Abstract
Let be a group and a nonempty subset of . Then, is product-free if for all . We say is a locally maximal product-free set if is product-free and not properly contained in any other product-free set. It is natural to ask whether it is possible to determine the smallest possible size of a locally maximal product-free set in . Alternatively, given a positive integer , one can ask the following: what is the largest integer such that there is a group of order with a locally maximal product-free set of size ? The groups containing locally maximal product-free sets of sizes and are known, and it has been conjectured that . The purpose of this paper is to prove this conjecture and hence show that the list of known locally maximal product-free sets of size 3 is complete. We also report some experimental observations about the sequence .
1. Introduction
Let be a group and a nonempty subset of . Then, is product-free if for all . For example, if is a subgroup of then is a product-free set for any . Traditionally these sets have been studied in abelian groups and have therefore been called sum-free sets (see, e.g., [1, 2]). Since we are working with arbitrary groups it makes more sense to say “product-free” in this context. We say is a locally maximal product-free set if is product-free and not properly contained in any other product-free set. We use the term locally maximal rather than maximal because the majority of the literature in this area uses maximal to mean maximal by cardinality (e.g., [3, 4]).
There are some obvious questions from the definition: given a group , what is the maximum cardinality of a product-free set in and what are the maximal (by cardinality) product-free sets? How many product-free sets are there in ? Given that each product-free set is contained in a locally maximal product-free set, what are the locally maximal product-free sets? What are the possible sizes of locally maximal product-free sets? Most of the work on product-free sets has been done in the abelian group case, particularly for and . The number of sum-free sets in the integers has been studied by, for example, Cameron and Erdös [1] and Green [5], who with Ruzsa also studied the density and number of sum-free sets in abelian groups [6]. The number of sum-free sets of is . The number of sum-free sets of an arbitrary abelian group is . See [7] for further work in this direction. Petrosyan [8] determined the asymptotic behaviour of the number of product-free sets in groups of even order. Green and Ruzsa in [6] also determined the maximal size of a sum-free set in an arbitrary abelian group. For the nonabelian case Kedlaya [9] showed that there exists a constant such that the largest product-free set in a group of order is of size at least . See also [10]. Gowers in his work on quasirandom groups proved that if the smallest nontrivial representation of a group is of dimension then the largest product-free set in is of size at most (Theorem and commentary at the start of Section of [11]). Much less is known about the minimum sizes of locally maximal product-free sets. This question was first asked in [3] and later in [12], where the authors ask what is the minimum size of a locally maximal product-free set in a group of order ? A good bound for this minimum size is still not known. Small locally maximal product-free sets when is an elementary abelian 2-group are of interest in finite geometry, because they correspond to complete caps in PG(). Locally maximal sum-free sets for elementary abelian 2-groups of order up to 64 were classified in [13]. In [14], all groups containing locally maximal product-free sets of sizes and were classified. Some general results were also obtained. Furthermore, there was a classification (Theorem of [14]) of groups containing locally maximal product-free sets of size for which not every subset of size in generates . Each of these groups has order of at most . Conjecture of [14] was that if is a group of order greater than , then does not contain a locally maximal product-free set of size . A list was given of all locally maximal product-free sets in groups of orders up to . So the conjecture asserts that this list is the complete list of all such sets. (This list is given in the current paper as Table 1; we include it both for ease of reference and because information from it is needed in the proofs of our results.) The main result of this paper is the following and its immediate corollary.
Theorem 1. Suppose is a locally maximal product-free set of size 3 in a group , such that every two-element subset of generates . Then .
Corollary 2. If a group contains a locally maximal product-free set of size 3, then and the only possibilities for and are listed in Table 1.
Proof. If not every two-element subset of generates , then, by Theorem of [14], . We may therefore assume that every two-element subset of generates . Then by Theorem 1. Now Table 1 is a list of all locally maximal product-free sets of size occurring in groups of order up to (a version of this table appeared in [14] as a list of such sets in groups of order up to ). Since we have shown that all locally maximal product-free sets of size 3 occur in groups of order up to , this table now constitutes a complete list of possibilities.
More generally, given a positive integer , one can ask the following: what is the largest integer such that there is a group of order with a locally maximal product-free set of size ? Using GAP [15] we have tested all groups of order up to 100 when , and the results suggest that the sequence begins , which means the sequence begins . This is rather intriguing and it would be interesting to know what the sequence actually is.
We finish this section by establishing the notation to be used in the rest of the paper and giving some basic results from [14]. For subsets and of a group , we use the standard notation for the product of and . That is, , . By definition, a nonempty set is product-free if and only if . In order to investigate locally maximal product-free sets, we introduce some further notations. For a set , we define the following sets:For a singleton set , we usually write instead of .
For a positive integer , we will denote by the alternating group of degree , by the cyclic group of order , by the dihedral group of order , and by the dicyclic group of order given by , , .
We now state the results from [14] that we will use.
Lemma 3 (see [14, Lemma ]). Suppose is a product-free set in the group . Then is a locally maximal product-free set if and only if .
The next result lists, in order, Proposition , Theorem , Propositions , , and , and Corollary of [14].
Theorem 4. Let be a locally maximal product-free set in a group . Then (i) is normal in and is either trivial or an elementary abelian 2-group;(ii);(iii)if is not an elementary abelian 2-group and , then ;(iv)every element of has even order, and all odd powers of lie in ;(v)if there exists and integers such that , then divides ;(vi)if , then .
We require one final fact.
Theorem 5 (see [14, Theorem ]). Up to isomorphism, the only instances of locally maximal product-free sets of size 3 of a group where are given in Table 1.
2. Proof of Theorem 1
Proposition 6. Suppose is a locally maximal product-free set of size 3 in . If is cyclic, then .
Proof. Write . First note that since is abelian, ; moreover ; so . Also . Thus By Lemma 3, ; so . Elements of cyclic groups have at most two square roots. Therefore . By Table 1, must now be one of , , , , , , , or . Theorem 4(iv) tells us that every element of has even order and all odd powers of lie in . This means that, for , , , or , we have and so . In particular, .
It remains to consider , , , and . For , the unique locally maximal product-free set of size is . Now if or is contained in , then consists of powers of a single element; so, by Theorem 4(v), divides . If neither nor is in , then , and so by Theorem 4(iii), therefore, divides . In there is a unique (up to group automorphisms) locally maximal product-free set of size , and it is , where is any element of order . If contains or , then contains all odd powers of that element by Theorem 4(iv), and hence contains , a contradiction. Therefore and so divides . Next, we consider . Recall that elements of must have even order. If contains any element of order 10, then contains all five odd powers of this element, which is impossible by Theorem 4(iv). This leaves only the involution of as a possible element of . Hence again and divides . Finally we look at . If contains any element of order , then , a contradiction. If contains an element of order 6 then contains all three of its odd powers, so . But then , contradicting the assumption that . Therefore, can only contain elements of order or . Up to group automorphism, we see from Table 1 that every locally maximal product-free set of size in with is one of or for some generator of . Each of these sets contains exactly one element of order or . Therefore, in every case, and so divides . This completes the proof.
Note that the bound on in Proposition 6 is attainable. For example, in there is a locally maximal product-free set of size , with .
Proposition 7. Suppose is a locally maximal product-free set of size in such that every -element subset of generates . Then either or contains exactly one involution.
Proof. First suppose contains no involutions. If , then Theorem 4(vi) tells us that has order of at most 37, and then, by Theorem 5, is one of the possibilities listed in Table 1. In particular . If , then for some , . But then , so is cyclic. Now by Proposition 6 we get . Next, suppose that contains at least two involutions (, ), with the third element being . Then, since every 2-element subset of generates , we have that is dihedral and is a locally maximal product-free set in . Let , so . The nontrivial coset of the subgroup is product-free of size . So if lies in this coset, then we have and . If does not lie in this coset then for some , and from the relations in a dihedral group , , , and . The coset consists of involutions, which cannot lie in . Thus by Lemma 3. A straightforward calculation shows that This means , and consists of two generating involutions , plus a power of their product , with the property that any two-element subset of generates . A glance at Table 1 shows there are no locally maximal product-free sets of this form in for . Therefore the only possibility is that , with consisting of the three reflections in . By Theorem 4(i), the index of in is a power of . By Theorem 4(ii), . Thus . Suppose for contradiction that . Now , and since consists of involutions, the elements of have order 4. So contains two elements of order and three elements of order 2 and the remaining nonidentity elements have order . Then the elements of whose order is a power of 2 must lie in three Sylow -subgroups of order , with trivial pairwise intersection. Each of these groups therefore has a unique involution and elements of order , all of which square to the given involution. But no group of order has fourteen elements of order . Hence , and so . Therefore either or contains exactly one involution.
Before we establish the next result, we first make a useful observation. Suppose , where and is an involution. Then a straightforward calculation shows that
Lemma 8. Suppose is a locally maximal product-free set of size in , every -element subset of generates , and contains exactly one involution. Then either or , where , have order and is an involution.
Proof. Suppose , where is an involution and , are not. Consider . Recall that . If then which implies that either has order or is cyclic (because, e.g., if then ). Thus if then either has order 3 or (by Proposition 6) . Now suppose that . The elements of are given in (4). If then, by remembering that , we deduce that is cyclic, generated by either or . For example, implies . Similarly, if , then is cyclic. Since has order of at least 3, we cannot have . If , then would not be product-free. For instance, implies that , and hence . The only remaining possibility is , meaning that has order 3. The same argument with shows that also has order .
We can now prove Theorem 1, which states that if is a locally maximal product-free set of size 3 in a group , such that every two-element subset of generates , then .
Proof of Theorem 1. Suppose is a locally maximal product-free set of size 3 in such that every two-element subset of generates . Then, by Lemma 8, either or , where , have order and is an involution. In the latter case, we observe that is an involution and so must be contained in . Using (4) we work through the possibilities. Obviously it is impossible for to be equal to any of , or because these elements are not of order . If any of , or were involutions, then it would imply that was generated by two involutions whose product has order 3. For example, if were an involution then . That is, would be dihedral of order . But there is no product-free set in containing two elements of order 3, because if , are the elements of order 3 in then and . So the remaining possibilities for are , and . Now implies , whereas implies and implies , each of which contradicts the fact that is product-free. We are now left with the cases , , and (which, if it is an involution, equals ). If , then , but the only product-free set of size 3 in contains no elements of order 3, so this is impossible. Therefore . If , then , so , which has order 3. If , then , again of order 3. So we see that . This is a well known presentation of the alternating group . As is the only element of whose order is even, we see that , and hence . Therefore in all cases .
3. A Table of All Locally Maximal Product-Free Sets of Size 3
Though Table 1 is essentially the same as the one in [14], we have taken the opportunity here to correct a typographical error in the entry for the (unnamed) group of order . The data was obtained using simple GAP programs [15] and additionally verified by hand for the smaller groups.
Competing Interests
The authors declare that they have no competing interests.
Acknowledgments
Chimere Anabanti is supported by a Birkbeck Ph.D. Scholarship.