**Zero divisors and units with small supports in group algebras
of torsion-free groups**
††footnotetext: 2010 Mathematics Subject Classification. 20C07; 16S34.
Keywords and Phrases. Kaplansky’s zero divisor conjecture, Kaplansky’s unit conjecture, group ring, torsion-free group, zero divisor.
Alireza Abdollahi and Zahra Taheri
Abstract. Kaplansky’s zero divisor conjecture (unit conjecture, respectively) states that for a torsion-free group G and a field F, the group ring F[G] has no zero divisors (has no units with supports of size greater than 1). In this paper, we study possible zero divisors and units in F[G] whose supports have size 3.
For any field F and all torsion-free groups G, we prove that if αβ=0 for some non-zero α,β∈F[G] such that ∣supp(α)∣=3, then ∣supp(β)∣≥10. If F=F2 is the field with 2 elements, the latter result can be improved so that ∣supp(β)∣≥20. This improves a result in [J. Group Theory, 16 (2013), no. 5, 667-693].
Concerning the unit conjecture, we prove that if αβ=1 for some α,β∈F[G] such that ∣supp(α)∣=3, then ∣supp(β)∣≥9. The latter improves a part of a result in [Exp. Math., 24 (2015), 326-338] to arbitrary fields.
1. Introduction and Results
Let R be a ring. A non-zero element α of R is called a zero divisor if αβ=0 or βα=0 for
some non-zero element β∈R. Let G be a group. Denote by R[G] the group ring of G over R. If R contains a zero divisor, then clearly so does R[G]. Also, if G contains a non-identity torsion element x of finite order n, then R[G] contains zero divisors α=1−x and β=1+x+⋯+xn−1, since αβ=0. Around 1950, Irving Kaplansky conjectured that existence of a zero divisor in a group ring depends only on the existence of such elements in the ring or non-trivial torsions in the group by stating one of the most challenging problems in the field of group rings [12].
Conjecture 1.1** (Kaplansky’s zero divisor conjecture).**
Let F be a field and G be a torsion-free group. Then F[G] does not contain a zero divisor.
Another famous problem, namely the unit conjecture, also proposed by Kaplansky [12], states that:
Conjecture 1.2** (Kaplansky’s unit conjecture).**
Let F be a field and G be a torsion-free group. Then F[G] has
no non-trivial units (i.e., units which are not non-zero scalar multiples of group elements).
It can be shown that the zero divisor conjecture is true if the unit conjecture has an affirmative solution (see [17, Lemma 13.1.2]).
Over the years, some partial results have been obtained on Conjecture 1.1 and it has been confirmed for special classes of groups which are torsion-free. One of the first known special families which satisfy Conjecture 1.1 are unique product groups [17, Chapter 13], in particular ordered groups. Furthermore, by the fact that Conjecture 1.1 is known to hold valid for amalgamated free products when the group ring of the subgroup over which the amalgam is formed satisfies the Ore condition [16], it is proved by Formanek [9] that supersolvable groups are another families which satisfy Conjecture 1.1. Another result, concerning large major sorts of groups for which Conjecture 1.1 holds in the affirmative, is obtained for elementary amenable groups [14]. The latter result covers the cases in which the group is polycyclic-by-finite, which was firstly studied in [4] and [7], and then extended in [22]. Some other affirmative results are obtained on congruence subgroups in [15] and [8], and certain hyperbolic groups [5].
Nevertheless, Conjecture 1.1 has not been confirmed for any fixed field and it seems that confirming the conjecture even for the smallest finite field F2 with two elements is still out of reach.
The support of an element α=∑x∈Gaxx of R[G], denoted by supp(α), is the set {x∈G∣ax=0}. For any field F and any torsion-free group G, it is known that F[G] does not contain a zero divisor whose support is of size at most 2 (see [20, Theorem 2.1]), but it is not known a similar result for group algebra elements with the support of size 3. By describing a combinatorial structure, named matched rectangles, Schweitzer [20] showed that if αβ=0 for α,β∈F2[G]∖{0} when ∣supp(α)∣=3, then ∣supp(β)∣>6. Also, with a computer-assisted approach, he showed that if ∣supp(α)∣=3, then ∣supp(β)∣>16. In the following we provide a little more details of graph theory which are used in this paper.
A graph G=(VG,EG,ψG) consists of a non-empty set VG, a possibly empty set EG and if EG=∅ a function ψG:EG→VG(1)∪VG(2)∪VG×VG, where VG(i) denotes the set of all i-element subsets of VG for i=1,2. The elements of VG and EG are called vertices and edges of the graph G, respectively. An edge e∈EG is called an undirected loop if ψG(e)∈VG(1). The edge e is called directed if ψG(e)∈VG×VG and it is called a directed loop if ψG(e)=(v,v) for some v∈VG. The graph G is called undirected if ψG(EG)∩(VG×VG)=∅.
We say a vertex u is adjacent to a vertex v, denoted by u∼v, if ψG(e)={u,v}, (u,v) or (v,u) for some e∈EG; otherwise, we say u is not adjacent to v, denoted by u∼v. In the latter case the vertices u and v are called the endpoints of the edge e and we say e joins its endpoints. If a vertex v is an endpoint of an edge e, we say v is adjacent to e or also e is adjacent to v. Two edges e1 and e2 are called adjacent , denoted by e1∼e2, if the sets of their endpoints have non-empty intersection. The graph G is called loopless whenever ψG(EG)∩VG(1)=∅ and ψG(EG)∩{(v,v)∣v∈VG}=∅. We say that the graph G has multi-edge if ψG is not injective. A simple graph is an undirected loopless graph having no multi-edge. So, the graph G is simple if EG is empty or ψG is an injective function from VG to VG(2). The degree of a vertex v of G, denoted by deg(v), is the number of edges adjacent to v. If all vertices of a graph have the same degree k, we say that the graph is k-regular and as a special case, a cubic graph is the one which is 3-regular. A path graph P is a simple graph with the vertex set {v0,v1,…,vn} such that vj−1∼vj for j=1,…,n. The length of a path is the number of its edges. A cycle of length n, denoted by Cn, is a path of length n in which its initial vertices (v0 and vn by the above notation) are identified. The cycles C3 and C4 are called triangle and square, respectively. We say an undirected graph is connected if there is a path between every pair of distinct vertices. A subgraph of a graph G is a graph H such that VH⊆VG, EH⊆EG and ψH is the restriction of ψG to EH. In a graph G, an induced subgraph I on a set of vertices W⊆VG is a subgraph in which VI=W and EI={e∈EG∣ the endpoints of e are in W}. An isomorphism between two undirected and loopless graphs G and H is a pair of bijections ϕV:VG→VH and ϕE:EG→EH preserving adjacency and non-adjacency i.e., for any pair of vertices u,v∈VG, ϕV(u)∼ϕV(v)⇔u∼v and for any pair of edges e1,e2∈EG, ϕE(e1)∼ϕE(e2)⇔e1∼e2. Two undirected and loopless graphs G and H are called isomorphic (denoted by G≅H) if there is an isomorphism between them. We say that an undirected and loopless graph H is a forbidden subgraph of a graph G if there is no subgraph isomorphic to H in G. An undirected graph is bipartite if its vertices can be partitioned into two sets (called partite sets) in such a way that no edge joins two vertices in the same set. A complete bipartite graph is a simple bipartite graph in which each vertex in one partite set is adjacent to all the vertices in the other one. A complete bipartite graph whose partite sets have cardinalities r and s is denoted by Kr,s. A Cayley graph Cay(G,S) for a group G and a subset S of G with 1∈S=S−1, is the graph whose vertex set is G and two vertices g and h are adjacent if gh−1∈S.
Let G be an arbitrary torsion-free group and let α∈F[G] be a possible zero divisor such that ∣supp(α)∣=3 and αβ=0 for some non-zero β∈F[G]. In this paper, we study the minimum possible size of the support of such an element β. Suppose that F=F2. In [20, Definition 4.1] a graph K(M) is associated to the non-degenerate 3×∣supp(β)∣ matched rectangle M corresponding to α and β and it is proved in [20, Theorem 4.2] that the graph K(M) is a simple cubic one without triangles. Here we define a graph Z(a,b) associated to any pair (a,b) of non-zero elements of a group algebra (not necessarily over a field with 2 elements and not necessarily on a torsion-free group) such that ab=0 (see Definition 2.1). We call Z(a,b) the zero-divisor graph of (a,b). The zero divisor graph Z(α,β) is isomorphic to K(M). Here we study subgraphs of some zero-divisor graphs. Our main results on Conjecture 1.1 are the followings.
Theorem 1.3**.**
None of the graphs in Figure 1 can be isomorphic to a subgraph of any zero-divisor graph of length 3 over any field and on any torsion-free group.
Theorem 1.4**.**
Let α and β be non-zero elements of the group algebra of any torsion-free group over an arbitrary field. If ∣supp(α)∣=3 and αβ=0 then ∣supp(β)∣≥10.
Theorem 1.5**.**
None of the graphs in Table LABEL:tab-forbiddens can be isomorphic to a subgraph of any zero-divisor graph of length 3 over F2 and on any torsion-free group.
By Theorem 1.3, the graphs in Figure 1 are forbidden subgraphs of any zero-divisor graph of length 3 over F and on any torsion-free group. Also by Theorem 1.5, the graphs in Table LABEL:tab-forbiddens are forbidden subgraphs of any zero-divisor graph of length 3 over F2 and on any torsion-free group. In Appendix of [1], some details of our computations needed in the proof of Theorem 1.5 are given for the reader’s convenience.
The following result improves parts (iii) and (iv) of [20, Theorem 1.3].
Theorem 1.6**.**
Let α and β be non-zero elements of the group algebra of any torsion-free group over the field with two elements. If ∣supp(α)∣=3 and αβ=0 then ∣supp(β)∣≥20.
The best known result on Conjecture 1.2, which has the purely group-theoretic approach, is concerned with unique product groups [17, 18].
The latter result covers ordered groups, in particular torsion-free nilpotent groups. Nevertheless, it is still unknown whether or not Conjecture 1.2 do hold true for supersolvable torsion-free groups. For any field F and any torsion-free group G, it is known that F[G] does not contain a unit element whose support is of size at most 2 (see [6, Theorem 4.2]), but it is not known a similar result for group algebra elements with the support of size 3. Dykema et al. [6] have shown that there exist no γ,δ∈F2[G] such that γδ=1, where ∣supp(γ)∣=3 and ∣supp(δ)∣≤11. Concerning Conjecture 1.2, we prove the following result which improves a part of the result in [6] to arbitrary fields.
Theorem 1.7**.**
Let γ and δ be elements of the group algebra of any torsion-free group over an arbitrary field. If ∣supp(γ)∣=3 and γδ=1 then ∣supp(δ)∣≥9.
It is known that a group algebra of a field F over a torsion-free group G contains a zero divisor if and only if it contains a non-zero element whose square is zero (see [17, Lemma 13.1.2]). Using the latter fact, it is mentioned in [20, p. 691] that it is sufficient to check Conjecture 1.1 only for the case that ∣supp(α)∣=∣supp(β)∣, but in the construction that, given a zero divisor produces an element of square zero, it is not clear how the length changes. We clarify the latter by the following.
Proposition 1.8**.**
Let G be a torsion-free group and F be a field with the property that there exists a positive integer k such that F[G] has no non-zero element α with ∣supp(α)∣≤k and α2=0. Then there exist no non-zero elements α1,α2∈F[G] such that α1α2=0 and ∣supp(α1)∣∣supp(α2)∣≤k.
2. Zero-divisor graphs and Unit graphs
Definition 2.1**.**
For any pair of non-zero elements (α,β) of a group algebra over a field F and a group G such that αβ=0, we assign a graph Z(α,β) to (α,β) called the zero-divisor graph of (α,β) as follows: the vertex set is supp(β), the edge set is
[TABLE]
and ψZ(α,β):EZ(α,β)→VZ(α,β)(2) is defined by
[TABLE]
for all {(h,h′,g,g′),(h′,h,g′,g)}∈EZ(α,β).
We call Z(α,β) a zero-divisor graph of length ∣supp(α)∣ over the field F and on the group G.
**
Remark 2.2**.**
- (1)
We note that the graph Z(α,β) is an undirected graph with no loops but it may happen that a zero-divisor graph has multi-edge; that is, in general a zero-divisor graph may not be a simple graph.
2. (2)
Being non-zero of β ensures that the vertex set is non-empty
and the zero-divisor condition αβ=0 and the conditions α=0 and β=0 imply that the edge set of the graph is non-empty.
3. (3)
One may use instead Z(α,β) the notations such as Z(α,β,∣supp(α)∣,∣supp(β)∣),
ZF(α,β) or ZF[G](α,β), … to indicate the support sizes of α and β, the underlying field or the underlying group algebra, ….
4. (4)
From the notation Z(α,β) we understand that αβ=0 for two non-zero elements of a group algebra and so xαβy=0 for all group elements x and y. So we may consider the zero-divisor graph
Z(xα,βy) which is isomorphic to Z(α,β) (see below, Lemma 2.4).
5. (5)
In this paper we will only study the zero-divisor graphs of pairs whose first component has support size 3 and mostly the field is F2 and the underlying group G is always torsion-free.
6. (6)
If ∣supp(α)∣=3 and the underlying group G is torsion-free, the zero-divisor graph is simple so that between two distinct vertices there is at most one edge (see below Proposition 3.7).
7. (7)
If we choose β of minimum support size with respect to the property αβ=0, the graph Z(α,β) will be connected (see below Lemma 2.6).
Example 2.3**.**
- (1)
Let G1=⟨x⟩ be the cyclic group of order 7 and let α=1+x2+x3+x4 and β=1+x+x5 in F2[G1], then αβ=0. Figures 2 and 3 show the graphs Z(α,β) and Z(β,α), respectively.
Lemma 2.4**.**
Z(α,β)≅Z(xα,βy)* for all group elements x and y.*
Proof.
Consider the maps ϕV:supp(β)→supp(βy) and ϕE:EZ(α,β)→EZ(xα,βy) defined by ϕV(g)=gy for all g∈supp(β) and
[TABLE]
for all {(h,h′,g,g′),(h′,h,g′,g)}∈EZ(α,β). Then it is easy to see that (ϕV,ϕE) is an isomorphism from Z(α,β) to Z(xα,βy).
∎
Lemma 2.5**.**
Suppose that α and β are non-zero elements of a group algebra such that αβ=0.
- (1)
Suppose that if αβ′=0 for some non-zero element β′ of the group algebra, then ∣supp(β′)∣≥∣supp(β)∣. Then supp(β)g−1⊆⟨xsupp(α)⟩ for all g∈supp(β) and all group elements x.
2. (2)
Suppose that if α′β=0 for some non-zero element α′ of the group algebra, then ∣supp(α′)∣≥∣supp(α)∣. Then
h−1supp(α)⊆⟨supp(β)y⟩
for all h∈supp(α) and all group elements y.
Proof.
(1) Let H=⟨xsupp(α)⟩ and suppose that {t1,t2,…,tk} be a subset of right coset representatives of H in the group such that supp(β)g−1∩Hti=∅ for all i∈{1,…,k} and supp(β)g−1⊆∪i=1kHti. Since xαβg−1=0 and Hti∩Htj=∅ for all distinct i,j∈{1,2,…,k}, (xα)(∑g′∈Htiβg′g′)=0 for all i∈{1,2,…,k}, where β=∑g′∈Gβg′g′. Now it follows from the hypothesis that k=1 so that supp(β)g−1⊆Ht1. Hence supp(β)g−1⊆H for all g∈supp(β).
(2) Consider the subgroup K=⟨supp(β)y⟩ and take a finite subset S of left coset representatives of K in the group such that h−1supp(α)∩sK=∅ for all s∈S and h−1supp(α)⊆∪s∈SsK. By a similar argument as in part (1), ∣S∣=1 and
the inclusion is proved.
∎
Lemma 2.6**.**
Let Z(α,β) be a zero-divisor graph such that β has minimum possible support size among all non-zero elements γ with αγ=0. Then Z(α,β) is connected.
Proof.
Suppose, for a contradiction, that VZ(α,β)=supp(β) is partitioned into two non-empty subsets V1 and V2 such that there is no edge between V1 and V2. It follows that supp(αβ1)∩supp(αβ2)=∅, where β=∑g∈Gβgg and βi=∑g∈Viβgg (i=1,2). Since αβ1+αβ2=0,
[TABLE]
Since ∣supp(βi)∣<∣supp(β)∣, the equalities 2.1 give contradiction. This completes the proof.
∎
Definition 2.7**.**
For any pair of elements (α,β) of a group algebra over a field F and on a group G such that αβ=1, we assign to (α,β) a graph U(α,β) called the unit graph of (α,β) as follows: the vertex set is supp(β), the edge set is
[TABLE]
and if EU(α,β)=∅, the function ψU(α,β):EU(α,β)→VU(α,β)(2) is defined by
[TABLE]
for all {(h,h′,g,g′),(h′,h,g′,g)}∈EU(α,β).
We call U(α,β) a unit graph of length ∣supp(α)∣ over the field F and on the group G.
**
Remark 2.8**.**
- (1)
Note that the main difference between the definition of the zero-divisor graph Z(α,β) and the unit graph U(α,β) is in the conditions on α and β and there is no difference in defining the vertex and edge sets.
2. (2)
Any unit graph is an undirected graph with no loops but it may happen that a unit graph has multi-edge; that is, in general a unit graph may not be a simple graph.
3. (3)
The vertex set of U(α,β) is non-empty, since β is non-zero.
4. (4)
If αβ=1, then there exists h∈supp(α) such that h−1∈supp(β). The latter allows us to assume 1∈supp(α)∩supp(β) whenever we are studying the unit graph U(α,β) in view of Lemma 2.10, below.
5. (5)
One may use instead U(α,β) the notations such as U(α,β,∣supp(α)∣,∣supp(β)∣),
UF(α,β) or UF[G](α,β), … to indicate the support sizes of α and β, the underlying field or the underlying group algebra, ….
6. (6)
In this paper we will only study unit graphs of pairs whose first component has support size 3 and the underlying group G is always torsion-free.
7. (7)
If ∣supp(α)∣=3 and the underlying group G is torsion-free, the unit graph is simple so that between two distinct vertices there is at most one edge (see below Proposition 3.9).
8. (8)
If we choose β of minimum support size with respect to the property αβ=1, the graph U(α,β) will be connected (see below Lemma 2.11).
Example 2.9**.**
- (1)
Let G2=⟨x⟩ be the cyclic group of order 3 and F3 be the field with 3 elements. Then α=−1+x−x2 is a non-trivial unit with inverse β=1+x in F3[G2]. Figures 4 and 5 show the graphs U(α,β) and U(β,α), respectively.
- (2)
Let G3=⟨x⟩ be the cyclic group of order 8 and F be an arbitrary field whose characteristic is not 2. Then α=−1−x+x3+2⋅x4+x5−x7 is a non-trivial unit with inverse β=−1+x−x3+2⋅x4−x5+x7 in F[G3]. Figure 6 shows the graphs U(α,β) and U(β,α) which are isomorphic.
Lemma 2.10**.**
U(α,β)≅U(x−1α,βx)* for all group elements x.*
Proof.
We note that (x−1α)(βx)=1 and so one may speak of U(x−1α,βx). The proof is similar to that of Lemma 2.4.
∎
Lemma 2.11**.**
Let U(α,β) be a unit graph such that β has minimum possible support size among all elements γ with αγ=1. Then U(α,β) is connected.
Proof.
Suppose, for a contradiction, that VU(α,β)=supp(β) is partitioned into two non-empty subsets V1 and V2 such that there is no edge between V1 and V2. It follows that supp(αβ1)∩supp(αβ2)=∅, where β=∑g∈Gβgg and βi=∑g∈Viβgg (i=1,2). Since αβ1+αβ2=1,
[TABLE]
Since ∣supp(βi)∣<∣supp(β)∣, each of the equalities 2.2 (if happens) is a contradiction. This completes the proof.
∎
Lemma 2.12**.**
Suppose that α and β are elements of a group algebra such that αβ=1.
- (1)
Suppose that if αβ′=1 for some element β′ of the group algebra, then ∣supp(β′)∣≥∣supp(β)∣. Then ⟨supp(β)⟩⊆⟨supp(α)⟩.
2. (2)
Suppose that if α′β=1 for some element α′ of the group algebra, then ∣supp(α′)∣≥∣supp(α)∣. Then ⟨supp(α)⟩⊆⟨supp(β)⟩.
Proof.
(1) Let H=⟨supp(α)⟩ and suppose that {t1,t2,…,tk} be a subset of right coset representatives of H in the group such that supp(β)∩Hti=∅ for all i∈{1,…,k} and supp(β)⊆∪i=1kHti. Since
[TABLE]
for all distinct i,j∈{1,2,…,k}, it follows that there exists i∈{1,…,k} such that Hti=H and
[TABLE]
for all ℓ∈{1,2,…,k}, where β=∑g∈Gβgg. Now it follows from the hypothesis that k=i=1 so that ⟨supp(β)⟩⊆H.
(2) Consider the subgroup K=⟨supp(β)⟩ and take a finite subset S of left coset representatives of K in the group such that supp(α)∩sK=∅ for all s∈S and supp(α)⊆∪s∈SsK.
The rest of the proof is similar to the part (1).
∎
We finish this section with following question:
Question 2.13**.**
Which graphs can be isomorphic to a zero-divisor or unit graph?
3. Zero-divisor graphs and Unit graphs for elements whose supports are of size 3
Corollary 3.1**.**
Suppose that α and β are non-zero elements of a group algebra of a torsion-free group G such that ∣supp(α)∣=3, αβ=0 and if αβ′=0 for some non-zero element β′ of the group algebra, then ∣supp(β′)∣≥∣supp(β)∣. Then ⟨h−1supp(α)⟩=⟨supp(β)g−1⟩ for all h∈supp(α) and g∈supp(β).
Proof.
It follows from [20, Theorem 2.1] that
if α′β=0 for some non-zero element α′ of the group algebra, then ∣supp(α′)∣≥∣supp(α)∣=3. Now Lemma 2.5 completes the proof.
∎
Corollary 3.2**.**
Suppose that α and β are elements of a group algebra of a torsion-free group G such that ∣supp(α)∣=3, αβ=1 and if αβ′=1 for some element β′ of the group algebra, then ∣supp(β′)∣≥∣supp(β)∣. Then ⟨supp(α)⟩=⟨supp(β)⟩.
Proof.
It follows from [6, Theorem 4.2] that
if α′β=1 for some non-zero element α′ of the group algebra, then ∣supp(α′)∣≥∣supp(α)∣=3. Now Lemma 2.12 completes the proof.
∎
Remark 3.3**.**
Suppose that α and β are non-zero elements of a group algebra of a torsion-free group G such that ∣supp(α)∣=3, αβ=0 and if αβ′=0 for some non-zero element β′ of the group algebra, then ∣supp(β′)∣≥∣supp(β)∣. In studying the zero-divisor graph Z(α,β), in view of Lemma 2.4 and Corollary 3.1 one may assume that 1∈supp(α)∩supp(β) and G=⟨supp(α)⟩=⟨supp(β)⟩.
**
Remark 3.4**.**
Suppose that α and β are elements of a group algebra of a torsion-free group G such that ∣supp(α)∣=3, αβ=1 and if αβ′=1 for some element β′ of the group algebra, then ∣supp(β′)∣≥∣supp(β)∣. In studying the unit graph U(α,β), by part (4) of Remark
2.8 and Corollary 3.2 one may assume that 1∈supp(α)∩supp(β) and G=⟨supp(α)⟩=⟨supp(β)⟩.
**
Lemma 3.5**.**
Let A be a subset of size 3 of a torsion-free group. If S:={h1−1h2∣h1,h2∈A,h1=h2} is of size at most 5 then there exists h∈A such that ⟨h−1A⟩ is an infinite cyclic group.
Proof.
Let A={h1,h2,h3}. Since ∣S∣≤5, it follows that hi−1hj=hi′−1hj′ for some (i,j)=(i′,j′) and (i,i′)=(j,j′).
It follows that (i′,j′) is equal to (j,i), (j,k) or (k,i), where k∈{1,2,3}∖{i,j}. If (i′,j′)=(j,i), then (hi−1hj)2=1 and since the group is torsion-free, hi=hj, a contradiction. If (i′,j′)=(j,k), then hi−1hk=(hi−1hj)2 and if (i′,j′)=(k,i) then
hk−1hj=(hi−1hj)2. Thus S={hi−1hj,(hi−1hj)−1,(hi−1hj)2,(hi−1hj)−2}. It follows that ⟨hi−1A⟩=⟨hi−1hj⟩ is an infinite cyclic group.
∎
Lemma 3.6**.**
Let α be a non-zero element in the group algebra of a torsion-free group such that αβ=0 for some non-zero element β of the group algebra. If ∣supp(α)∣=3, then S={h−1h′∣h,h′∈supp(α),h=h′} has size 6.
Proof.
Let A:=supp(α) and suppose, for a contradiction, that ∣S∣≤5. It follows from
Lemma 3.5 that H:=⟨h−1A⟩ is an infinite cyclic group for some h∈A. Now assume that ∣supp(β)∣ is minimum with respect to the property αβ=0. It follows from Lemma 2.5 that h−1α,βg−1 belong to the group algebra on H for any arbitrary element g∈supp(β).
Now (h−1α)(βg−1)=0 contradicts the fact that the group algebra of the infinite cyclic group has no zero-divisors (see [18, Theorem 26.2]). This completes the proof.
∎
Proposition 3.7**.**
Let α be a non-zero element in the group algebra of a torsion-free group such that αβ=0 for some non-zero element β of the group algebra. If ∣supp(α)∣=3 and S:={h−1h′∣h,h′∈supp(α),h=h′}, then
Z(α,β) is isomorphic to the induced subgraph of the Cayley graph Cay(G,S) on supp(β).
Proof.
Let Γ be the induced subgraph of the Cayley graph Cay(G,S) on supp(β).
The map ϕE:EZ(α,β)→EΓ defined by {(h,h′,g,g′),(h′,h,g′,g)}↦{g,g′} is a bijective map; for by Lemma 3.6, ∣S∣=6 which implies that h−1h′=h1−1h1′ if and only if (h,h′)=(h1,h1′), whenever the entries of the latter pairs are distinct and belong to supp(α). Now take ϕV be the identity map on supp(β), then (ϕV,ϕE) is an isomorphism from Z(α,β) to Γ.
∎
Lemma 3.8**.**
Let α be an element in the group algebra of a torsion-free group G such that αβ=1 for some element β of the group algebra. If ∣supp(α)∣=3, then S={h−1h′∣h,h′∈supp(α),h=h′} has size 6.
Proof.
Let A:=supp(α) and suppose, for a contradiction, that ∣S∣≤5. It follows from
Lemma 3.5 that H:=⟨h−1A⟩ is an infinite cyclic group for some h∈A. Here we have (h−1α)(βh)=1. Now assume that ∣supp(β)∣ is minimum with respect to the property αβ=1.
It follows from Lemma 2.12 that h−1α,βh belong to the group algebra on H.
Now (h−1α)(βh)=1 contradicts the fact that the group algebra of an infinite cyclic group has no non-trivial units (see [18, Theorem 26.2]). This completes the proof.
∎
Proposition 3.9**.**
Let α be an element in the group algebra of a torsion-free group such that αβ=1 for some element β of the group algebra. If ∣supp(α)∣=3 and S:={h−1h′∣h,h′∈supp(α),h=h′}, then
U(α,β) is isomorphic to the induced subgraph of the Cayley graph Cay(G,S) on supp(β).
Proof.
The proof is similar to that of Proposition 3.7.
∎
Definition 3.10**.**
Let Γ be a zero-divisor graph or a unit graph on a pair of elements (α,β) in a group algebra such that ∣supp(α)∣=3. Let C be a cycle of length k in Γ as Figure 7 and suppose that {g1,g2,g3,…,gk}⊆supp(β) is the vertex set of C such that gi∼gi+1 for all i∈{1,2,…,k−1} and g1∼gk. By an arrangement ℓ of the vertex set C, we mean a sequence of all vertices as x1,…,xk such that xi∼xi+1 for all i∈{1,2,…,k−1} and x1∼xk.
There exist a1,b1,a2,b2,…,ak,bk∈supp(α) satisfying the following relations:
[TABLE]
We assign the 2k-tuple TCℓ=[a1,b1,a2,b2,…,ak,bk] to C corresponding to the above arrangement ℓ of the vertex set of C. We denote by R(TCℓ) the above set R of relations.
It can be derived from the relations 3.1 that
r(TCℓ):=(a1−1b1)(a2−1b2)⋯(ak−1bk) is equal to 1. It follows from Lemma 3.6 or Lemma 3.8 that if TCℓ′=[a1′,b1′,…,ak′,bk′] is the 2k-tuple of C corresponding to another arrangement ℓ′ of the vertex set of C, then TCℓ′ is one of the following 2k-tuples:
[TABLE]
The set of all such 2k-tuples will be denoted by T(C). Also, R(C)={R(T)∣T∈T(C)}.
**
Definition 3.11**.**
Let Γ be a zero-divisor graph or a unit graph on a pair of elements (α,β) in a group algebra such that ∣supp(α)∣=3. Let C be a cycle of length k in Γ. Since r(T1)=1 if and only if r(T2)=1, for all T1,T2∈T(C), a member of {r(T)∣T∈T(C)} is given as a representative and denoted by r(C). Also, r(C)=1 is called the relation of C.
**
Definition 3.12**.**
Let Γ be a zero-divisor graph or a unit graph on a pair of elements (α,β) in a group algebra such that ∣supp(α)∣=3. Let C and C′ be two cycles of length k in Γ. We say that these two cycles are equivalent, if T(C)∩T(C′)=∅.
**
Remark 3.13**.**
Let Γ be a zero-divisor graph or a unit graph on a pair of elements (α,β) in a group algebra such that ∣supp(α)∣=3. If C and C′ are two equivalent cycles of length k in Γ, then T(C)=T(C′).
Remark 3.14**.**
Let Γ be a zero-divisor graph or a unit graph on a pair of elements (α,β) in a group algebra such that ∣supp(α)∣=3. If C is a cycle of length 3 (a triangle) in Γ and T∈T(C), then T=[a1,b1,a2,b2,a3,b3] with exactly one of the following conditions:
[TABLE]
[TABLE]
We note that if the condition 3.3 is satisfied, then {a1,b1,a2,b2,a3,b3}={a1,b1,b2}=supp(α).
**
Definition 3.15**.**
Let Γ be a zero-divisor graph or a unit graph on a pair of elements (α,β) in a group algebra such that ∣supp(α)∣=3. Suppose that C is a triangle in Γ with T=[a1,b1,a2,b2,a3,b3]∈T(C). If the condition 3.2 or 3.3 is satisfied, we call C a triangle of type (I) or type (II), respectively.
**
We need the following remark in the sequel and we apply it without referring.
Remark 3.16**.**
For integers m and n, the Baumslag-Solitar group BS(m,n) is the group given by the presentation ⟨a,b∣bamb−1=an⟩. Baumslag-Solitar groups are HNN-extensions of an infinite cyclic group and so they are torsion-free by [21, Theorem 6.4.5]. Since Baumslag-Solitar groups are one-relator, it follows from [3] that they are locally indicable. Now [17, Lemmas 1.8 (iii) and 1.9] imply that they satisfy both Conjectures 1.1 and 1.2. It follows from the first part of the proof of [20, Theorem 3.1] that every torsion-free quotient of BS(1,n) is either abelian or it is isomorphic to BS(1,n) itself. Therefore every torsion-free quotient of B(1,n) satisfies both Conjectures 1.1 and 1.2.**
Lemma 3.17**.**
Suppose that α and β are non-zero elements of a group algebra of a torsion-free group G such that ∣supp(α)∣=3, αβ=0 and if αβ′=0 for some non-zero element β′ of the group algebra, then ∣supp(β′)∣≥∣supp(β)∣. Then there is no triangle of type (I) in Z(α,β).
Proof.
By Remark 3.3, one may assume that 1∈supp(α) and G=⟨supp(α)⟩. Let supp(α)={1,h2,h3}. Suppose, for a contradiction, that C is a triangle of type (I) in Z(α,β). Then it is easy to see that there are 13 non-equivalent cases for C, and r(C) corresponding to such cases are one of the members of the following set:
[TABLE]
Therefore, G is the group generated by h2 and h3 with a relation a=1 for some a∈A. We will arrive to a contradiction since such a group G has at least one of the following properties:
- (1)
It is an abelian group,
2. (2)
It is a quotient of BS(1,k) or BS(k,1) where k∈{−1,1},
3. (3)
It has a non-trivial torsion element.
This completes the proof.
∎
Lemma 3.18**.**
Suppose that α and β are elements of a group algebra of a torsion-free group G such that ∣supp(α)∣=3, αβ=1 and if αβ′=1 for some element β′ of the group algebra, then ∣supp(β′)∣≥∣supp(β)∣. Then there is no triangle of type (I) in U(α,β).
Proof.
The proof is similar to that of Lemma 3.17.
∎
Theorem 3.19**.**
Suppose that α and β are non-zero elements of a group algebra of a torsion-free group such that ∣supp(α)∣=3, αβ=0 and if αβ′=0 for some non-zero element β′ of the group algebra, then ∣supp(β′)∣≥∣supp(β)∣. Then Z(α,β) contains no subgraphs isomorphic to the graph in Figure 9 i.e. two triangles with one edge in common.
Proof.
Suppose, for a contradiction, that Z(α,β) contains two triangles with one edge in common as Figure 9 for some distinct elements gi,gj,gk,gl∈supp(β). Then Z(α,β) contains two triangles C and C′ with vertex sets {gi,gj,gk} and {gi,gl,gk}. Therefore by Remark 3.14 and Lemma 3.17, C and C′ are triangles of type (II) and there are RC∈R(C) and RC′∈R(C′) as a1gi=b1gk=c1gj and a1gi=b1gk=c2gl, respectively, where {a1,b1,c1}={a1,b1,c2}=supp(α). Hence, c1=c2 which implies gj=gl, a contradiction. This completes the proof.
∎
Theorem 3.20**.**
Suppose that α and β are elements of a group algebra of a torsion-free group such that ∣supp(α)∣=3, αβ=1 and if αβ′=1 for some element β′ of the group algebra, then ∣supp(β′)∣≥∣supp(β)∣. Then U(α,β) contains no subgraphs isomorphic to the graph in Figure 9 i.e. two triangles with one edge in common.
Proof.
The proof is similar to that of Theorem 3.19.
∎
Remark 3.21**.**
Suppose that α and β are non-zero elements of a group algebra of a torsion-free group such that ∣supp(α)∣=3, αβ=0 and if αβ′=0 for some non-zero element β′ of the group algebra, then ∣supp(β′)∣≥∣supp(β)∣. By Theorem 3.19, if C is a cycle of length 4 (an square) in Z(α,β) as Figure 10 with vertex set gi,gj,gk,gl∈supp(β), then gi∼gk and gj∼gl. So, if T∈T(C), then T=[a1,b1,a2,b2,a3,b3,a4,b4] with the condition a1=b1=a2=b2=a3=b3=a4=b4=a1.
**
Remark 3.22**.**
Suppose that α and β are elements of a group algebra of a torsion-free group such that ∣supp(α)∣=3, αβ=1 and if αβ′=1 for some element β′ of the group algebra, then ∣supp(β′)∣≥∣supp(β)∣. By Theorem 3.20, if C is an square in U(α,β) as Figure 10 with vertex set gi,gj,gk,gl∈supp(β), then gi∼gk and gj∼gl. So, if T∈T(C), then T=[a1,b1,a2,b2,a3,b3,a4,b4] with the condition a1=b1=a2=b2=a3=b3=a4=b4=a1.
**
It is proved in [20, Theorem 4.2] that a triangle is a forbidden subgraph for any zero-divisor graph of length 3 over the field F2 on any torsion-free group (see below, Theorem 4.1). To finish this section, we consider zero-divisor graphs and unit graphs of length 3 over a field F and on any torsion-free group containing a subgraph isomorphic to an square. We have not been able to prove that squares are forbidden subgraphs for such latter graphs even for the case that F=F2. However we show that by existence of squares, zero-divisor graphs (unit graphs, respectively) of length 3 over a field F and on any torsion-free group give us certain slightly significant relations on elements of the support of a possible zero-divisor (on elements of the support of a possible unit) (see below, Theorems 3.25 and 3.26).
Theorem 3.23**.**
Suppose that α and β are non-zero elements of a group algebra of a torsion-free group G such that ∣supp(α)∣=3, αβ=0 and if αβ′=0 for some non-zero element β′ of the group algebra, then ∣supp(β′)∣≥∣supp(β)∣. If Z(α,β) contains an square C, then there are 9 non-equivalent cases for C and r(C)=1 is one of the relations 5, 7, 14, 17, 21, 22, 25, 26 or 29 in Table 2, for {h2,h3}=supp(α)∖{1}.
Proof.
By Remark 3.3, one may assume that 1∈supp(α) and G=⟨supp(α)⟩. Let supp(α)={1,h2,h3} and C be an square in Z(α,β). By Remark 3.21, if T∈T(C) then T=[a1,b1,a2,b2,a3,b3,a4,b4] with the condition a1=b1=a2=b2=a3=b3=a4=b4=a1. Therefore by using GAP [10], there are 36 non-equivalent cases for C. The relations of such non-equivalent cases are listed in the column labelled by R of Table 2. It is easy to see that each of such relations, except the 9 cases marked by “∗”s in the column labelled by E of Table 2, gives a contradiction because the group G generated by h2 and h3 with one of such relations has at least one of the following properties:
- (1)
It is an abelian group,
2. (2)
It is a quotient of BS(1,k) or BS(k,1) where k∈{−2,−1,1,2},
3. (3)
It has a non-trivial torsion element.
Each relation which leads to being G an abelian group or G having a non-trivial torsion element is marked by an A or a T in the column labelled by E, respectively. Also, if G is a quotient of a Baumslag-Solitar group, then it is denoted by BS(1,k) or BS(k,1) for k∈{−2,−1,1,2}, in the column E. This completes the proof.
∎
Theorem 3.24**.**
Suppose that α and β are elements of a group algebra of a torsion-free group G such that ∣supp(α)∣=3, αβ=1 and if αβ′=1 for some element β′ of the group algebra, then ∣supp(β′)∣≥∣supp(β)∣. If U(α,β) contains an square C, then there are 9 non-equivalent cases for C and r(C)=1 is one of the relations 5, 7, 14, 17, 21, 22, 25, 26 or 29 in Table 2, for {h2,h3}=supp(α)∖{1}.
Proof.
The proof is similar to that of Theorem 3.23.
∎
Theorem 3.25**.**
Suppose that α and β are non-zero elements of a group algebra of a torsion-free group G over a field F such that ∣supp(α)∣=3, αβ=0 and if αβ′=0 for some non-zero element β′ of the group algebra, then ∣supp(β′)∣≥∣supp(β)∣. If Z(α,β) contains an square C, then there exist non-trivial group elements x and y such that x2=y3 and either {1,x,y} or {1,y,y−1x} is the support of a zero divisor in F[G].
Proof.
By Remark 3.3, one may assume that 1∈supp(α) and G=⟨supp(α)⟩. Let α=α1⋅1+α2h2+α3h3. By Theorem 3.23, r(C) is one of the relations 5, 7, 14, 17, 21, 22, 25, 26 or 29 in Table 2. So, we have the followings:
- (5)
h22h3h2−1h3=1: Let x=h2−1h3 and y=h2−1. So, x2=y3. Also since αβ=0, we have h2−1(αβ)=0. So, α2⋅1+α3x+α1y is a zero divisor with the support {1,x,y}.
2. (7)
h22h3−2h2=1: Let x=h3 and y=h2. So x2=y3 and α1⋅1+α2y+α3x is a zero divisor with the support {1,x,y}.
3. (14)
h2h3h2−2h3=1: Let x=h2h3 and y=h2. So x2=y3 and α1⋅1+α2y+α3y−1x=α1⋅1+α2h2+α3h3 is a zero divisor with the support {1,y,y−1x}.
4. (17)
h2(h3h2−1)2h3=1: Let x=h2−1 and y=h3h2−1. So x2=y3. Also since αβ=0, we have αh2−1h2β=0. So, α2⋅1+α1x+α3y is a zero divisor with the support {1,x,y}.
5. (21)
h2h3−3h2=1: By interchanging h2 and h3 in (7) and with the same discussion, the statement is true.
6. (22)
h2h3−2h2h3=1: By interchanging h2 and h3 in (14) and with the same discussion, the statement is true.
7. (25)
h2h3−1h2h32=1: By interchanging h2 and h3 in (5) and with the same discussion, the statement is true.
8. (26)
h2h3−1h2h3h2−1h3=1: Let x=h2h3−1h2 and y=h3−1h2. So, x2=y3. Also since α1⋅1+α2h2+α3h3=α2xy−1+α3xy−2+α1x2y−3, we have x−1(α2xy−1+α3xy−2+α1x2y−3)yy−1β=0. Therefore, α2⋅1+α3y−1+α1xy−2 is also a zero divisor with the support of size 3. Furthermore, (α3⋅1+α2y+α1y−1x)(y−3β)=(y−1(α2⋅1+α3y−1+α1xy−2)y2)(y−3β)=0. Hence, α3⋅1+α2y+α1y−1x is a zero divisor with the support {1,y,y−1x}.
9. (29)
(h2h3−1)2h2h3=1: By interchanging h2 and h3 in (17) and with the same discussion, the statement is true.
This completes the proof.
∎
Theorem 3.26**.**
Suppose that α and β are elements of a group algebra of a torsion-free group G over a field F such that ∣supp(α)∣=3, αβ=1 and if αβ′=1 for some element β′ of the group algebra, then ∣supp(β′)∣≥∣supp(β)∣. If U(α,β) contains an square, then there exist non-trivial group elements x and y such that x2=y3 and either {1,x,y} or {1,y,y−1x} is the support of a unit in F[G].
Proof.
The proof is similar to that of Theorem 3.25.
∎
In the following, we discuss about the existence of two squares in a zero-divisor graph or a unit graph of length 3 over an arbitrary field and on a torsion-free group.
Lemma 3.27**.**
Suppose that α and β are non-zero elements of a group algebra of a torsion-free group G such that ∣supp(α)∣=3, αβ=0 and if αβ′=0 for some non-zero element β′ of the group algebra, then ∣supp(β′)∣≥∣supp(β)∣. Every two squares of Z(α,β) are equivalent.
Proof.
By Remark 3.3, one may assume that 1∈supp(α) and G=⟨supp(α)⟩. Let supp(α)={1,h2,h3}. By Theorem 3.23, if there exist two squares in Z(α,β), then these two cycles must be one of the 9 non-equivalent cases with the relations 5, 7, 14, 17, 21, 22, 25, 26 or 29 in Table 2. We may choose two relations similar to or different from each other. When choosing two relations different from each other, there are (29)=36 cases. Using GAP [10], each group with two generators h2 and h3, and two of the relations of the 36 latter cases is finite and solvable, that is a contradiction. So, if there exist two squares in the graph Z(α,β), then such cycles must be equivalent.
∎
Lemma 3.28**.**
Suppose that α and β are elements of a group algebra of a torsion-free group G such that ∣supp(α)∣=3, αβ=1 and if αβ′=1 for some element β′ of the group algebra, then ∣supp(β′)∣≥∣supp(β)∣. Every two squares of U(α,β) are equivalent.
Proof.
The proof is similar to that of Lemma 3.27.
∎
Theorem 3.29**.**
Suppose that α and β are non-zero elements of a group algebra of a torsion-free group G such that ∣supp(α)∣=3, αβ=0 and if αβ′=0 for some non-zero element β′ of the group algebra, then ∣supp(β′)∣≥∣supp(β)∣. Then Z(α,β) contains no subgraph isomorphic to the complete bipartite graph K2,3.
Proof.
Suppose, for a contradiction, that Z(α,β) contains K2,3 as a subgraph. Then it contains two squares C and C′ with two edges in common as Figure 11, for some distinct elements gi,gj,gk,gl,gm∈supp(β). So, there are TC∈T(C) and TC′∈T(C′) such that TC=[a1,b1,a2,b2,a3,b3,a4,b4] and TC′=[a1,b1,a2,b2,a3′,b3′,a4′,b4′]. Also by Remark 3.21, the following conditions are satisfied:
[TABLE]
Since the graph with the vertex set {gi,gm,gk,gl} in K2,3 is also an square, by Remark 3.21 the following conditions are also satisfied:
[TABLE]
[TABLE]
By Lemma 3.27, the cycles C and C′ are equivalent. So, TC′ must be in T(C). In the following, we show that this gives contradictions.
- (1)
Let TC′=[a1,b1,a2,b2,a3,b3,a4,b4]: So a3=a3′, that is a contradiction with 3.5.
2. (2)
Let TC′=[a4,b4,a1,b1,a2,b2,a3,b3]: Therefore, TC=[a1,b1,a1,b1,a3,b3,a1,b1] and TC′=[a1,b1,a1,b1,a1,b1,a3,b3]. By 3.5 and 3.6, we have a3=a1 and b3=b1. Also, in such 8-tuples we have a3=b1, b3=a1 and a1=b1. Therefore, a3=b3 since a3,b3∈supp(α), that is a contradiction.
3. (3)
Let TC′=[a3,b3,a4,b4,a1,b1,a2,b2]: So a3=a1=a3′, that is a contradiction with 3.5.
4. (4)
Let TC′=[a2,b2,a3,b3,a4,b4,a1,b1]: Therefore, TC=[a1,b1,a1,b1,a1,b1,a4,b4] and TC′=[a1,b1,a1,b1,a4,b4,a1,b1]. By 3.5 and 3.6, we have a4=a1 and b4=b1. Also, in such 8-tuples we have a4=b1, b4=a1 and a1=b1. Therefore, a4=b4 since a4,b4∈supp(α), that is a contradiction.
5. (5)
Let TC′=[b1,a1,b4,a4,b3,a3,b2,a2]: So a1=b1, that is a contradiction with 3.4.
6. (6)
Let TC′=[b2,a2,b1,a1,b4,a4,b3,a3]: So b1=a2, that is a contradiction with 3.4.
7. (7)
Let TC′=[b3,a3,b2,a2,b1,a1,b4,a4]: So a2=b2, that is a contradiction with 3.4.
8. (8)
Let TC′=[b4,a4,b3,a3,b2,a2,b1,a1]: So b2=a3, that is a contradiction with 3.4.
This completes the proof.
∎
Theorem 3.30**.**
Suppose that α and β are elements of a group algebra of a torsion-free group G such that ∣supp(α)∣=3, αβ=1 and if αβ′=1 for some element β′ of the group algebra, then ∣supp(β′)∣≥∣supp(β)∣. Then U(α,β) contains no subgraph isomorphic to the complete bipartite graph K2,3.
Proof.
The proof is similar to that of Theorem 3.29.
∎
In the next three sections, we discuss about zero-divisor graphs of length 3 over F2 on any torsion-free group and give some forbidden subgraphs for such graphs.
4. **Zero-divisor graphs of length 3 over F2 on any torsion-free group
and some of their subgraphs containing an square**
Throughout this section suppose that α and β are non-zero elements of the group algebra F2[G] of a torsion-free group G such that ∣supp(α)∣=3, αβ=0 and if αβ′=0 for some non-zero element β′ of the group algebra, then ∣supp(β′)∣≥∣supp(β)∣. By Remark 3.3, one may assume that 1∈supp(α) and G=⟨supp(α)⟩. Let supp(α)={1,h2,h3} and n:=∣supp(β)∣. The following theorem is obtained in [20].
Theorem 4.1** (Theorem 4.2 of [20]).**
The zero-divisor graph Z(α,β) is a connected simple cubic one containing no subgraph isomorphic to a triangle.
Proof.
The connectedness follows from Lemma 2.6. Also by Proposition 3.7, the graph is simple.
Furthermor, for (h,g)∈supp(α)×supp(β), there is a unique (h′,g′)∈supp(α)×supp(β) such that (h,g)=(h′,g′) and hg=h′g′. So, Z(α,β) is a cubic graph.
Suppose, for a contradiction, that C is a triangle in Z(α,β). If C is of type (II) and T∈T(C), then by Remark 3.14, T=[a1,b1,a2,b2,a3,b3] with the condition a1=b1=a2=b2=a3=b3=a1 where {a1,b1,b2}=supp(α). So, there are distinct elements a,b,c∈supp(β) such that a1a=b1b=b2c, that is a contradiction. Therefore, C is of type (I) and so by Lemma 3.17, there is no triangle in Z(α,β).
∎
Remark 4.2**.**
It follows from Theorem 4.1 that n=∣supp(β)∣, which is the number of vertices of Z(α,β), is always an even number because the number of vertices of any simple cubic graph is even.**
Theorem 4.3**.**
Suppose that Z(α,β) contains two squares with exactly one edge in common. Then exactly one of the relations 14, 22 or 26 of Table 2 is satisfied in G.
So, there are TC∈T(C) and TC′∈T(C′) such that TC=[a1,b1,a2,b2,a3,b3,a4,b4] and TC′=[a1,b1,a2,b2,a3′,b3′,a4′,b4′].
Proof.
Suppose that the graph Z(α,β) contains two squares C and C′ with exactly one common edge as Figure 12, for some distinct elements gi,gj,gk,gl,gm,gp∈supp(β). So, there are TC∈T(C) and TC′∈T(C′) such that TC=[a1,b1,a2,b2,a3,b3,a4,b4] and TC′=[a1,b1,a2′,b2′,a3′,b3′,a4′,b4′], and R(TC) and R(TC′) are as follows:
[TABLE]
By Remark 3.21, we have
[TABLE]
Now we prove that a2=a2′ and b4=b4′. Suppose, for a contradiction, that a2=a2′. So by 4.1, b2gk=a2′gj=b2′gm. Since αβ=0 in F2[G], there are ga∈supp(β)∖{gj,gk,gm} and ha∈supp(α)∖{a2′,b2,b2′} such that b2gk=a2′gj=b2′gm=haga, a contradiction because ∣supp(α)∣=3 and {a2′,b2,b2′}=supp(α). Hence,
[TABLE]
Also with the same discussion such as above,
[TABLE]
By Lemma 3.27, the cycles C and C′ are equivalent. So, TC′ must be in T(C). In the following, we explain each cases in details.
- (1)
Let TC′=[a1,b1,a2,b2,a3,b3,a4,b4]: So a2=a2′, that is a contradiction with 4.3.
2. (2)
Let TC′=[a4,b4,a1,b1,a2,b2,a3,b3]: So, we have
[TABLE]
By 4.2, 4.3 and 4.4, a2=a1, a2=b1, b3=a1, b3=b1 and a1=b1. So, a2=b3 because b3∈supp(α)={a1,b1,a2}. So, TC=[a1,b1,a2,b2,a3,a2,a1,b1] and TC′=[a1,b1,a1,b1,a2,b2,a3,a2]. By 4.2, b2=a2, a3=a2 and b2=a3. Also, supp(α)={a1,b1,a2}. So, there are exactly two cases for b2,a3∈supp(α). In the following, we show that each of such cases gives a contradiction.
- i)
Let b2=a1 and a3=b1: So, TC=[a1,b1,a2,a1,b1,a2,a1,b1] and the relation of C is a1−1b1a2−1a1b1−1a2a1−1b1=1. Since 1∈supp(α)={a1,b1,a2}, the possible cases are as follows:
- a)
Let a1=1: a2−1b1a2=b1 2 and so G is a quotient of BS(1,2), a contradiction.
2. b)
Let b1=1: a2−1a1a2=a1 2 and so G is a quotient of BS(1,2), a contradiction.
3. c)
Let a2=1: a1−1b1a1b1−1a1−1b1=1. If x=a1−1b1 and y=b1−1, then y−1xy=x2 and G is a quotient of BS(1,2), a contradiction.
- ii)
Let b2=b1 and a3=a1: So, TC=[a1,b1,a2,b1,a1,a2,a1,b1] and the relation of C is a1−1b1a2−1b1a1−1a2a1−1b1=1. With the same discussion as item (i), the possible cases are as follows:
- a)
Let a1=1: a2−1b1a2=b1−2 and so G is a quotient of BS(1,−2), a contradiction.
2. b)
Let b1=1: a2−1a1a2=a1−2 and so G is a quotient of BS(1,−2), a contradiction.
3. c)
Let a2=1: a1−1b1 2a1−2b1=1. If x=b1−1a1 and y=b1−1, then y−1xy=x−2 and G is a quotient of BS(1,−2), a contradiction.
Therefore, TC′=[a4,b4,a1,b1,a2,b2,a3,b3].
3. (3)
Let TC′=[a3,b3,a4,b4,a1,b1,a2,b2]: So, we have
[TABLE]
By 4.2, a2=b1. In the following, we show that a2=a1.
Suppose, for a contradiction, that a2=a1. So, TC=[a1,b1,a1,b2,a1,b1,a4,b4] and TC′=[a1,b1,a4,b4,a1,b1,a1,b2]. By 4.2, 4.3 and 4.4, b1=a1, b1=a4, b4=a1, b4=a4 and a1=a4. So, b1=b4 because b4∈supp(α)={a1,b1,a4}. Now by 4.2, 4.3 and 4.4, b2=a1, b2=b1, a4=a1, a4=b1 and a1=b1. Hence, b2=a4 because b2∈supp(α)={a1,b1,a4}. So, TC=[a1,b1,a1,b2,a1,b1,b2,b1] and the relation of C is a1−1b1a1−1b2a1−1b1b2−1b1=1. Since 1∈supp(α)={a1,b1,b2}, the possible cases are as follows:
- a)
Let a1=1: b2−1b1−2b2=b1 and so G is a quotient of BS(−2,1), a contradiction.
2. b)
Let b1=1: b2−1a1−2b2=a1 and so G is a quotient of BS(−2,1), a contradiction.
3. c)
Let b2=1: a1−1b1a1−2b1 2=1. If x=b1a1−1 and y=a1−1, then y−1x−2y=x and G is a quotient of BS(−2,1), a contradiction.
Therefore, a2=a1.
By 4.2, 4.3 and 4.4, a1=b1, a4=a2, a4=b1 and b1=a2. So, a1=a4 because a1∈supp(α)={b1,a2,a4}. Now by 4.2, 4.3 and 4.4, a2=a1, a2=b2, b4=a1, b4=b2 and a1=b2. So, a2=b4 because b4∈supp(α)={a1,a2,b2}. Also, b2=a1, b2=a2, b1=a1, b1=a2 and a1=a2. Therefore, b1=b2 because b1∈supp(α)={a1,a2,b2}. Hence, TC=[a1,b1,a2,b1,a1,b1,a1,a2] and the relation of C is a1−1b1a2−1b1a1−1b1a1−1a2=1. Since 1∈supp(α)={a1,b1,a2}, the possible cases are as follows:
- a)
Let a1=1: a2−1b1−2a2=b1 and so G is a quotient of BS(−2,1), a contradiction.
2. b)
Let b1=1: a2−1a1−2a2=a1 and so G is a quotient of BS(−2,1), a contradiction.
3. c)
Let a2=1: a1−1b1 2a1−1b1a1−1=1. If x=a1b1−1 and y=a1−1, then y−1x−2y=x and G is a quotient of BS(−2,1), a contradiction.
Therefore, TC′=[a3,b3,a4,b4,a1,b1,a2,b2].
4. (4)
Let TC′=[a2,b2,a3,b3,a4,b4,a1,b1]: So, we have
[TABLE]
By 4.2, 4.3 and 4.4, we have a3=a1, a3=b1, b4=a1, b4=b1 and a1=b1. Therefore, a3=b4 because b4∈supp(α)={a1,b1,a3}. By 4.2, b3=a3, a4=a3 and b3=a4. Also, supp(α)={a1,b1,a3}. So, there are exactly two cases for b3,a4∈supp(α). In the following, we show that each of such cases gives a contradiction.
- i)
Let b3=a1 and a4=b1: So, TC=[a1,b1,a1,b1,a3,a1,b1,a3] and the relation of C is a1−1b1a1−1b1a3−1a1b1−1a3=1. Since 1∈supp(α)={a1,b1,a3}, the possible cases are as follows:
- a)
Let a1=1: a3−1b1a3=b1 2 and so G is a quotient of BS(1,2), a contradiction.
2. b)
Let b1=1: a3−1a1a3=a1 2 and so G is a quotient of BS(1,2), a contradiction.
3. c)
Let a3=1: a1−1b1a1b1−1a1−1b1=1. If x=a1−1b1 and y=b1−1, then y−1xy=x2 and G is a quotient of BS(1,2), a contradiction.
- ii)
Let b3=b1 and a4=a1: So, TC=[a1,b1,a1,b1,a3,b1,a1,a3] and the relation of C is a1−1b1a1−1b1a3−1b1a1−1a3=1. With the same discussion as item (i), the possible cases are as follows:
- a)
Let a1=1: a3−1b1a3=b1−2 and so G is a quotient of BS(1,−2), a contradiction.
2. b)
Let b1=1: a3−1a1a3=a1−2 and so G is a quotient of BS(1,−2), a contradiction.
3. c)
Let a3=1: a1−1b1 2a1−2b1=1. If x=b1−1a1 and y=b1−1, then y−1xy=x−2 and G is a quotient of BS(1,−2), a contradiction.
Therefore, TC′=[a2,b2,a3,b3,a4,b4,a1,b1].
5. (5)
Let TC′=[b1,a1,b4,a4,b3,a3,b2,a2]: So a1=b1, that is a contradiction with 4.2.
6. (6)
Let TC′=[b4,a4,b3,a3,b2,a2,b1,a1]: So a1=b4, that is a contradiction with 4.2.
7. (7)
Let TC′=[b2,a2,b1,a1,b4,a4,b3,a3]: So b1=a2, that is a contradiction with 4.2.
8. (8)
Let TC′=[b3,a3,b2,a2,b1,a1,b4,a4]: So, we have
[TABLE]
We have a2=a1 or a2=a1. In the following, we explain each cases in details.
- A)
Let a2=a1: So, TC=[a1,b1,a1,b2,b1,a1,a4,b4] and TC′=[a1,b1,b2,a1,b1,a1,b4,a4]. By 4.2, b2=a1, b2=b1 and a1=b1. So, {a1,b1,b2}=supp(α). Also, a4=a1, b4=a1 and a4=b4. Therefore, there are exactly two cases for a4,b4∈supp(α). In the following, we show that a4=b1 and b4=b2.
Suppose, for a contradiction, that a4=b2 and b4=b1. So, TC=[a1,b1,a1,b2,b1,a1,b2,b1] and the relation of C is a1−1b1a1−1b2b1−1a1b2−1b1=1. Since 1∈supp(α)={a1,b1,b2}, the possible cases are as follows:
- a)
Let a1=1: b2−1b1 2b2=b1 and so G is a quotient of BS(2,1), a contradiction.
2. b)
Let b1=1: b2−1a1 2b2=a1 and so G is a quotient of BS(2,1), a contradiction.
3. c)
Let b2=1: a1−1b1a1−1b1−1a1b1=1. If x=b1a1−1 and y=a1−1, then y−1x2y=x and G is a quotient of BS(2,1), a contradiction.
Therefore, a4=b1 and b4=b2. So, TC=[a1,b1,a1,b2,b1,a1,b1,b2] and the relation of C is a1−1b1a1−1b2b1−1a1b1−1b2=1. Since 1∈supp(α)={a1,b1,b2}, the possible cases are as follows:
- a)
Let a1=1: b1b2b1−2b2=1, where {b1,b2}={h2,h3}.
2. b)
Let b1=1: a1b2a1−2b2=1, where {a1,b2}={h2,h3}.
3. c)
Let b2=1: b1a1−1b1a1b1−1a1=1, where {a1,b1}={h2,h3}.
So, a2=a1 implies that TC=[a1,b1,a1,b2,b1,a1,b1,b2], TC′=[a1,b1,b2,a1,b1,a1,b2,b1] and {a1,b1,b2}=supp(α). Also, exactly one of the relations 14, 22 or 26 of Table 2 is satisfied in G.
2. B)
Let a2=a1: By 4.2, a2=a1, a2=b1 and a1=b1. So, {a1,b1,a2}=supp(α). Also, b2=b1 and b2=a2. So, b2=a1, TC=[a1,b1,a2,a1,b1,a1,a4,b4] and TC′=[a1,b1,a1,a2,b1,a1,b4,a4]. Also by 4.2, a4=a1, b4=a1 and a4=b4. Therefore, there are exactly two cases for a4,b4∈supp(α). In the following, we show that a4=a2 and b4=b1.
Suppose, for a contradiction, that a4=b1 and b4=a2. So, TC=[a1,b1,a2,a1,b1,a1,b1,a2] and the relation of C is a1−1b1a2−1a1b1−1a1b1−1a2=1. Since 1∈supp(α)={a1,b1,a2}, the possible cases are as follows:
- a)
Let a1=1: a2−1b1 2a2=b1 and so G is a quotient of BS(2,1), a contradiction.
2. b)
Let b1=1: a2−1a1 2a2=a1 and so G is a quotient of BS(2,1), a contradiction.
3. c)
Let a2=1: a1−1b1a1b1−1a1b1−1=1. If x=a1b1−1 and y=b1−1, then y−1x2y=x and G is a quotient of BS(2,1), a contradiction.
Therefore, a4=a2 and b4=b1. So, TC=[a1,b1,a2,a1,b1,a1,a2,b1] and the relation of C is a1−1b1a2−1a1b1−1a1a2−1b1=1. Since 1∈supp(α)={a1,b1,a2}, the possible cases are as follows:
- a)
Let a1=1: b1a2b1−2a2=1, where {b1,a2}={h2,h3}.
2. b)
Let b1=1: a1a2a1−2a2=1, where {a1,a2}={h2,h3}.
3. c)
Let a2=1: b1a1−1b1a1b1−1a1=1, where {a1,b1}={h2,h3}.
So, a2=a1 implies that TC=[a1,b1,a2,a1,b1,a1,a2,b1], TC′=[a1,b1,a1,a2,b1,a1,b1,a2] and {a1,b1,a2}=supp(α). Also, exactly one of the relations 14, 22 or 26 of Table 2 is satisfied in G.
This completes the proof.
∎
Remark 4.4**.**
Suppose that Z(α,β) contains two squares C and C′ with exactly one common edge as Figure 12, for some distinct elements gi,gj,gk,gl,gm,gp∈supp(β). Let TC and TC′ be 8-tuples of C and C′, respectively, with R(TC) and R(TC′) as follows:
[TABLE]
where a1,b1,a2,b2,a3,b3,a4,b4,a2′,b2′,a3′,b3′,a4′,b4′∈supp(α).
By the proof of Theorem 4.3, the possible cases for {TC,TC′} and the relation of C and C′ are:
-
{TC,TC′}={[h2,1,h2,h3,1,h2,1,h3],[h2,1,h3,h2,1,h2,h3,1]} and the relation of C and C′ is relation 14 of Table 2.
2. 2)
{TC,TC′}={[1,h2,1,h3,h2,1,h2,h3],[1,h2,h3,1,h2,1,h3,h2]} and the relation of C and C′ is relation 14 of Table 2.
3. 3)
{TC,TC′}={[h3,1,h3,h2,1,h3,1,h2],[h3,1,h2,h3,1,h3,h2,1]} and the relation of C and C′ is relation 22 of Table 2.
4. 4)
{TC,TC′}={[1,h3,1,h2,h3,1,h3,h2],[1,h3,h2,1,h3,1,h2,h3]} and the relation of C and C′ is relation 22 of Table 2.
5. 5)
{TC,TC′}={[h2,h3,h2,1,h3,h2,h3,1],[h2,h3,1,h2,h3,h2,1,h3]} and the relation of C and C′ is relation 26 of Table 2.
6. 6)
{TC,TC′}={[h3,h2,h3,1,h2,h3,h2,1],[h3,h2,1,h3,h2,h3,1,h2]} and the relation of C and C′ is relation 26 of Table 2.
Theorem 4.5**.**
The zero-divisor graph Z(α,β) is isomorphic to none of the graphs Ln and Mn in Figure 13.
Proof.
Let n be the number of vertices of the graph Ln or Mn. Then by Figure 13, the number of cycles of length 4 in Ln or Mn is equal to n/2. Also, each two consecutive squares have a common edge. So, if Z(α,β) contains no two squares with exactly one common edge, then the latter graph is isomorphic to none of the graphs Ln and Mn.
Suppose that Z(α,β) contains a subgraph as Figure 14, for some elements a1,b1,…,am+1,bm+1∈supp(β), in which the number of consecutive C4 cycles is denoted by m≥2. We denote by C1,C2,…,Cm the consecutive squares in Figure 14, from the left to the right respectively. For i∈{1,2,…,m−1}, let TCii and TCi+1i be 8-tuples of Ci and Ci+1, respectively, with R(TCii) and R(TCi+1i) as follows:
[TABLE]
where c1,d1,c2,d2,c3,d3,c4,d4,c2′,d2′,c3′,d3′,c4′,d4′∈supp(α).
We claim that for each i, i∈{1,2,…,m−1}, TCii=TC11 and TCi+1i=TC21. Note that by Remark 4.4, there are 6 possible cases for {TCii,TCi+1i}, where i∈{1,2,…,m−1}, and in particular for {TC11,TC21}. For each one of the latter cases, we prove our claim by induction on m:
Case 1: Let {TC11,TC21}={A,B}, where A=[h2,1,h2,h3,1,h2,1,h3] and B=[h2,1,h3,h2,1,h2,h3,1]. So, there are two subcases as follows:
- a)
Let TC11=A and TC21=B. If m=2, then the statement is obviously true. Suppose that the statement is true for m−1. Therefore, TCm−2m−2=A and TCm−1m−2=B. So by 4.5, R(TCm−1m−2) is:
[TABLE]
Hence, TCm−1m−1=A and so by Remark 4.4, TCmm−1=B.
2. b)
Let TC11=B and TC21=A. If m=2, then the statement is obviously true. Suppose that the statement is true for m−1. Therefore, TCm−2m−2=B and TCm−1m−2=A. So by 4.5, R(TCm−1m−2) is:
[TABLE]
Hence, TCm−1m−1=B and so by Remark 4.4, TCmm−1=A.
Cases 2-6: The proof of our claim for the remaining five possible cases of {TC11,TC21} are similar to the proof for case 1.
Therefore for each i∈{1,2,…,m−1}, TCii=TC11 and TCi+1i=TC21.
Now suppose, for a contradiction, that Z(α,β) is isomorphic to the graph Ln. So if m=n/2, then Z(α,β) contains a subgraph as Figure 14, where {a1,b1,…,am,bm}=supp(β), am+1=a1 and bm+1=b1. We denote by C1,C2,…,Cm the consecutive squares in Figure 14, from the left to the right respectively. With the above discussion, TCii=TC11 and TCi+1i=TC21, for i∈{1,2,…,m−1}. By Remark 4.4, there are 6 possible cases for {TC11,TC21}. Here we show that for each one of the latter cases, there is a contradiction:
Case 1: Let {TC11,TC21}={A,B}, where A=[h2,1,h2,h3,1,h2,1,h3] and B=[h2,1,h3,h2,1,h2,h3,1]. So, there are two subcases as follows:
- a)
Let TC11=A and TC21=B. Then TCii=A and TCi+1i=B, for i∈{1,2,…,m−1}. So by 4.5, a1=h3a2,a2=h3a3,…,am=h3a1. Therefore, a1=h3ma1 and so h3m=1, a contradiction because G is torsion-free and ∣supp(α)∣=3.
2. b)
Let TC11=B and TC21=A. Then TCii=B and TCi+1i=A, for i∈{1,2,…,m−1}. So by 4.5, h3a1=a2,h3a2=a3,…,h3am=a1. Therefore, a1=h3ma1 and so h3m=1, a contradiction because G is torsion-free and ∣supp(α)∣=3.
Cases 2-6: The proof for the remaining five possible cases of {TC11,TC21} are similar to the proof for case 1.
Therefore, the graph Z(α,β) cannot be isomorphic to the graph Ln.
Now suppose, for a contradiction, that Z(α,β) is isomorphic to the graph Mn. So if m=n/2, then Z(α,β) contains a subgraph as Figure 14, where {a1,b1,…,am,bm}=supp(β), am+1=b1 and bm+1=a1. We denote by C1,C2,…,Cm the consecutive squares in Figure 14, from the left to the right respectively. With the above discussion, TCii=TC11 and TCi+1i=TC21, for i∈{1,2,…,m−1}. By Remark 4.4, there are 6 possible cases for {TC11,TC21}. Here we show that for each one of the latter cases, there is a contradiction:
Case 1: Let {TC11,TC21}={A,B}, where A=[h2,1,h2,h3,1,h2,1,h3] and B=[h2,1,h3,h2,1,h2,h3,1]. So, there are two subcases as follows:
- a)
Let TC11=A and TC21=B. Then TCii=A and TCi+1i=B, for i∈{1,2,…,m−1}. So by 4.5, a1=h3a2,a2=h3a3,…,am−1=h3am, am=h3b1 and b1=h2a1. Therefore, a1=h3mh2a1 and so h3mh2=1. Hence, h2=h3−m and so G is abelian, a contradiction because abelian torsion-free groups satisfy the Conjecture 1.1 (see [18, Theorem 26.2]).
2. b)
Let TC11=B and TC21=A. Then TCii=B and TCi+1i=A, for i∈{1,2,…,m−1}. So by 4.5, h3a1=a2,h3a2=a3,…,h3am−1=am, h3am=b1 and b1=h2a1. Therefore, a1=h2−1h3ma1 and so h2−1h3m=1. Hence, h2=h3m and so G is abelian, a contradiction because abelian torsion-free groups satisfy the Conjecture 1.1.
Cases 2-6: The proof for the remaining five possible cases of {TC11,TC21} are similar to the proof for case 1.
Therefore, the graph Z(α,β) cannot be isomorphic to the graph Mn. This completes the proof.
∎
5. Forbidden subgraphs of Zero-divisor graphs over F2 on any torsion-free group
In Section 3, we studied the existence of triangles and squares in zero-divisor graphs of length 3 over an arbitrary field and on any torsion-free group and we showed that C3 and K2,3 are two forbidden subgraphs of such graphs.
Throughout this section suppose that α and β are non-zero elements of the group algebra F2[G] of a torsion-free group G such that ∣supp(α)∣=3, αβ=0 and if αβ′=0 for some non-zero element β′ of the group algebra, then ∣supp(β′)∣≥∣supp(β)∣. By Remark 3.3, one may assume that 1∈supp(α) and G=⟨supp(α)⟩. Let supp(α)={1,h2,h3} and n=∣supp(β)∣.
With the same discussion such as about C3 and C4 cycles, we can study cycles of other lengths in the graph Z(α,β) by using their relations. In this section, by using cycles up to lengths 7 and their relations, we find additional forbidden subgraphs of the graph Z(α,β). The procedure of finding such graphs is similar to the procedure of finding previous examples. So, the frequent tedious details are omitted. In Appendix of [1], some details of our computations are given for the reader’s convenience. Forbidden subgraphs of Z(α,β) are listed in Table LABEL:tab-forbiddens. In the following, we give some details about such subgraphs.
5.1. K2,3
By Theorem 3.29, Z(α,β) contains no K2,3 as a subgraph.
5.2. C4−−C5
It can be seen that there are 121 different cases for the relations of the cycles C4 and C5 in the subgraph C4−−C5. Using GAP [10], we see that in 111 cases of these 121 cases, the groups which are obtained are finite and solvable, a contradiction. So, there are just 10 cases which may lead to the existence of a subgraph isomorphic to the graph C4−−C5 in Z(α,β). It can be seen that each of such cases gives a contradiction and so, the graph Z(α,β) contains no subgraph isomorphic to the graph C4−−C5.
We note that each group with two generators h2 and h3 and two relations which is one of the 10 latter cases is a quotient of B(1,k), for some integer k, or has a torsion element.
5.3. C4−−C6
It can be seen that there are 658 different cases for the relations of the cycles C4 and C6 in the subgraph C4−−C6. Using GAP [10], we see that in 632 cases of these 658 cases, the groups which are obtained are finite or solvable, a contradiction. So, there are just 20 cases which may lead to the existence of a subgraph isomorphic to the graph C4−−C6 in Z(α,β). It can be seen that each of such cases gives a contradiction and so, the graph Z(α,β) contains no subgraph isomorphic to the graph C4−−C6.
We note that each group with two generators h2 and h3 and two relations which is one of the 20 latter cases is a quotient of B(1,k), for some integer k, has a torsion element or is a cyclic group.
5.4. C4−C5(−C5−)
It can be seen that there are 42 different cases for the relations of a cycle C4 and two cycles C5 in the subgraph C4−C5(−C5−). Using GAP [10], we see that in 38 cases of these 42 cases, the groups which are obtained are finite and solvable, a contradiction. So, there are just 4 cases which may lead to the existence of a subgraph isomorphic to the graph C4−C5(−C5−) in Z(α,β). It can be seen that each of such cases gives a contradiction and so, the graph Z(α,β) contains no subgraph isomorphic to the graph C4−C5(−C5−).
We note that each group with two generators h2 and h3 and three relations which is one of the 4 latter cases is a quotient of B(1,k), for some integer k, has a torsion element.
5.5. C4−C5(−C4−)
It can be seen that there are 4 different cases for the relations of two cycles C4 and a cycle C5 in the subgraph C4−C5(−C4−). Using GAP [10], we see that in all of such cases, the groups which are obtained are finite and solvable, a contradiction. So, the graph Z(α,β) contains no subgraph isomorphic to the graph C4−C5(−C4−).
5.6. C4−C5(−C6−−)
It can be seen that there are 126 different cases for the relations of a cycle C4, a cycle C5 and a cycle C6 in the subgraph C4−C5(−C6−−). Using GAP [10], we see that in 122 cases of these 126 cases, the groups which are obtained are finite and solvable, a contradiction. So, there are just 4 cases which may lead to the existence of a subgraph isomorphic to the graph C4−C5(−C6−−) in Z(α,β). It can be seen that each of such cases gives a contradiction and so, the graph Z(α,β) contains no subgraph isomorphic to the graph C4−C5(−C6−−).
We note that each group with two generators h2 and h3 and three relations which is one of the 4 latter cases is a quotient of B(1,k), for some integer k.
5.7. C4−C5(−C6−)
It can be seen that there are 462 different cases for the relations of a cycle C4, a cycle C5 and a cycle C6 in the subgraph C4−C5(−C6−). Using GAP [10], we see that in 436 cases of these 462 cases, the groups which are obtained are finite or solvable, a contradiction. So, there are just 22 cases which may lead to the existence of a subgraph isomorphic to the graph C4−C5(−C6−) in Z(α,β). It can be seen that each of such cases gives a contradiction and so, the graph Z(α,β) contains no subgraph isomorphic to the graph C4−C5(−C6−).
We note that each group with two generators h2 and h3 and three relations which is one of the 22 latter cases is a quotient of B(1,k), for some integer k, is a cyclic group or is a solvable group.
5.8. C4−C5(−C7−−)
It can be seen that there are 648 different cases for the relations of a cycle C4, a cycle C5 and a cycle C7 in the subgraph C4−C5(−C7−−). Using GAP [10], we see that in 608 cases of these 648 cases, the groups which are obtained are finite and solvable, a contradiction. So, there are just 40 cases which may lead to the existence of a subgraph isomorphic to the graph C4−C5(−C7−−) in Z(α,β). It can be seen that each of such cases gives a contradiction and so, the graph Z(α,β) contains no subgraph isomorphic to the graph C4−C5(−C7−−).
We note that each group with two generators h2 and h3 and three relations which is one of the 40 latter cases is a quotient of B(1,k), for some integer k, has a torsion element, is a cyclic group or is a solvable group.
5.9. C5−−C5(−−C5)
It can be seen that there are 192 different cases for the relations of three cycles C5 in the subgraph C5−−C5(−−C5). Using GAP [10], we see that in 188 cases of these 192 cases, the groups which are obtained are finite and solvable, a contradiction. So, there are just 4 cases which may lead to the existence of a subgraph isomorphic to the graph C5−−C5(−−C5) in Z(α,β). It can be seen that each of such cases gives a contradiction and so, the graph Z(α,β) contains no subgraph isomorphic to the graph C5−−C5(−−C5).
We note that each group with two generators h2 and h3 and three relations which is one of the 4 latter cases is a quotient of B(1,k), for some integer k.
5.10. C5−−C5(−−C6)
It can be seen that there are 1006 different cases for the relations of two cycles C5 and a cycle C6 in the subgraph C5−−C5(−−C6). Using GAP [10], we see that in 986 cases of these 1006 cases, the groups which are obtained are finite and solvable, a contradiction. So, there are just 20 cases which may lead to the existence of a subgraph isomorphic to the graph C5−−C5(−−C6) in Z(α,β). It can be seen that each of such cases gives a contradiction and so, the graph Z(α,β) contains no subgraph isomorphic to the graph C5−−C5(−−C6).
We note that each group with two generators h2 and h3 and three relations which is one of the 20 latter cases is a quotient of B(1,k), for some integer k, is a cyclic group or is a solvable group.
5.11. C4−C6(−−C7−−)(C7−1)
It can be seen that there are 176 different cases for the relations of a cycle C4, two cycles C7 and a cycle C6 in the subgraph C4−C6(−−C7−−)(C7−1). Using GAP [10], we see that in all of such cases, the groups which are obtained are finite and solvable, a contradiction. So, the graph Z(α,β) contains no subgraph isomorphic to the graph C4−C6(−−C7−−)(C7−1).
5.12. C4−C6(−−C7−−)(−−C5−)
It can be seen that there are 28 different cases for the relations of a cycle C4, a cycle C6, a cycle C7 and a cycle C5 in the subgraph C4−C6(−−C7−−)(−−C5−). Using GAP [10], we see that in 24 cases of these 28 cases, the groups which are obtained are finite and solvable, a contradiction. So, there are just 4 cases which may lead to the existence of a subgraph isomorphic to the graph C4−C6(−−C7−−)(−−C5−) in Z(α,β). It can be seen that each of such cases gives a contradiction and so, the graph Z(α,β) contains no subgraph isomorphic to the graph C4−C6(−−C7−−)(−−C5−).
We note that each group with two generators h2 and h3 and four relations which is one of the 4 latter cases is a cyclic group.
5.13. C4−C6(−C6−−)(−C4−)
It can be seen that there is no case for the relations of two cycles C4 and two cycles C6 in the subgraph C4−C6(−C6−−)(−C4−). It means that the graph Z(α,β) contains no subgraph isomorphic to the graph C4−C6(−C6−−)(−C4−).
5.14. C4−C6(−C6−−)(−−C5−)
It can be seen that there are 22 different cases for the relations of a cycle C4, two cycles C6 and a cycle C5 in the subgraph C4−C6(−C6−−)(−−C5−). Using GAP [10], we see that in all of such cases, the groups which are obtained are finite and solvable, a contradiction. So, the graph Z(α,β) contains no subgraph isomorphic to the graph C4−C6(−C6−−)(−−C5−).
5.15. C4−C6(−C6−−)(C6−−−)
It can be seen that there are 66 different cases for the relations of a cycle C4 and three cycles C6 in the subgraph C4−C6(−C6−−)(C6−−−). Using GAP [10], we see that in 62 cases of these 66 cases, the groups which are obtained are finite and solvable, a contradiction. So, there are just 4 cases which may lead to the existence of a subgraph isomorphic to the graph C4−C6(−C6−−)(C6−−−) in Z(α,β). It can be seen that each of such cases gives a contradiction and so, the graph Z(α,β) contains no subgraph isomorphic to the graph C4−C6(−C6−−)(C6−−−).
We note that each group with two generators h2 and h3 and four relations which is one of the 4 latter cases is a quotient of B(1,k), for some integer k.
5.16. C4−C6(−C6−−)(−−−C4)
It can be seen that there is no case for the relations of two cycles C4 and two cycles C6 in the subgraph C4−C6(−C6−−)(−−−C4). It means that the graph Z(α,β) contains no subgraph isomorphic to the graph C4−C6(−C6−−)(−−−C4).
5.17. C5−C5(−−C6−−)
It can be seen that there are 440 different cases for the relations of two cycles C5 and a cycle C6 in the subgraph C5−C5(−−C6−−). Using GAP [10], we see that in 404 cases of these 440 cases, the groups which are obtained are finite and solvable, a contradiction. So, there are just 36 cases which may lead to the existence of a subgraph isomorphic to the graph C5−C5(−−C6−−) in Z(α,β). It can be seen that each of such cases gives a contradiction and so, the graph Z(α,β) contains no subgraph isomorphic to the graph C5−C5(−−C6−−).
We note that each group with two generators h2 and h3 and three relations which is one of the 36 latter cases is a quotient of B(1,k), for some integer k, has a torsion element or is a cyclic group.
5.18. C5−C5(−C6−−)(C6−−−)
It can be seen that there are 56 different cases for the relations of two cycles C5 and two cycles C6 in the subgraph C5−C5(−C6−−)(C6−−−). Using GAP [10], we see that in 54 cases of these 56 cases, the groups which are obtained are finite or solvable, a contradiction. So, there are just 2 cases which may lead to the existence of a subgraph isomorphic to the graph C5−C5(−C6−−)(C6−−−) in Z(α,β). It can be seen that each of such cases gives a contradiction and so, the graph Z(α,β) contains no subgraph isomorphic to the graph C5−C5(−C6−−)(C6−−−).
We note that each group with two generators h2 and h3 and four relations which is one of the 2 latter cases is a cyclic group.
5.19. C5−C5(−C6−−)(−−C6−1)
It can be seen that there are 56 different cases for the relations of two cycles C5 and two cycles C6 in the subgraph C5−C5(−C6−−)(−−C6−1). Using GAP [10], we see that in all of such cases, the groups which are obtained are finite and solvable, a contradiction. So, the graph Z(α,β) contains no subgraph isomorphic to the graph C5−C5(−C6−−)(−−C6−1).
5.20. C5−C5(−C6−−)(−C5−−)
It can be seen that there are 14 different cases for the relations of three cycles C5 and one cycle C6 in the subgraph C5−C5(−C6−−)(−C5−−). Using GAP [10], we see that in all of such cases, the groups which are obtained are finite and solvable, a contradiction. So, the graph Z(α,β) contains no subgraph isomorphic to the graph C5−C5(−C6−−)(−C5−−).
5.21. C6−−−C6(C6−−−C6)
It can be seen that there are 46 different cases for the relations of four C6 cycles in the subgraph C6−−−C6(C6−−−C6). Using GAP [10], we see that in 30 cases of these 46 cases, the groups which are obtained are finite or solvable, a contradiction. So, there are just 16 cases which may lead to the existence of a subgraph isomorphic to the graph C6−−−C6(C6−−−C6) in Z(α,β). It can be seen that each of such cases gives a contradiction and so, the graph Z(α,β) contains no subgraph isomorphic to the graph C6−−−C6(C6−−−C6).
We note that each group with two generators h2 and h3 and four relations which is one of the 16 latter cases is a quotient of B(1,k), for some integer k, or has a torsion element.
5.22. C6−−−C6(C6)(C6)(C6)
It can be seen that there are 10 different cases for the relations of five C6 cycles in the subgraph C6−−−C6(C6)(C6)(C6). Using GAP [10], we see that in 6 cases of these 10 cases, the groups which are obtained are finite or solvable, a contradiction. So, there are just 4 cases which may lead to the existence of a subgraph isomorphic to the graph C6−−−C6(C6)(C6)(C6) in Z(α,β). It can be seen that each of such cases gives a contradiction and so, the graph Z(α,β) contains no subgraph isomorphic to the graph C6−−−C6(C6)(C6)(C6).
We note that each group with two generators h2 and h3 and five relations which is one of the 4 latter cases is a quotient of B(1,k), for some integer k, or is a cyclic group.
5.23. C5(−−C6−−)C5(−−−C6)
It can be seen that there are 134 different cases for the relations of two cycles C5 and two cycles C6 in the subgraph C5(−−C6−−)C5(−−−C6). Using GAP [10], we see that in 130 cases of these 134 cases, the groups which are obtained are finite or solvable, a contradiction. So, there are just 4 cases which may lead to the existence of a subgraph isomorphic to the graph C5(−−C6−−)C5(−−−C6) in Z(α,β). It can be seen that each of such cases gives a contradiction and so, the graph Z(α,β) contains no subgraph isomorphic to the graph C5(−−C6−−)C5(−−−C6).
We note that each group with two generators h2 and h3 and four relations which is one of the 4 latter cases is a quotient of B(1,k), for some integer k, or is a cyclic group.
5.24. C6−−C6(C6−−C6)
It can be seen that there are 5119 different cases for the relations of four C6 cycles in the subgraph C6−−C6(C6−−C6). Using GAP [10], we see that in 4983 cases of these 5119 cases, the groups which are obtained are finite or solvable, a contradiction. So, there are just 136 cases which may lead to the existence of a subgraph isomorphic to the graph C6−−C6(C6−−C6) in Z(α,β). It can be seen that each of such cases gives a contradiction and so, the graph Z(α,β) contains no subgraph isomorphic to the graph C6−−C6(C6−−C6).
We note that each group with two generators h2 and h3 and four relations which is one of the 136 latter cases is a quotient of B(1,k), for some integer k, has a torsion element or is a cyclic group.
5.25. C6−−−C6(C6−−C6)
It can be seen that there are 1594 different cases for the relations of four C6 cycles in the subgraph C6−−−C6(C6−−C6). Using GAP [10], we see that in 1446 cases of these 1594 cases, the groups which are obtained are finite or solvable, a contradiction. So, there are just 148 cases which may lead to the existence of a subgraph isomorphic to the graph C6−−−C6(C6−−C6) in Z(α,β). It can be seen that each of such cases gives a contradiction and so, the graph Z(α,β) contains no subgraph isomorphic to the graph C6−−−C6(C6−−C6).
We note that each group with two generators h2 and h3 and four relations which is one of the 148 latter cases is a quotient of B(1,k), for some integer k, has a torsion element or is a cyclic group.
5.26. C6−−−C6(−C5−)
It can be seen that there are 1482 different cases for the relations of two cycles C6 and a cycle C5 in the subgraph C6−−−C6(−C5−). Using GAP [10], we see that in 1358 cases of these 1482 cases, the groups which are obtained are finite and solvable, a contradiction. So, there are just 124 cases which may lead to the existence of a subgraph isomorphic to the graph C6−−−C6(−C5−) in Z(α,β). It can be seen that each of such cases gives a contradiction and so, the graph Z(α,β) contains no subgraph isomorphic to the graph C6−−−C6(−C5−).
We note that each group with two generators h2 and h3 and three relations which is one of the 124 latter cases is a quotient of B(1,k), for some integer k, has a torsion element or is a cyclic group.
5.27. C4−C6(−−C7−−)(−−−C6)
It can be seen that there are 124 different cases for the relations of a cycle C4, two cycles C6 and a cycle C7 in the subgraph C4−C6(−−C7−−)(−−−C6). Using GAP [10], we see that in 112 cases of these 124 cases, the groups which are obtained are finite and solvable, a contradiction. So, there are just 12 cases which may lead to the existence of a subgraph isomorphic to the graph C4−C6(−−C7−−)(−−−C6) in Z(α,β). It can be seen that each of such cases gives a contradiction and so, the graph Z(α,β) contains no subgraph isomorphic to the graph C4−C6(−−C7−−)(−−−C6).
We note that each group with two generators h2 and h3 and four relations which is one of the 12 latter cases is a quotient of B(1,k), for some integer k.
5.28. C4−C6(−−C7−−)(C4)(C4)
It can be seen that there are 8 different cases for the relations of three cycles C4, a cycle C7 and a cycle C6 in the subgraph C4−C6(−−C7−−)(C4)(C4). Using GAP [10], we see that in all of such cases, the groups which are obtained are finite and solvable, a contradiction. So, the graph Z(α,β) contains no subgraph isomorphic to the graph C4−C6(−−C7−−)(C4)(C4).
5.29. C6−−−C6(−C5−−)
It can be seen that there are 418 different cases for the relations of two cycles C6 and a cycle C5 in the subgraph C6−−−C6(−C5−−). Using GAP [10], we see that in 358 cases of these 418 cases, the groups which are obtained are finite and solvable, a contradiction. So, there are just 60 cases which may lead to the existence of a subgraph isomorphic to the graph C6−−−C6(−C5−−) in Z(α,β). It can be seen that each of such cases gives a contradiction and so, the graph Z(α,β) contains no subgraph isomorphic to the graph C6−−−C6(−C5−−).
We note that each group with two generators h2 and h3 and three relations which is one of the 60 latter cases is a quotient of B(1,k), for some integer k, has a torsion element, is a cyclic group or is a solvable group.
5.30. C6−−C6(−−C5−)(−C5−)
It can be seen that there are 62 different cases for the relations of two cycles C5 and two cycles C6 in the subgraph C6−−C6(−−C5−)(−C5−). Using GAP [10], we see that in 56 cases of these 62 cases, the groups which are obtained are finite or solvable, a contradiction. So, there are just 6 cases which may lead to the existence of a subgraph isomorphic to the graph C6−−C6(−−C5−)(−C5−) in Z(α,β). It can be seen that each of such cases gives a contradiction and so, the graph Z(α,β) contains no subgraph isomorphic to the graph C6−−C6(−−C5−)(−C5−).
We note that each group with two generators h2 and h3 and four relations which is one of the 6 latter cases has a torsion element or is a cyclic group.
5.31. C6−−C6(−−C5−)(C6−−−)
It can be seen that there are 76 different cases for the relations of a cycle C5 and three cycles C6 in the subgraph C6−−C6(−−C5−)(C6−−−). Using GAP [10], we see that in 64 cases of these 76 cases, the groups which are obtained are finite or solvable, a contradiction. So, there are just 12 cases which may lead to the existence of a subgraph isomorphic to the graph C6−−C6(−−C5−)(C6−−−) in Z(α,β). It can be seen that each of such cases gives a contradiction and so, the graph Z(α,β) contains no subgraph isomorphic to the graph C6−−C6(−−C5−)(C6−−−).
We note that each group with two generators h2 and h3 and four relations which is one of the 12 latter cases is a quotient of B(1,k), for some integer k, has a torsion element or is a cyclic group.
5.32. C5(−−C6−−)C5(C6)
It can be seen that there are 120 different cases for the relations of two cycles C5 and two cycles C6 in the subgraph C5(−−C6−−)C5(C6). Using GAP [10], we see that in 104 cases of these 120 cases, the groups which are obtained are finite or solvable, a contradiction. So, there are just 16 cases which may lead to the existence of a subgraph isomorphic to the graph C5(−−C6−−)C5(C6) in Z(α,β). It can be seen that each of such cases gives a contradiction and so, the graph Z(α,β) contains no subgraph isomorphic to the graph C5(−−C6−−)C5(C6).
We note that each group with two generators h2 and h3 and four relations which is one of the 16 latter cases is a quotient of B(1,k), for some integer k, has a torsion element or is a cyclic group.
5.33. C5(−−C6−−)C5(C7)
It can be seen that there are 248 different cases for the relations of two cycles C5, a cycle C6 and a cycle C7 in the subgraph C5(−−C6−−)C5(C7). Using GAP [10], we see that in 220 cases of these 248 cases, the groups which are obtained are finite or solvable, a contradiction. So, there are just 28 cases which may lead to the existence of a subgraph isomorphic to the graph C5(−−C6−−)C5(C7) in Z(α,β). It can be seen that each of such cases gives a contradiction and so, the graph Z(α,β) contains no subgraph isomorphic to the graph C5(−−C6−−)C5(C7).
We note that each group with two generators h2 and h3 and four relations which is one of the 28 latter cases is a quotient of B(1,k), for some integer k, has a torsion element or is a cyclic group.
5.34. C5−C5(−−C7−−)(−−C5)
It can be seen that there are 394 different cases for the relations of a cycle C7 and three cycles C5 in the subgraph C5−C5(−−C7−−)(−−C5). Using GAP [10], we see that in 352 cases of these 394 cases, the groups which are obtained are finite or solvable, a contradiction. So, there are just 42 cases which may lead to the existence of a subgraph isomorphic to the graph C5−C5(−−C7−−)(−−C5) in Z(α,β). It can be seen that each of such cases gives a contradiction and so, the graph Z(α,β) contains no subgraph isomorphic to the graph C5−C5(−−C7−−)(−−C5).
We note that each group with two generators h2 and h3 and four relations which is one of the 42 latter cases is a quotient of B(1,k), for some integer k, has a torsion element, is a cyclic group or is a solvable group.
5.35. C5−C5(−−C7−−)(−C5−)
It can be seen that there are 138 different cases for the relations of a cycle C7 and three cycles C5 in the subgraph C5−C5(−−C7−−)(−C5−). Using GAP [10], we see that in 132 cases of these 138 cases, the groups which are obtained are finite or solvable, a contradiction. So, there are just 6 cases which may lead to the existence of a subgraph isomorphic to the graph C5−C5(−−C7−−)(−C5−) in Z(α,β). It can be seen that each of such cases gives a contradiction and so, the graph Z(α,β) contains no subgraph isomorphic to the graph C5−C5(−−C7−−)(−C5−).
We note that each group with two generators h2 and h3 and four relations which is one of the 6 latter cases has a torsion element, is a cyclic group or is a solvable group.
5.36. C5−C5(−C6−−)(−−C6−2)
It can be seen that there are 22 different cases for the relations of two cycles C5 and two cycles C6 in the subgraph C5−C5(−C6−−)(−−C6−2). Using GAP [10], we see that in all of such cases, the groups which are obtained are finite and solvable, a contradiction. So, the graph Z(α,β) contains no subgraph isomorphic to the graph C5−C5(−C6−−)(−−C6−2).
5.37. C4−C4(−C7−)(C4)
It can be seen that there are 32 different cases for the relations of a cycle C7 and three cycles C4 in the subgraph C4−C4(−C7−)(C4). Using GAP [10], we see that in all of such cases, the groups which are obtained are finite and solvable, a contradiction. So, the graph Z(α,β) contains no subgraph isomorphic to the graph C4−C4(−C7−)(C4).
5.38. C5−−C5(−C5−−)
It can be seen that there are 64 different cases for the relations of three cycles C5 in the subgraph C5−−C5(−C5−−). Using GAP [10], we see that in 58 cases of these 64 cases, the groups which are obtained are finite and solvable, a contradiction. So, there are just 6 cases which may lead to the existence of a subgraph isomorphic to the graph C5−−C5(−C5−−) in Z(α,β). It can be seen that each of such cases gives a contradiction and so, the graph Z(α,β) contains no subgraph isomorphic to the graph C5−−C5(−C5−−).
We note that each group with two generators h2 and h3 and three relations which is one of the 6 latter cases is a quotient of B(1,k), for some integer k or is a cyclic group.
5.39. C6−−−C6(−C4)
It can be seen that there are 420 different cases for the relations of two cycles C6 and a cycle C4 in the subgraph C6−−−C6(−C4). Using GAP [10], we see that in 398 cases of these 420 cases, the groups which are obtained are finite or solvable, a contradiction. So, there are just 22 cases which may lead to the existence of a subgraph isomorphic to the graph C6−−−C6(−C4) in Z(α,β). It can be seen that each of such cases gives a contradiction and so, the graph Z(α,β) contains no subgraph isomorphic to the graph C6−−−C6(−C4).
We note that each group with two generators h2 and h3 and three relations which is one of the 22 latter cases is a cyclic group.
5.40. C6−−C6(C4)
It can be seen that there are 279 different cases for the relations of two cycles C6 and a cycle C4 in the subgraph C6−−C6(C4). Using GAP [10], we see that in 268 cases of these 279 cases, the groups which are obtained are finite or solvable, a contradiction. So, there are just 11 cases which may lead to the existence of a subgraph isomorphic to the graph C6−−C6(C4) in Z(α,β). It can be seen that each of such cases gives a contradiction and so, the graph Z(α,β) contains no subgraph isomorphic to the graph C6−−C6(C4).
We note that each group with two generators h2 and h3 and three relations which is one of the 11 latter cases is a quotient of B(1,k), for some integer k, or is a cyclic group.
5.41. C4−C6(−C4)(−C4)
It can be seen that there are 36 different cases for the relations of a cycle C6 and three cycles C4 in the subgraph C4−C6(−C4)(−C4). Using GAP [10], we see that in all of such cases, the groups which are obtained are solvable, a contradiction. So, the graph Z(α,β) contains no subgraph isomorphic to the graph C4−C6(−C4)(−C4).
5.42. C4−C6(−−C7−−)(−C5−)
It can be seen that there are 62 different cases for the relations of a cycle C4, a cycle C6, a cycle C7 and a cycle C5 in the subgraph C4−C6(−−C7−−)(−C5−). Using GAP [10], we see that in 58 cases of these 62 cases, the groups which are obtained are solvable, a contradiction. So, there are just 4 cases which may lead to the existence of a subgraph isomorphic to the graph C4−C6(−−C7−−)(−C5−) in Z(α,β). It can be seen that each of such cases gives a contradiction and so, the graph Z(α,β) contains no subgraph isomorphic to the graph C4−C6(−−C7−−)(−C5−).
We note that each group with two generators h2 and h3 and four relations which is one of the 4 latter cases is a cyclic group.
5.43. C5−C5(−C6−−)(−−C5−)
It can be seen that there are 14 different cases for the relations of three cycles C5 and a cycle C6 in the subgraph C5−C5(−C6−−)(−−C5−). Using GAP [10], we see that in all of such cases, the groups which are obtained are solvable, a contradiction. So, the graph Z(α,β) contains no subgraph isomorphic to the graph C5−C5(−C6−−)(−−C5−).
5.44. C4−C6(−−C7−−)(C7−2)
It can be seen that there are 168 different cases for the relations of a cycle C4, a cycle C6 and two cycles C7 in the subgraph C4−C6(−−C7−−)(C7−2). Using GAP [10], we see that in 152 cases of these 168 cases, the groups which are obtained are solvable, a contradiction. So, there are just 16 cases which may lead to the existence of a subgraph isomorphic to the graph C4−C6(−−C7−−)(C7−2) in Z(α,β). It can be seen that each of such cases gives a contradiction and so, the graph Z(α,β) contains no subgraph isomorphic to the graph C4−C6(−−C7−−)(C7−2).
We note that each group with two generators h2 and h3 and four relations which is one of the 16 latter cases is a quotient of B(1,k), for some integer k, or is a cyclic group.
In the following theorem, we summarize our results about forbidden subgraphs of a zero-divisor graph of length 3 over F2 on any torsion-free group.
Theorem 5.1**.**
Suppose that α and β are non-zero elements of a group algebra over F2 on any torsion-free group G such that ∣supp(α)∣=3, αβ=0 and if αβ′=0 for some non-zero element β′ of the group algebra, then ∣supp(β′)∣≥∣supp(β)∣. Then Z(α,β) is a triangle-free graph which contains none of the other 46 graphs in Table LABEL:tab-forbiddens as a subgraph.
6. **The possible number of vertices of a zero-divisor graph
of length 3 over F2 on any torsion-free group**
By Remark 4.2, the number of vertices of a zero-divisor graph of length 3 over F2 on any torsion-free group must be an even positive integer n≥4. Also by Theorem 4.1, such a latter graph is a connected simple cubic one containing no subgraph isomorphic to a triangle. Furthermore, we found 46 other forbidden subgraphs of such a graph in Sections 4 and 5.
Using Sage Mathematics Software [19] and its package Nauty-geng, all non-isomorphic connected cubic triangle-free graphs with the size of the vertex sets n can be found. In this section by using Sage Mathematics Software, we give some results about checking each of the mentioned forbidden subgraphs in all of non-isomorphic connected cubic triangle-free graphs with the size of vertex sets n≤20. By using such results, we show that n must be greater than or equal to 20. Also, some results in the case n=20 is given.
Table LABEL:tab-graphs lists all results about the number of non-isomorphic connected cubic triangle-free graphs with the size of vertex sets n≤20 which contain each of the forbidden subgraphs. The results in this table, from top to bottom, are presented in such a way that by checking each of the forbidden subgraphs in a row, the number of graphs containing these subgraph are omitted from the total number and the existence of the next forbidden subgraph is checked among the remaining ones.
The discussion above and the results of Table LABEL:tab-graphs are summarized in the following theorem.
Theorem 6.1**.**
The vertex set size of a zero-divisor graph of length 3 over F2 on any torsion-free group must be greater than or equal to 20. Furthermore, there are just 1120 graphs with vertex set size equal to 20 which may be isomorphic to such latter graphs.
Corollary 6.2**.**
Let α and β be non-zero elements of the group algebra of any torsion-free group over F2. If ∣supp(α)∣=3 and αβ=0 then ∣supp(β)∣≥20.
7. Possible zero divisors with supports of size 3 in F[G]
Throughout this section let α be a non-zero element in the group algebra of a torsion-free group G over a field F such that ∣supp(α)∣=3 and αβ=0 for some non-zero element β of the group algebra. It is known that ∣supp(β)∣≥3 (see [20, Theorem 2.1]). In this section, we show that ∣supp(β)∣ must be at least 10. Here, one may assume that β has minimum possible support size among all elements γ with αγ=0. Therefore by Remark 3.3, 1∈supp(α) and G=⟨supp(α)⟩. Let supp(α)={h1,h2,h3}, supp(β)={g1,g2,…,gn} and n=∣supp(β)∣. If A={1,2,3}×{1,2,…,n}, then for all (i,j)∈A there must be an (i′,j′)∈A such that i=i′, j=j′ and higj=hi′gj′ because αβ=0.
Theorem 7.1** (Corollary of [13]).**
Let G be an arbitrary group and let B and C be finite non-empty subsets of G. Suppose that each non-identity element g of G has a finite or infinite order greater than or equal to ∣B∣+∣C∣−1. Then ∣BC∣≥∣B∣+∣C∣−1.
Theorem 7.2** (Corollary 11 of [11]).**
If C is a finite generating subset of a nonabelian torsion-free group G such that 1∈C and ∣C∣≥4, then ∣BC∣≥∣B∣+∣C∣+1 for all B⊂G with ∣B∣≥3.
Abelian torsion-free groups satisfy the Conjecture 1.1 (see [18, Theorem 26.2]). So, G must be a nonabelian torsion-free group. Also, 1∈supp(α) and G=⟨supp(α)⟩. Therefore by Theorem 7.2, 3n≥∣supp(α)supp(β)∣≥4+n.
Theorem 7.3** (Proposition 4.12 of [6]).**
There exist no γ,δ∈F2[G] such that γδ=1, where ∣supp(γ)∣=3 and ∣supp(δ)∣≥13 is an odd integer.
- (1)
Let n=3. By Theorem 7.1, 3n≥∣supp(α)supp(β)∣≥∣supp(α)∣+∣supp(β)∣−1 because G is torsion-free. Let ∣supp(β)∣=3. Then 9≥∣supp(α)supp(β)∣≥5. Since 9−5=4, there is an (i,j)∈A such that higj=hi′gj′ for all (i′,j′)∈A where i=i′ and j=j′, a contradiction with αβ=0. So, ∣supp(β)∣ must be at least 4.
2. (2)
Let n=4. Then by Theorem 7.2, 12≥∣supp(α)supp(β)∣≥8. Since 12−8=4, there is an (i,j)∈A such that higj=hi′gj′ for all (i′,j′)∈A where i=i′ and j=j′, a contradiction with αβ=0. So, ∣supp(β)∣ must be at least 5.
3. (3)
Let n=5. Then by Theorem 7.2, 15≥∣supp(α)supp(β)∣≥9. Since 15−9=6, there is an (i,j)∈A such that higj=hi′gj′ for all (i′,j′)∈A where i=i′ and j=j′, a contradiction with αβ=0. So, ∣supp(β)∣ must be at least 6.
4. (4)
Let n=6. Then by Theorem 7.2, 18≥∣supp(α)supp(β)∣≥10. Since 18−10=8, there is an (i,j)∈A such that higj=hi′gj′ for all (i′,j′)∈A where i=i′ and j=j′, a contradiction with αβ=0. So, ∣supp(β)∣ must be at least 7.
5. (5)
Let n=7. Then by Theorem 7.2, 21≥∣supp(α)supp(β)∣≥11. Since 21−11=10, there is an (i,j)∈A such that higj=hi′gj′ for all (i′,j′)∈A where i=i′ and j=j′, a contradiction with αβ=0. So, ∣supp(β)∣ must be at least 8.
6. (6)
Let n=8. Then by Theorem 7.2, 24≥∣supp(α)supp(β)∣≥12. Let ∣supp(α)supp(β)∣>12. Then ∣supp(α)supp(β)∣≥13. Since 24−13=11, there is an (i,j)∈A such that higj=hi′gj′ for all (i′,j′)∈A where i=i′ and j=j′, a contradiction with αβ=0. So, ∣supp(α)supp(β)∣=12 and because αβ=0, there is a partition π of A with all sets containing two elements, such that if (i,j) and (i′,j′) belong to the same set of π, then higj=hi′gj′. Let α′=∑a∈supp(α)a and β′=∑b∈supp(β)b. So, α′,β′∈F2[G], ∣supp(α′)∣=3 and ∣supp(β′)∣=8 and with the above discussion we have α′β′=0, that is a contradiction (see Corollary 6.2). Therefore, ∣supp(α)supp(β)∣=12 and so ∣supp(β)∣ must be at least 9.
7. (7)
Let n=9. Then by Theorem 7.2, 27≥∣supp(α)supp(β)∣≥13. Let ∣supp(α)supp(β)∣>13. Then ∣supp(α)supp(β)∣≥14. Since 27−14=13, there is an (i,j)∈A such that higj=hi′gj′ for all (i′,j′)∈A where i=i′ and j=j′, a contradiction with αβ=0. So, ∣supp(α)supp(β)∣=13 and because αβ=0, there is a partition π of A with one set of size 3 and all other sets containing two elements, such that if (i,j) and (i′,j′) belong to the same set of π, then higj=hi′gj′. With the discussion above, (∑a∈supp(α)a)(∑b∈supp(β)b)x−1=1 where x=higj for some (i,j) belongs to the set of size 3 in π. Hence, there are γ,δ∈F2[G] such that γδ=1, where γ=∑a∈supp(α)a, δ=∑b∈supp(β)bx−1, ∣supp(γ)∣=3 and ∣supp(δ)∣=∣supp(β)∣=9, that is a contradiction with Theorem 7.3. Therefore, ∣supp(α)supp(β)∣=13 and so ∣supp(β)∣ must be at least 10.
With the discussion above, we have the following theorem.
Theorem 7.4**.**
Let α be a non-zero element in the group algebra of a torsion-free group over an arbitrary field such that ∣supp(α)∣=3 and αβ=0, for some non-zero element β of the group algebra. Then ∣supp(β)∣≥10.
Proposition 7.5**.**
If F[G] has no non-zero element γ with ∣supp(γ)∣≤k such that γ2=0, then there exist no non-zero elements γ1,γ2∈F[G] such that γ1γ2=0 and ∣supp(γ1)∣∣supp(γ2)∣≤k.
Proof.
Suppose, for a contradiction, that γ1,γ2∈F[G]∖{0} such that γ1γ2=0 and ∣supp(γ1)∣∣supp(γ2)∣≤k. We may assume that 1∈supp(γ1)∩supp(γ2), since (a−1γ1)(γ2b−1)=0 for any a∈supp(γ1) and b∈supp(γ2).
Suppose, for a contradiction, that γ2xγ1=0 for all x∈G. Then it follows from
[18, Lemma 1.3] that θ(γ2)θ(γ1)=0, where θ is the projection θ:F[G]→F[Δ] given by
β=∑x∈Gfxx↦θ(β)=∑x∈Δfxx, where Δ is the subgroup of all
elements of G having a finite number of conjugates in G (see [18, p. 3]). Now it follows from [18, Lemma 2.2] and [18, Lemma 2.4] that θ(γ1)=0 or θ(γ2)=0, which are both contradiction since 1∈supp(γ1)∩supp(γ2).
Therefore, there exists an element x∈G such that β=γ2xγ1=0. Now
[TABLE]
and
[TABLE]
which is a contradiction. This completes the proof.
∎
8. Possible units with supports of size 3 in F[G]
It is known that a group algebra over any torsion-free group does not contain a unit element whose support is of size at most 2 (see [6, Theorem 4.2]), but it is not known a similar result for group algebra elements with the supports of size 3.
Throughout this section let γ be an element in the group algebra of a torsion-free group G over a field F such that ∣supp(γ)∣=3 and γδ=1 for some element δ of the group algebra. In this section, we show that ∣supp(δ)∣ must be at least 9. Here, one may assume that δ has minimum possible support size among all elements α with γα=1. Therefore by Remark 3.4, 1∈supp(γ) and G=⟨supp(γ)⟩. Let supp(γ)={h1,h2,h3}, supp(δ)={g1,g2,…,gn} and n=∣supp(δ)∣.
Suppose that A={1,2,3}×{1,2,…,n}. Since γδ=1, there must be at least one (i,j)∈A such that higj=1. By renumbering, we may assume that (i,j)=(1,1). Replacing γ by h1−1γ and δ by δg1−1 we may assume that h1=g1=1.
There is a partition π of A such that (i,j) and (i′,j′) belong to the same set of π if and only if higj=hi′gj′ and because of the relation γδ=1, for all E∈π we have
[TABLE]
Let E1 be the set in π which contains (1,1).
8.1. The support of δ is of size at least 4
By Theorem 7.1, 3n≥∣supp(γ)supp(δ)∣≥∣supp(γ)∣+∣supp(δ)∣−1 because G is torsion-free. Let n=3. Then with the above discussion, 9≥∣supp(γ)supp(δ)∣≥5 and so, there are at least 4 sets different from E1 in π, namely E2,E3,E4,E5. Since γδ=1, each of such sets must have at least two elements such that ∑(i,j)∈Ekγiδj=0 for all k∈{2,3,4,5}, but since 9−5=4, ∣E1∣=1 and ∣Ek∣=2 for all k∈{2,3,4,5}. Therefore E1={(1,1)} and for all k∈{2,3,4,5}, Ek={(i,j),(i′,j′)} where higj=hi′gj′ for some (i,j),(i′,j′)∈A such that i=i′ and j=j′. Let γ′=∑a∈supp(γ)a and δ′=∑b∈supp(δ)b. So, γ′,δ′∈F2[G], ∣supp(γ′)∣=3 and ∣supp(δ′)∣=3 and with the above discussion we have γ′δ′=1, that is a contradiction with Theorem 7.3. Therefore, n=3.
8.2. The support of δ must be of size greater than or equal to 8
Abelian torsion-free groups satisfy the Conjecture 1.2 (see [18, Theorem 26.2]). So, G must be a nonabelian torsion-free group. Therefore by Theorem 7.2, if n≥4, then ∣supp(γ)supp(δ)∣≥∣supp(γ)∣+∣supp(δ)∣+1. Also, it is easy to see that ∣supp(γ)∣∣supp(δ)∣≥∣supp(γ)supp(δ)∣. Hence, 3n≥∣supp(γ)supp(δ)∣≥4+n.
- (1)
Let n=4. Then with the above discussion, 12≥∣supp(γ)supp(δ)∣≥8 and so, there are at least 7 sets different from E1 in π, namely E2,E3,…,E8. Since γδ=1, each of such sets must have at least two elements such that ∑(i,j)∈Ekγiδj=0 for all k∈{2,3,…,8}, but since 12−8=4, ∣Ek∣≤1 for some k∈{2,3,…,8}, a contradiction. Therefore, n=4.
2. (2)
Let n=5. Then with the discussion above 15≥∣supp(γ)supp(δ)∣≥9 and so, there are at least 8 sets different from E1 in π, namely E2,E3,…,E9. Since γδ=1, each of such sets must have at least two elements such that ∑(i,j)∈Ekγiδj=0 for all k∈{2,3,…,9}, but since 15−9=6, ∣Ek∣≤1 for some k∈{2,3,…,9}, a contradiction. Therefore, n=5.
3. (3)
Let n=6. Then with the discussion above 18≥∣supp(γ)supp(δ)∣≥10 and so, there are at least 9 sets different from E1 in π, namely E2,E3,…,E10. Since γδ=1, each of such sets must have at least two elements such that ∑(i,j)∈Ekγiδj=0 for all k∈{2,3,…,10}, but since 18−10=8, ∣Ek∣≤1 for some k∈{2,3,…,10}, a contradiction. Therefore, n=6.
4. (4)
Let n=7. Then with the discussion above 21≥∣supp(γ)supp(δ)∣≥11 and so, there are at least 10 sets different from E1 in π, namely E2,E3,…,E11. Since γδ=1, each of such sets must have at least two elements such that ∑(i,j)∈Ekγiδj=0 for all k∈{2,3,…,11}, but since 21−11=10, ∣E1∣=1 and ∣Ek∣=2 for all k∈{2,3,…,11}. Therefore E1={(1,1)} and for all k∈{2,3,…,11}, Ek={(i,j),(i′,j′)} where higj=hi′gj′ for some (i,j),(i′,j′)∈A such that i=i′ and j=j′. Let γ′=∑a∈supp(γ)a and δ′=∑b∈supp(δ)b. So, γ′,δ′∈F2[G], ∣supp(γ′)∣=3 and ∣supp(δ′)∣=7 and with the above discussion we have γ′δ′=1, that is a contradiction with Theorem 7.3. Therefore, n=7.
5. (5)
Let n=8. Then with the discussion above 24≥∣supp(γ)supp(δ)∣≥12. Let ∣supp(γ)supp(δ)∣>12. Then ∣supp(γ)supp(δ)∣≥13 and so, there are at least 12 sets different from E1 in π, namely E2,E3,…,E13. Since γδ=1, each of such sets must have at least two elements such that ∑(i,j)∈Ekγiδj=0 for all k∈{2,3,…,13}, but since 24−13=11, ∣Ek∣≤1 for some k∈{2,3,…,13}, a contradiction. So, ∣supp(γ)supp(δ)∣=12. Therefore, there are 11 sets different from E1 in π, namely E2,E3,…,E12. Since γδ=1, there are two cases for the number of elements in such sets.
- (a)
∣E1∣=2 and for all k∈{2,3,…,12}, Ek={(i,j),(i′,j′)} where higj=hi′gj′ for some (i,j),(i′,j′)∈A such that i=i′ and j=j′. Let γ′=∑a∈supp(γ)a and δ′=∑b∈supp(δ)b. So, γ′,δ′∈F2[G], ∣supp(γ′)∣=3 and ∣supp(δ′)∣=8 and with the above discussion we have γ′δ′=0, that is a contradiction (see Corollary 6.2).
2. (b)
E1={(1,1)} and ∣El∣=3 for exactly one l∈{2,3,…,12} and for all k∈{2,3,…,12}∖{l}, Ek={(i,j),(i′,j′)} where higj=hi′gj′ for some (i,j),(i′,j′)∈A such that i=i′ and j=j′.
Theorem 8.1**.**
Let γ and δ be elements of the group algebra of any torsion-free group over an arbitrary field. If ∣supp(γ)∣=3 and γδ=1 then ∣supp(δ)∣≥8.
8.3. The support of δ must be of size greater than or equal to 9
In the following, we focus on the unit graph of γ and δ over F, U(γ,δ). By Proposition 3.9, U(γ,δ) is simple. Also by Theorems 3.20 and 3.30, U(γ,δ) contains no C3−C3 or K2,3 as a subgraph.
By item (5) of Subsection 8.2, if n=8, then ∣supp(γ)supp(δ)∣=12 and there are exactly 11 sets different from E1 in π, namely E2,E3,…,E12. Also, E1={(1,1)} and ∣El∣=3 for exactly one l∈{2,3,…,12}, and for all k∈{2,3,…,12}∖{l}, Ek={(i,j),(i′,j′)} such that i=i′, j=j′ and higj=hi′gj′.
Let El={(i1,j1),(i2,j2),(i3,j3)}. Therefore, hi1gj1=hi2gj2=hi3gj3 and so there is a triangle in U(γ,δ) with the vertex set {gj1,gj2,gj3} and there is no other triangle in the latter graph. Let (2,1),(3,1)∈/El. Then by the way we have chosen Ek for k∈{1,2,3,…,12}, the degree of gj1, gj2 and gj3 are equal to 4. So, there must be 6 other vertices different from gj1, gj2 and gj3 in the vertex set of U(γ,δ) because there is no other triangle in the latter graph. This gives a contradiction because the size of the vertex set of U(γ,δ) is n=8. Hence, El={(a,1),(i,j),(i′,j′)} where a∈{2,3} and {ha,hi,hi′}=supp(γ). Since ∣E1∣=1, 1=h1g1=hmgn for all (m,n)∈A∖E1. So, deg(g1)=3 and by renumbering, we may assume that U(γ,δ) has the graph H in Figure 15 as a subgraph and there is no other vertex in U(γ,δ). In H, g5∼g2 or g5∼g2.
Let g5∼g2. Since (a,2),(a,5),(a,6),(a,7),(a,8)∈/El,E1 for all a∈{1,2,3} and ∣Ek∣=2 for all k∈{2,3,…,12}∖{l}, deg(g2)=deg(g5)=deg(g6)=deg(g7)=deg(g8)=3. Since U(γ,δ) is a simple graph which contains exactly one triangle, g6∼gi for all i∈{1,3,4,5,6}. So, without loss of generality we may assume that g6∼g7 since deg(g6)=3. Suppose, for a contradiction, that g6∼g8. Then there is a subgraph isomorphic to K2,3 in U(γ,δ) with the vertex set {g3,g4,g6,g7,g8}, a contradiction. So, g6∼g2. With a same discussion we have g7∼g2 and U(γ,δ) has the graph H1 in Figure 16 as a subgraph. Since deg(g2)=deg(g5)=deg(g6)=deg(g7)=deg(g8)=3 and deg(g1)=deg(g3)=deg(g4)=4, we must have g5∼g8 with a double edge, a contradiction because U(γ,δ) is simple. Therefore, g5∼g2.
Since U(γ,δ) is a simple graph which contains exactly one triangle, g5∼gi for all i∈{1,2,3,4,5,6}. So, g5∼g7 or g5∼g8 because deg(g5)=3. As we can see in Figure 15, without loss of generality we may assume that g5∼g7 and U(γ,δ) has the graph H2 in Figure 16 as a subgraph. Since U(γ,δ) is a simple graph which contains exactly one triangle, g6∼gi for all i∈{1,3,4,5,6}. So, g6∼g2, g6∼g7 or g6∼g8 because deg(g6)=3. If g6∼g2 or g6∼g7 then there is a subgraph isomorphic to K2,3 in U(γ,δ) as we can see in the graph H2 of Figure 16, a contradiction. Therefore we must have g6∼g8 with a double edge, a contradiction because U(γ,δ) is simple.
Hence with the above discussion, n=8 and by Theorem 8.1, we have the following result.
Theorem 8.2**.**
Let γ and δ be elements of the group algebra of any torsion-free group over an arbitrary field. If ∣supp(γ)∣=3 and γδ=1 then ∣supp(δ)∣≥9.
Acknowledgements
The authors are grateful to the referee for his/her valuable suggestions and comments. The first author was supported in part by Grant No. 95050219 from School of Mathematics, Institute for Research in Fundamental Sciences (IPM). The first author was additionally financially supported by the Center of Excellence for Mathematics at the University of Isfahan.
