Finite groups with star-free noncyclic graphs

Xuanlong Ma; Gary L. Walls; Kaishun Wang

doi:10.1515/math-2019-0071

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Finite groups with star-free noncyclic graphs

Xuanlong Ma , Gary L. Walls und Kaishun Wang

Veröffentlicht/Copyright: 8. August 2019

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Aus der Zeitschrift Open Mathematics Band 17 Heft 1

Abstract

For a finite noncyclic group G, let Cyc(G) be the set of elements a of G such that 〈a, b〉 is cyclic for each b of G. The noncyclic graph of G is a graph with the vertex set G ∖ Cyc(G), having an edge between two distinct vertices x and y if 〈x, y〉 is not cyclic. In this paper, we classify all finite noncyclic groups whose noncyclic graphs are K_1,n-free, where K_1,n is a star and 3 ≤ n ≤ 6.

Keywords: Noncyclic graph; finite group; star-free

MSC 2010: 05C25; 20B05

1 Introduction

All groups considered in this paper are finite. Let G be a noncyclic group. The cyclicizer Cyc(G) of G is the set

{a∈G:〈a,b〉 is cyclic for each b∈G},

which is a normal cyclic subgroup of G (see [1]). Graphs associated with groups and other algebraic structures have been actively investigated, since they have valuable applications (cf. [2, 3, 4, 5]) and are related to automata theory (cf. [5, 6]).

The noncyclic graph Γ_G of G is the graph whose vertex set is G ∖ Cyc(G), and two distinct vertices are adjacent if they do not generate a cyclic subgroup. In 2007, Abdollahi and Hassanabadi [7] introduced the concept of a noncyclic graph and established some basic graph theoretical properties of noncyclic graphs. In [8], Abdollahi and Hassanabadi investigated the clique number of a noncyclic graph. Recently, Costa et. al [9] studied the Eulerian properties of noncyclic graphs of finite groups. Aalipour et. al [10] studied the relationship between the complement graph of a noncyclic graph and two well-studied graphs-power graphs [11, 12, 13, 14, 15, 16, 17] and commuting graphs [18]. Finite groups whose noncyclic graphs have genus one were classified by Selvakumar and Subajini [19] and, independently, by Ma [20]. Moreover, the full automorphism group of a noncyclic graph was determined in [21].

A graph is said to be Γ-free if it has no induced subgraphs isomorphic to Γ. Forbidden graph characterization appears in many contexts; for instance, forbidden subgraph problem (Turán-type problem), or extremal graph theory where lower and upper bounds can be obtained for various numerical invariants of the corresponding graphs. Some graphs obtained from groups with small forbidden induced subgraphs have been studied in the literature. For example, Doostabadi et al. [22] studied the finite groups with K_1,3-free power graphs. Akhlaghi and Tong-Viet [23] studied the finite groups with K₄-free prime graphs, where K₄ is the complete graph of order 4. Kayacan [24] classified the finite groups with K_3,3-free intersection graphs of subgroups, where K_3,3 is the complete bipartite graph with each partition of size 3. In [25], Das and Nongsiang classified K₃-free commuting graphs of finite non-abelian groups.

In this paper, we study noncyclic graphs of finite groups. In Sect. 2, we classify all finite groups G with a unique involution and π_e(G) = {2, 3, 4, 6}, where π_e(G) is the set of natural numbers consisting of orders of non-identity elements of G. In Sect. 3, we classify all finite noncyclic groups whose noncyclic graphs are K_1,n-free, where 3 ≤ n ≤ 6.

2 A result on finite groups

An element of order 2 in a group is called an involution. The exponent of G is the least common multiple of the orders of the elements of G. We denote the cyclic group of order n and the quaternion group of order 8 by ℤ_n and Q₈, respectively. Also Znm is used for the m-fold direct product of the cyclic group ℤ_n with itself.

In this section we prove the following a result about finite groups, which will be used to classify finite groups with K_1,5-free noncyclic graphs.

Theorem 2.1

Let G be a finite group having a unique involution and that π_e(G) = {2, 3, 4, 6}. Then, either G ≅ SL(2, 3) or G ≅ Z3n ⋊ ℤ₄, where ℤ₄ acts on Z3n by inversion.

Let G be a finite group and p a prime number dividing |G|. Denote by Syl_p(G) and O_p(G) the set of all Sylow p-subgroups of G and the largest normal p-subgroup of G, respectively. Note that O_p(G) = ⋂_{P∈Syl_p(G)} P. Let n_p = |Syl_p(G)| and P ∈ Syl_p(G). Recall that n_p = |G: N_G(P)| ≡ 1 (mod p) and n_p is a divisor of |G : P|.

Lemma 2.2

Let G be a finite group and suppose that n_p = p + 1 for some prime number p. Then for any two distinct P_i, P_j ∈ Syl_p(G), P_i ∩ P_j = O_p(G).

Proof

Let m = n_p and L = Syl_p(G) = {P₁, P₂, …, P_m}. Now in order to prove the required result, it suffices to prove the equality P₁ ∩ P₂ = O_p(G).

Let R = P₁ ∩ P₂ and let R act on L by conjugation. Note that for all i, we have that (R ∩ N_G(P_i))P_i = P_i(R ∩ N_G(P_i)) and (R ∩ N_G(P_i))P_i is a p-subgroup of N_G(P_i), so R ∩ N_G(P_i) ⊆ P_i. It follows that R_{P_i} = R ∩ N_G(P_i) = R ∩ P_i, where R_{P_i} is the stabilizer of P_i in R. Also, since R = P₁ ∩ P₂, we deduce that |Orbit_R(P₁)| = |Orbit_R(P₂)| = 1, where Orbit_R(P_i) is the R-orbit containing P_i. Note that every R-orbit has length 1 or p. Since |L| = p + 1, we have that every R-orbit has length 1. This implies that R = R_{P_i} for all i. It follows that for each i ≥ 3, P₁ ∩ P₂ = P₁ ∩ P₂ ∩ P_i, that is, P₁ ∩ P₂ ⊆ P_i. Thus, P₁ ∩ P₂ ⊆ ⋂i=3m P_i and so P₁ ∩ P₂ = ⋂_{P∈Syl_p(G)} P = O_p(G), as desired. □

Note that for any prime number p, a p-group with a unique subgroup of order p is either a cyclic group or a generalized quaternion group (see [26, Theorem 5.4.10 (ii)]).

Proof of Theorem 1.2

Suppose that |G| = 2^t ⋅ 3ⁿ for some t ≥ 1, n ≥ 1. Let Q and P be a Sylow 2-subgroup and a Sylow 3-subgroup of G, respectively. Since G has a unique involution and 8 ∉ π_e(G), we know that Q ∈ {Q₈, ℤ₄}. Since G has no elements of order 9, we deduce that P has exponent 3. Denote by x the unique involution of G. Then x ∈ Z(G), the center of G.

Q = Q₈.

Let 〈a〉, 〈b〉, and 〈c〉 be the three cyclic subgroups of Q of order 4, and that ab = c. Then a² = b² = c² = x. Since in this case |G| = 8 ⋅ 3ⁿ, we have that n₃ is a divisor of 8. This implies that n₃ = 1 or 4.

Suppose that O₃(G) ≠ 1. Then let a, b, c act on O₃(G) by conjugation. Neither of them can fix any non-identity elements of O₃(G), since G has no elements of order 12. Thus, a, b, c act as fixed-point-free automorphisms of O₃(G). Since a² = b² = c² ∈ Z(G), we have that a, b, c act as fixed-point-free automorphisms of order 2. Now by Burnside’s result (see [26, Theorem 1.4, page 336] or [27]), we know that O₃(G) is abelian and for each non-trivial element g ∈ O₃(G), we have that g^a = g⁻¹ = g^b = g^c. It then follows that g^c = g^ab = g, and hence |abg| = 12, a contradiction. Therefore, we conclude O₃(G) = 1.

Now we know that n₃ = 4. By Lemma 2.2 we have that for any two distinct P_i, P_j ∈ Syl₃(G), P_i ∩ P_j = 1. It follows that the number of elements of order 3 is 4(3ⁿ − 1). Also, since every element of order 3 and x can generate a cyclic subgroup of order 6, the number of elements of order 6 is 4(3ⁿ − 1). Now all that remains is to count the number of elements of order 4.

Let w be an element of order 4 in G. Then, there is a Q₁ ∈ Syl₂(G) so that w ∈ Q₁. Note that Q₁ ≅ Q₈. It follows that Q₁ ⊆ N_G(〈w〉). If there exists an element y of order 3 such that 〈w〉^y = 〈w〉, then 〈w〉 is normal in 〈w〉〈y〉 and |〈w〉〈y〉| = 12, and so by the N/C lemma we have C_G(〈w〉) = N_G(〈w〉), which implies that 〈w〉〈y〉 ≅ ℤ₁₂, a contradiction. It follows that Q₁ = N_G(〈w〉). Thus, every element of order 4 is contained in a unique Sylow 2-subgroup of G. It means that the number of elements of order 4 is 6n₂.

Suppose that N_G(Q) = Q. Then n₂ = 3ⁿ. Counting all the elements of G gives that

8⋅3n=6⋅3n+8(3n−1)+2.

This implies that 3ⁿ = 1, contrary to the order of G. Thus, we have Q ⊂ N_G(Q).

Suppose that |P ∩ N_G(Q)| ≥ 9. Then there exist w₁, w₂ ∈ P ∩ N_G(Q) so that 〈w₁, w₂〉 is an abelian group of order 9. Now both w₁ and w₂ act on Q by conjugation. We conclude that both w₁ and w₂ act as 3-cycles on {〈a〉, 〈b〉, 〈c〉} because otherwise 12 ∈ π_e(G). But then there is an element u of order 3 that fixes some cyclic subgroup 〈v〉 of order 4, where u = w1w2i for some integer i. It follows that there exists an element of order 12 in 〈u〉〈v〉, a contradiction. Thus, we get |P ∩ N_G(Q)| ≤ 3.

Note by the modular law that N_G(Q) = G ∩ N_G(Q) = Q(P ∩ N_G(Q)). Since Q ⊂ N_G(Q), we have that P ∩ N_G(Q) is a subgroup of order 3 and |N_G(Q)| = 24. This forces that n₂ = 3ⁿ⁻¹. Now as above we get that

8⋅3n=6⋅3n−1+8(3n−1)+2,

which implies n = 1 and so |G| = 24. Note that in this case Q is normal in G. It is easy to see that G ≅ SL(2, 3).
Q = ℤ₄.

Let Q = 〈y〉. Since 〈x〉 P ⊆ N_G(P), we deduce |G : N_G(P)| ≠ 4. Note that n₃ is a divisor of 4. Then n₃ = 1 and so P is normal in G. Now as above y acts as a fixed-point-free automorphism of order 2 on P by conjugation. By Burnside’s result, P is abelian and so P ≅ Z3n for some n, and for all w ∈ P we have w^y = w⁻¹. It follows that G ≅ Z3n ⋊ ℤ₄, as desired. □

3 Main results

In this section we classify all finite groups with K_1,n-free noncyclic graphs, where 3 ≤ n ≤ 6.

In the remainder of this paper, we always use G to denote a finite noncyclic group with the identity element e. Euler’s totient function is denoted by ϕ. A proper cyclic subgroup 〈x〉 is said to be maximal in G if 〈x〉 ⊆ 〈y〉 implies that 〈x〉 = 〈y〉, where y is an element of G. We first begin with the following two lemmas which will be used frequently in the sequel.

Lemma 3.1

Suppose that 〈g〉 is a maximal cyclic subgroup of G. Then Γ_G has an induced subgraph isomorphic to K_1,ϕ(|g|).

Proof

Let n = ϕ(|g|) and let {g₁, g₂, …, g_n} be all generators of 〈g〉. Note that G is noncyclic. Pick an element a in G ∖ 〈g〉. Since 〈g〉 is maximal cyclic, we deduce that 〈a, g_i〉 is not cyclic for each i ∈ {1, 2, …, n}. This implies that {g₁, g₂, …, g_n, a} induces a subgraph isomorphic to K_1,n. □

For a graph Γ, we denote the sets of the vertices and the edges of Γ by V(Γ) and E(Γ), respectively. An independent set of Γ is is a subset of the vertices such that no two vertices in the subset represent an edge of Γ. The independence number of a graph Γ is the cardinality of the largest independent set and is denoted by α(Γ). The following result follows from [7, Proposition 4.6].

Lemma 3.2

α(Γ_G) = max{|g|: g ∈ G} − |Cyc(G)|.

Γ_G is complete if and only if G is an elementary abelian 2-group (see [7, Proposition 3.1]). So, as Γ_G is connected (see [7, Proposition 3.2]), we first note that Γ_G is K_1,2-free if and only if G is an elementary abelian 2-group.

A claw is another name for the complete bipartite graph K_1,3. We first classify the finite groups whose noncyclic graphs are claw-free.

Theorem 3.3

Γ_G is claw-free if and only if G is isomorphic to one of the following groups:

Q₈;
Z2n , n ≥ 2;
A noncyclic 3-group of exponent 3;
A noncyclic group G with π_e(G) = {2, 3}.

Proof

Since Γ_Q₈ ≅ K_2,2,2, we have that Γ_Q₈ is claw-free. Also, by Lemma 3.2 we see that the independence number of the noncyclic graph of every group in (b)–(d)is at most 2, and so each of the noncyclic graphs is claw-free.

Now we suppose that Γ_G is claw-free. It follows from Lemma 3.1 that for every maximal cyclic subgroup 〈g〉 of G, ϕ(|g|) ≤ 2. This implies that every cyclic subgroup of G has at most two generators. Thus, π_e(G) ⊆ {2, 3, 4, 6}.

Suppose that G has an element a of order 6. Note that G is noncyclic. Pick an element x in G ∖ 〈a〉. If |x| = 2, since 〈x, a³〉 is noncyclic, we have that {x, a, a³, a⁵} induces a subgraph isomorphic to K_1,3, which is impossible. If the order of x is 3 or 4, since 〈x, a²〉 is noncyclic, it follows that {x, a, a², a⁵} induces a subgraph isomorphic to K_1,3, which is also impossible. We conclude |x| = 6. Namely, every element of G ∖ 〈a〉 has order 6. However, in this case we have that both x² and x³ belong to 〈a〉. It follows that x ∈ 〈a〉, a contradiction.

Thus, we conclude that π_e(G) ⊆ {2, 3, 4}. Suppose that there exists an element g of order 4 in G. If there is x ∈ G ∖ 〈g〉 with |x| = 2 or 3, then 〈x, g²〉 is noncyclic, and so {x, g, g², g³} induces a subgraph isomorphic to K_1,3, a contradiction. Consequently, in this case G has a unique involution and π_e(G) = {2, 4}. By [26, Theorem 5.4.10 (ii)], we see that G is isomorphic to Q₈.

We now assume π_e(G) ⊆ {2, 3}. If π_e(G) = {2}, then G is an elementary abelian 2-group, as desired. If π_e(G) = {3}, then G is a 3-group of exponent 3, as desired. □

Theorem 3.4

Γ_G is K_1,4-free if and only if G is isomorphic to one of the following groups:

A noncyclic group G with π_e(G) ⊆ {2, 3, 4};
ℤ₆ × Z2m , m ≥ 1.

Proof

If G is isomorphic to a noncyclic group with π_e(G) ⊆ {2, 3, 4}, then α(Γ_G) ≤ 3 by Lemma 3.1, and so Γ_G is K_1,4-free, as desired. If G ≅ ℤ₆ × Z2m for some m ≥ 1, then |Cyc(G)| = 3 and we can obtain that Γ_G is a complete multipartite graph whose each partite set has size 3, which implies that in this case Γ_G is also K_1,4-free.

For the converse, suppose that Γ_G is K_1,4-free. Note that ϕ(n) is even for any integer n ≥ 3. By Lemma 3.1 we see that each cyclic subgroup of G has at most two generators. It follows that π_e(G) ⊆ {2, 3, 4, 6}. In order to get the desired result, we now suppose that G has an element g of order 6. Clearly, 〈g〉 is maximal cyclic. If there exists an element a in G ∖ 〈g〉 such that |a| = 3 or 4, then {a, g, g², g⁴, g⁵} induces a subgraph isomorphic to K_1,4, a contradiction. This means that G has a unique subgroup of order 3 and π_e(G) = {2, 3, 6}.

Let P and Q be a Sylow 2-subgroup and a Sylow 3-subgroup, respectively. Then G = P ⋉ Q, P is an elementary abelian 2-group of order great than 2 and Q ≅ ℤ₃. Pick an involution u in G. If 〈u, g²〉 is noncyclic, then {u, g, g², g⁴, g⁵} induces a subgraph isomorphic to K_1,4, a contradiction. Thus, we have that every element of P and every element of Q commute. It follows that

G=P×Q≅Z6×Z2m, m≥1,

as required. □

Theorem 3.5

Γ_G is K_1,5-free if and only if G is isomorphic to one of the following groups:

A noncyclic group G with π_e(G) ⊆ {2, 3, 4, 5};
ℤ₆ × Z2m , m ≥ 1;
ℤ₂ × Q, where Q is a noncyclic 3-group of exponent 3;
The special linear group SL(2, 3);
Z3n ⋊ ℤ₄, where ℤ₄ acts on Z3n by inversion and n ≥ 1.

Proof

If G is isomorphic to a group in (a), then α(Γ_G) ≤ 4 by Lemma 3.2, and so Γ_G is K_1,5-free. Moreover, by Theorem 3.4, ΓZ6×Z2m is K_1,5-free, where m ≥ 1. If G ≅ ℤ₂ × Q for some noncyclic 3-group Q of exponent 3, then |Cyc(G)| = 2 and we may check that Γ_G is a complete multipartite graph whose each partite set has size 4, which implies that in this case Γ_G is also K_1,5-free. Furthermore, if G is isomorphic to SL(2, 3) or a group in (e), then it is easy to see that Γ_G is a complete multipartite graph whose maximal partite set has size 4. Thus, Γ_G is K_1,5-free if G is one group of (e) and (d).

Conversely, suppose that Γ_G is K_1,5-free. It follows from Lemma 3.1 that π_e(G) ⊆ {2, 3, 4, 5, 6, 8, 10, 12}.

Suppose that g ∈ G with |g| = 12. If there exists an element a with order 5 or 8, then {a, g, g², g⁵, g⁷, g¹¹} induces a subgraph isomorphic to K_1,5, a contradiction. This implies that π_e(G) ⊆ {2, 3, 4, 6, 12}. If G has an element b of G ∖ 〈g〉 with |b| < 12, then {b, g, g⁵, g⁷, g¹¹, g^t} induces a subgraph isomorphic to K_1,5, where |g^t| = |b|. Thus, in this case one has G ≅ ℤ₁₂, a contradiction. This means that G has no elements of order 12. Similarly, we can get 10 ∉ π_e(G). If G has an element of order 8, then a similar argument implies that G is a 2-group and it has a unique involution and a unique cyclic subgroup of order 4, which implies that G is a generalized quaternion group having precisely two elements of order 4, a contradiction. Thus, we conclude π_e(G) ⊆ {2, 3, 4, 5, 6}.

In order to get the desired result, we suppose that G has an element h of order 6. Then it is easy to see that 5 ∉ π_e(G).

G has two distinct cyclic subgroups of order 6.

Assume that H₁ = 〈h〉 and H₂ = 〈h₂〉 are two distinct cyclic subgroups of order 6. In order to avoid that {h₂, h, h², h³, h⁴, h⁵} induces K_1,5, we may assume that |H₁ ∩ H₂| ≥ 2. In fact, any two distinct cyclic subgroups of order 6 have nontrivial intersection.
There exist two distinct cyclic subgroups of order 6 such that their intersection has order 3.

Without loss of generality, we may assume that |H₁ ∩ H₂| = 3. Suppose that G has an element x of order 4. If 〈h³, x〉 is not cyclic, then {x, h, h², h³, h⁴, h⁵} induces a subgraph isomorphic to K_1,5, a contradiction. We conclude that 〈h³, x〉 is cyclic, and so h³ = x². Similarly, we also can obtain h23 = x². It follows that x² ∈ H₁ ∩ H₂, which is impossible as |H₁ ∩ H₂| = 3. Hence, in this subcase π_e(G) ⊆ {2, 3, 6}. Since every generator of any maximal cyclic subgroup of order 2 or 3 is adjacent to each of 〈h〉 ∖ {e}, every cyclic subgroup of order 2 or 3 is not maximal. If 〈y〉 ≠ 〈h²〉 is a subgroup of order 3, and let 〈y〉 ⊆ 〈h₃〉 with |h₃| = 6, then |〈h₃〉 ∩ H_i| = 2 for i = 1, 2 and so h33=h3=h23 , which is impossible as H₁ ≠ H₂. This implies that G has a unique subgroup of order 3. Now we know that G ≅ Z2m ⋉ ℤ₃ for some integer m ≥ 2. Pick an involution u in G. Then since 〈u〉 is not maximal, there exists an element h′ of order 6 such that 〈u〉 ⊆ 〈h′〉. By the uniqueness of the subgroup of order 3, we see that 〈h′〉 ∩ 〈h〉 = 〈h²〉. It follows that 〈u, h²〉 is cyclic. Namely, every involution of G and h² commute. This implies that G ≅ Z2m × ℤ₃ for some integer m ≥ 2, as desired.
The intersection of each two distinct cyclic subgroups of order 6 has order 2.

In this case we first claim that G has a unique involution. Assume, to the contrary, that u is an involution of G such that u ≠ h³. Then 〈u, h²〉 is not cyclic, since there are no two cyclic subgroups of order 6 such that their intersection has order 3. This implies that {u, h, h², h³, h⁴, h⁵} induces a subgraph isomorphic to K_1,5, a contradiction. Thus, our claim is valid.

Now note that π_e(G) ⊆ {2, 3, 4, 6}. If 4 ∉ π_e(G), then G ≅ ℤ₂ × Q, where Q is a noncyclic 3-group of exponent 3. Thus, we may assume that π_e(G) = {2, 3, 4, 6}. Note that ℤ₃ ⋊ ℤ₄ has a unique cyclic subgroup of order 6, where ℤ₄ acts on ℤ₃ by inversion. By Theorem 2.1 we see that that G ≅ SL(2, 3) or G ≅ Z3n ⋊ ℤ₄, where ℤ₄ acts on Z3n by inversion, and n ≥ 2, as required.
G has a unique cyclic subgroup of order 6.

We first see that 〈h〉 is a normal subgroup of G. Note that π_e(G) ⊆ {2, 3, 4, 6}. If G has an element x in G ∖ 〈h〉 such that x ∈ C_G(h), the centralizer of h in G, then G has a subgroup isomorphic to ℤ₂ × ℤ₆ or ℤ₃ × ℤ₆, which contradicts the fact that G has precisely two elements of order 6. This implies that C_G(h) = 〈h〉. So G/〈h〉 is isomorphic to a subgroup of ℤ₂, and hence |G| = 6 or 12. It follows that G ≅ ℤ₃ ⋊ ℤ₄, where ℤ₄ acts on ℤ₃ by inversion, as desired. □

Theorem 3.6

Γ_G is K_1,6-free if and only if G is isomorphic to one of the following groups:

A noncyclic group G with π_e(G) ⊆ {2, 3, 4, 5, 6};
ℤ₁₀ × Z2m , m ≥ 1.

Proof

If G is isomorphic to a group in (a), then it follows from Lemma 3.2 that α(Γ_G) ≤ 5, and so Γ_G is K_1,6-free. If G ≅ ℤ₁₀ × Z2m for some m ≥ 1, then |Cyc(G)| = 5 and we may check that Γ_G is a complete multipartite graph whose each partite set has size 5, which implies Γ_G is K_1,6-free.

For the converse, suppose that Γ_G is K_1,6-free. Since ϕ(n) is even for n ≥ 3, by Lemma 3.1, we have that ϕ(|g|) ≤ 4 for any g ∈ G. It follows that π_e(G) ⊆ {2, 3, 4, 5, 6, 8, 10, 12}. An argument similar to the one used in the third paragraph of the proof of Theorem 3.5 shows that 8, 12 ∉ π_e(G). Consequently, we have π_e(G) ⊆ {2, 3, 4, 5, 6, 10}.

In order to get the desired result, we suppose that G has an element h of order 10. Then it is easy to see that π_e(G) = {2, 5, 10} and G has a unique subgroup of order 5. Thus, we may assume that G ≅ P ⋉ ℤ₅, where P is an elementary abelian 2-group of order at least 4. Pick any involution u in P. If 〈u, h²〉 is not cyclic, then {u, h², h⁴, h⁶, h⁸, h⁵, h} induces a subgraph isomorphic to K_1,6, a contradiction. Thus, every element in P and h² commute. It follows that G ≅ P × ℤ₅, that is, G ≅ ℤ₁₀ × Z2m for some m ≥ 1, as desired. □

Acknowledgements

We are grateful to the referee for many useful suggestions and comments.

Ma was supported by the National Natural Science Foundation of China (Grant No. 11801441), the Scientific Research Program funded by Shaanxi Provincial Education Department (Program No. 18JK0623), and the Young Talent fund of University Association for Science and Technology in Shaanxi, China (Program No. 20190507). Wang was supported by the National Natural Science Foundation of China (Grant No. 11671043).

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Received: 2018-10-07

Accepted: 2019-06-19

Published Online: 2019-08-08

This work is licensed under the Creative Commons Attribution 4.0 Public License.

Artikel in diesem Heft

https://doi.org/10.1515/math-2019-0071

Schlagwörter für diesen Artikel

Noncyclic graph; finite group; star-free

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