = ⟨ g , a ⟩ , contrary to b ∉ ⟨ g , a ⟩ . As a consequence, we have ⟨ a , b ⟩ ≅ D 6 , and so ( a b ) 3 = e . Therefore, we have

( g a b ) 3 = g a b ( g a b ) 2 = ( g a b ) ( g a ( b g b ) b a b ) = ( g a b ) ( g ( a g − 1 a ) a b a b ) = ( g a b ) g 2 ( a b ) 2 = g 3 ( a b ) 3 = g 3

and so ( g a b ) 7 = a b , which implies o ( ( g a b ) 3 ) = 7 and o ( ( g a b ) 7 ) = 3 , it follows that o ( g a b ) = 21 , a contradiction. We conclude that every involution of G belongs to ⟨ g , a ⟩ . It follows that G has precisely seven involutions a , g a , g 2 a , … , g 6 a . Suppose now, by contradiction, that there exists an element c of order 3. Then ⟨ g , c ⟩ = ⟨ g ⟩ ⟨ c ⟩ has order 21. By GAP [20], there are two groups of order 21 up to isomorphism, that is, Z 21 and SmallGroup ( 21 , 1 ) . Since G has no elements of order 21, we have ⟨ g , c ⟩ = ⟨ g , c : g 7 = c 3 = e , c − 1 g c = g 4 ⟩ ≅ SmallGroup ( 21 , 1 ) . Note that c a c − 1 is an involution. We may assume that c a c − 1 = g t a for some 0 ≤ t ≤ 6 . Note that a g a = g − 1 and c g c − 1 = g 2 . We have

c ( g − t a ) = c a ( a g − t a ) = c a ( g t ) = ( g t a c ) g t = ( g t a ) ( c g t ) = ( g t a ) ( g 2 t c ) = ( g t ) ( a g 2 t a ) ( a c ) = ( g t ) ( g − 2 t ) ( a c ) = ( g − t a ) c .

It follows that g − t a c is an element of order 6, a contradiction. Thus, in this case, π e ( G ) = { 1 , 2 , 7 } . Since every involution of G belongs to ⟨ g , a ⟩ , and G has precisely one subgroup ⟨ g ⟩ of order 7, we have that G = ⟨ g , a ⟩ ≅ D 14 , as desired.

Case 2

G has no involutions.

Then π e ( G ) = { 1 , 3 , 7 } . Let w ∈ G with o ( w ) = 3 . Note that a group of order 21 is isomorphic to either Z 21 or SmallGroup ( 21 , 1 ) . Hence, by Lemma 2.11, we have that

⟨ g , w ⟩ ≅ SmallGroup ( 21 , 1 ) ≅ ⟨ g , w : g 7 = w 3 = e , w − 1 g w = g 4 ⟩ .

Now let P be a Sylow 3-subgroup of G . Since π e ( G ) = { 1 , 3 , 7 } , we have G = P ⟨ g ⟩ . Suppose, by contradiction, that ∣ P ∣ > 3 . Then every non-trivial element of P has order 3. Note that the fact that a non-trivial p -group has non-trivial center. We have that there exist u , v ∈ P such that o ( u ) = o ( v ) = 3 , ⟨ u ⟩ ≠ ⟨ v ⟩ , and [ u , v ] = e . Note that ⟨ g , u ⟩ ≅ ⟨ g , v ⟩ ≅ SmallGroup ( 21 , 1 ) . We have u − 1 g u = v − 1 g v = g 4 . It follows that

( u 2 v g ) 3 = ( u 2 v ( g u ) u v g ) ( u 2 v g ) = ( u 2 v ( u g 4 ) u v g ) ( u 2 v g ) = ( v g 4 u v g ) ( u 2 v g ) = ( g v u v g ) ( u 2 v g ) = ( g u v − 1 g ) ( u 2 v g ) = ( u g 4 g 4 v − 1 ) ( u 2 v g ) = ( u g v − 1 ) ( u 2 v g ) = u g ( u − 1 g ) = u g g 4 u − 1 = u g 5 u − 1 ,

and so o ( ( u 2 v g ) 3 ) = o ( g 5 ) = 7 . Moreover, we have

( u 2 v g ) 7 = ( u g 5 u − 1 ) ( u g 5 u − 1 ) ( u 2 v g ) = ( u g 3 u − 1 ) ( u 2 v g ) = ( u g 3 u − 1 v ) ( u − 1 g ) = ( u g 3 u − 1 ) ( v g 4 ) u − 1 = ( u g 3 u − 1 ) ( g v ) u − 1 = u g 3 ( u − 1 g ) v u − 1 = u g 3 ( g 4 u − 1 ) v u − 1 = u 2 v ,

and so o ( ( u 2 v g ) 7 ) = 3 . It follows that u 2 v g has order 21, a contradiction. We conclude that ∣ P ∣ = 3 , and thus G has order 21 so have that G = ⟨ g , w ⟩ ≅ SmallGroup ( 21 , 1 ) , as desired.□

Lemma 3.4

Let G be a group with { 6 } ⊆ π e ( G ) ⊆ { 1 , 2 , 3 , 4 , 6 } . Then the following facts are equivalent:

γ ( P I ( G ) ) = 1 ;
G ≅ Z 6 or D 12 ;
γ ¯ ( P I ( G ) ) = 1 .

Proof

We first prove that (a) and (b) are equivalent. By Lemma 2.11, we have γ ( P I ( Z 6 ) ) = 1 . Moreover, by (1), it is easy to see that P I ( D 12 ) has exactly one block isomorphic to K 6 , and every other block is isomorphic to K 2 . As a consequence, γ ( P I ( D 12 ) ) = 1 by Theorem 2.6. Thus, we have that (b) implies (a).

Conversely, suppose that γ ( P I ( G ) ) = 1 . Let g ∈ G with o ( g ) = 6 . We first claim that G has a unique cyclic subgroup of order 6. By Lemma 2.2, we have that the number of cyclic subgroups of order 6 is not equal to 2. Suppose, by contradiction, that G has at least three distinct cyclic subgroups ⟨ x ⟩ , ⟨ y ⟩ , and ⟨ z ⟩ of order 6. Assume that ⟨ x ⟩ , and ⟨ y ⟩ satisfy ∣ ⟨ x ⟩ ∩ ⟨ y ⟩ ∣ = 1 or 2. Let U = { u ∈ ⟨ x ⟩ ∪ ⟨ y ⟩ : o ( u ) ≠ 2 } . Then the subgraph Δ induced by U is a union of two complete graphs of order 5 that share the identity vertex, as displayed in Figure 5. Thus, Δ has two distinct blocks isomorphic to K 5 . It follows that γ ( P I ( G ) ) ≥ γ ( Δ ) ≥ 2 by Theorem 2.6, a contradiction. The same argument holds for ⟨ x ⟩ and ⟨ z ⟩ or ⟨ y ⟩ and ⟨ z ⟩ . We conclude that ∣ ⟨ x ⟩ ∩ ⟨ y ⟩ ∩ ⟨ z ⟩ ∣ = 3 , which implies that the subgraph induced by ( ⟨ x ⟩ ∪ ⟨ y ⟩ ∪ ⟨ z ⟩ ) ⧹ { x 3 , y 3 , z 3 } is isomorphic to K 9 , a contradiction as γ ( K 9 ) = 3 . Consequently, G has a unique cyclic subgroup ⟨ g ⟩ of order 6 and has exactly two elements of order 6, that is, g and g − 1 . In particular, ⟨ g ⟩ is normal in G .

If G has an element a of order 3 such that a ∉ ⟨ g ⟩ , then ⟨ g , a ⟩ = ⟨ g ⟩ ⟨ a ⟩ has order 18. But by GAP [20], there is no group A of order 18 such that π e ( A ) ⊆ { 1 , 2 , 3 , 4 , 6 } and A has precisely two elements of order 6. We conclude that G has exactly two elements of order 3, that is, g 2 and g 4 .

Case 1

G has a unique involution.

If G has no elements of order 4, then by the above discussion, G ≅ Z 6 , as desired. Suppose next, by contradiction, that G has an element of order 4. If G has two distinct cyclic subgroups ⟨ b ⟩ and ⟨ c ⟩ of order 4, since G has a unique involution, we have ∣ ⟨ b ⟩ ∩ ⟨ c ⟩ ∣ = 2 . Note that ∣ ⟨ b ⟩ ∩ ⟨ c ⟩ ∩ ⟨ g ⟩ ∣ = 2 . Then the subgraph induced by ⟨ b ⟩ ∪ ⟨ c ⟩ ∪ ⟨ g ⟩ has a subgraph Λ is a union of a complete graph of order 5 and a complete graph of order 6 that share the identity vertex, as displayed in Figure 6. Thus, Λ has two distinct blocks isomorphic to K 5 and K 6 . Consequently, γ ( P I ( G ) ) ≥ γ ( Λ ) ≥ 2 by Theorem 2.6, a contradiction. We conclude that G has a unique cyclic subgroup of order 4. Note now that G has a unique involution, two distinct elements of order 4, two distinct elements of order 6, and two distinct elements of order 3. Since π e ( G ) ⊆ { 1 , 2 , 3 , 4 , 6 } , it follows that G has order 8, which is impossible since a group of order 8 has no elements of order 6.

Case 2

G has at least two involutions.

We first prove that G has no elements of order 4. Suppose, by contradiction, that G has an element h of order 4. Then ⟨ g , h ⟩ = ⟨ g ⟩ ⟨ h ⟩ has order 24 or 12. By GAP [20], there is no group B of order 24 such that π e ( B ) ⊆ { 1 , 2 , 3 , 4 , 6 } and B has precisely two elements of order 6, we have that ⟨ g , h ⟩ has order 12. It follows that the intersection of ⟨ g ⟩ and ⟨ h ⟩ has size 2. This implies that if ⟨ k ⟩ ⊆ G is of order 4 and different from ⟨ h ⟩ , then ∣ ⟨ k ⟩ ∩ ⟨ h ⟩ ∣ = 2 . Since the intersection of two distinct cyclic subgroups of order 4 has size 1 by the proof of Case 1, we have that G has exactly one cyclic subgroup ⟨ h ⟩ of order 4. Note that G has exactly two elements g 2 , g 4 of order 3. It follows that both ⟨ h ⟩ and ⟨ g 2 ⟩ are normal in G , and so h − 1 g − 2 h g 2 ∈ ⟨ h ⟩ ∩ ⟨ g 2 ⟩ = { e } . Thus, h and g 2 commute, and so o ( h g 2 ) = 12 , a contradiction since π e ( G ) ⊆ { 1 , 2 , 3 , 4 , 6 } . We conclude that G has no elements of order 4.

Thus, π e ( G ) = { 1 , 2 , 3 , 6 } . Let u be an involution of G with u ∉ ⟨ g ⟩ . Then ⟨ g , u ⟩ = ⟨ g ⟩ ⟨ u ⟩ has order 12. Note that G has a unique cyclic subgroup of order 6. We deduce that ⟨ g , u ⟩ ≅ D 12 by GAP [20]. In order to prove G = ⟨ g , u ⟩ , it suffices to show that every involution of G belongs to ⟨ g , u ⟩ . Suppose, by contradiction, that G has an involution v ∉ ⟨ g , u ⟩ . Then ⟨ g , v ⟩ ≅ D 12 , and so v g v = g − 1 .

Assume that [ u , v ] = e . Then ( g u v ) 2 = g u ( v g v ) u = g 2 , and so ( g u v ) 3 = g 3 u v ≠ e since v ∉ ⟨ g , u ⟩ . Note that ( g 3 u v ) 2 = g 3 u ( v g 3 v ) u = g 6 = e . It follows that o ( ( g u v ) 2 ) = 3 and o ( ( g u v ) 3 ) = 2 , and so g u v has order 6. Now it is easy to see that g u v ∉ { g , g 5 } , hence G has at least three distinct elements of order 6, a contradiction.

Assume next that [ u , v ] ≠ e . Note that ⟨ u , v ⟩ is a dihedral group. It follows from π e ( G ) = { 1 , 2 , 3 , 6 } that ⟨ u , v ⟩ ≅ D 6 or D 12 . If ⟨ u , v ⟩ ≅ D 12 , then g ∈ ⟨ u , v ⟩ , and so v ∈ ⟨ u , v ⟩ = ⟨ g , u ⟩ , a contradiction. As a consequence, we have ⟨ u , v ⟩ ≅ D 6 , which implies o ( u v ) = 3 . Note that G has exactly two elements g 2 , g 4 of order 3. Thus, we have u v = g i where i = 2 or 4. It follows that v = u g i ∈ ⟨ g , u ⟩ , a contradiction. Now we have that (a) and (b) are equivalent.

Note next that γ ¯ ( Δ ) = 2 and γ ¯ ( Λ ) = 2 by Theorems 2.3 and 2.6. Then similar to the above proof, we get that (b) and (c) are equivalent.□

$Figure 5 The subgraph Δ \Delta .$

Figure 5

The subgraph Δ .

$Figure 6 The subgraph Λ \Lambda .$

Figure 6

The subgraph Λ .

4 Proofs of Theorems 1.1 and 1.2

We are now ready to prove our main theorems.

Proof of Theorem 1.1

The sufficiency follows from Lemmas 3.1, 3.2, 3.3, and 3.4. Suppose next that γ ( P I ( G ) ) = 1 . Then Lemma 2.11 implies π e ( G ) ⊆ { 1 , 2 , 3 , 4 , 5 , 6 , 7 } . If π e ( G ) ⊆ { 1 , 2 , 3 , 4 } , then Lemma 3.1 implies the desired result. Thus, we may assume there that G has an element of order 5, 6, or 7. Suppose that G has an element a of order 5. If G has an element b of order 6 or 7, then the subgraph induced by ⟨ a ⟩ ∪ ⟨ b ⟩ has a subgraph Δ , which is isomorphic to a union of two complete graphs of order 5 that share the identity element of G . Thus, Δ has precisely two blocks isomorphic to K 5 , and so γ ( Δ ) = 2 by Theorems 2.3 and 2.6, a contradiction. Thus, { 5 } ⊆ π e ( G ) ⊆ { 1 , 2 , 3 , 4 , 5 } , and Lemma 3.2 implies the desired result. Similarly, we have the desired result if G contains an element of order 6 or 7 using Lemmas 3.3 and 3.4.□

The proof of Theorem 1.1 can be modified to prove Theorem 1.2. Thus, we omit the proof of Theorem 1.2 for the sake of brevity.

Acknowledgments

The authors are grateful to the referees for many detailed comments and thoughtful suggestions that have helped improve this paper substantially.

Funding information: This research was supported by the National Natural Science Foundation of China (Grant No. 11801441), the Natural Science Basic Research Program of Shaanxi (Program No. 2020JQ-761), and the Young Talent fund of University Association for Science and Technology in Shaanxi, China (Program No. 20190507).
Conflict of interest: Authors state no conflict of interest.

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Received: 2021-02-07

Revised: 2021-06-28

Accepted: 2021-07-07

Published Online: 2021-08-20

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https://doi.org/10.1515/math-2021-0071

Keywords for this article

intersection power graph; finite group; genus

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