Groups containing locally maximal product-free sets of size 4
Every locally maximal product-free set S in a finite group G satisfies G = S ∪ SS ∪ S⁻¹S ∪ SS⁻¹ ∪ √S, where SS = {xy | x, y ∈ S}, S⁻¹S = {x⁻¹y | x, y ∈ S}, SS⁻¹ = {xy⁻¹ | x, y ∈ S} and √S = {x ∈ G | x² ∈ S}. To better understand locally maximal product-free sets, Bertram asked whether every locally...
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| Cite this: | Groups containing locally maximal product-free sets of size 4 / C.S. Anabanti // Algebra and Discrete Mathematics. — 2021. — Vol. 31, № 2. — С. 167–194. — Бібліогр.: 12 назв. — англ. |
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Anabanti, C.S. 2023-03-11T15:51:12Z 2023-03-11T15:51:12Z 2021 Groups containing locally maximal product-free sets of size 4 / C.S. Anabanti // Algebra and Discrete Mathematics. — 2021. — Vol. 31, № 2. — С. 167–194. — Бібліогр.: 12 назв. — англ. 1726-3255 DOI:10.12958/adm1347 2020 MSC: 20D60, 05E15, 11B75 https://nasplib.isofts.kiev.ua/handle/123456789/188705 Every locally maximal product-free set S in a finite group G satisfies G = S ∪ SS ∪ S⁻¹S ∪ SS⁻¹ ∪ √S, where SS = {xy | x, y ∈ S}, S⁻¹S = {x⁻¹y | x, y ∈ S}, SS⁻¹ = {xy⁻¹ | x, y ∈ S} and √S = {x ∈ G | x² ∈ S}. To better understand locally maximal product-free sets, Bertram asked whether every locally maximal product-free set S in a finite abelian group satisfy |√S| ≤ 2|S|. This question was recently answered in the negation by the current author. Here, we improve some results on the structures and sizes of finite groups in terms of their locally maximal product-free sets. A consequence of our results is the classification of abelian groups that contain locally maximal product-free sets of size 4, continuing the work of Street, Whitehead, Giudici and Hart on the classification of groups containing locally maximal product-free sets of small sizes. We also obtain partial results on arbitrary groups containing locally maximal product-free sets of size 4, and conclude with a conjecture on the size 4 problem as well as an open problem on the general case. en Інститут прикладної математики і механіки НАН України Algebra and Discrete Mathematics Groups containing locally maximal product-free sets of size 4 Article published earlier |
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Groups containing locally maximal product-free sets of size 4 |
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Every locally maximal product-free set S in a finite group G satisfies G = S ∪ SS ∪ S⁻¹S ∪ SS⁻¹ ∪ √S, where SS = {xy | x, y ∈ S}, S⁻¹S = {x⁻¹y | x, y ∈ S}, SS⁻¹ = {xy⁻¹ | x, y ∈ S} and √S = {x ∈ G | x² ∈ S}. To better understand locally maximal product-free sets, Bertram asked whether every locally maximal product-free set S in a finite abelian group satisfy |√S| ≤ 2|S|. This question was recently answered in the negation by the current author. Here, we improve some results on the structures and sizes of finite groups in terms of their locally maximal product-free sets. A consequence of our results is the classification of abelian groups that contain locally maximal product-free sets of size 4, continuing the work of Street, Whitehead, Giudici and Hart on the classification of groups containing locally maximal product-free sets of small sizes. We also obtain partial results on arbitrary groups containing locally maximal product-free sets of size 4, and conclude with a conjecture on the size 4 problem as well as an open problem on the general case.
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1726-3255 |
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Groups containing locally maximal product-free sets of size 4 / C.S. Anabanti // Algebra and Discrete Mathematics. — 2021. — Vol. 31, № 2. — С. 167–194. — Бібліогр.: 12 назв. — англ. |
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AT anabantics groupscontaininglocallymaximalproductfreesetsofsize4 |
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2025-11-26T21:44:28Z |
| last_indexed |
2025-11-26T21:44:28Z |
| _version_ |
1850773956941840384 |
| fulltext |
“adm-n2” — 2021/7/19 — 10:26 — page 167 — #3
© Algebra and Discrete Mathematics RESEARCH ARTICLE
Volume 31 (2021). Number 2, pp. 167–194
DOI:10.12958/adm1347
Groups containing locally maximal
product-free sets of size 4
C. S. Anabanti
Communicated by I. Ya. Subbotin
Abstract. Every locally maximal product-free set S in a
finite group G satisfies G = S ∪ SS ∪ S−1S ∪ SS−1 ∪
√
S, where
SS = {xy | x, y ∈ S}, S−1S = {x−1y | x, y ∈ S}, SS−1 = {xy−1 |
x, y ∈ S} and
√
S = {x ∈ G | x2 ∈ S}. To better understand
locally maximal product-free sets, Bertram asked whether every
locally maximal product-free set S in a finite abelian group satisfy
|
√
S| 6 2|S|. This question was recently answered in the negation by
the current author. Here, we improve some results on the structures
and sizes of finite groups in terms of their locally maximal product-
free sets. A consequence of our results is the classification of abelian
groups that contain locally maximal product-free sets of size 4,
continuing the work of Street, Whitehead, Giudici and Hart on the
classification of groups containing locally maximal product-free sets
of small sizes. We also obtain partial results on arbitrary groups
containing locally maximal product-free sets of size 4, and conclude
with a conjecture on the size 4 problem as well as an open problem
on the general case.
1. Introduction
Let S be a non-empty subset of a finite group G. Then S is product-
free in G if there is no solution to the equation xy = z for x, y, z ∈ S;
equivalently, if S ∩ SS = ∅, where SS = {xy | x, y ∈ S}. For a finite
group G, a locally maximal product-free set in G is a product-free subset
2020 MSC: 20D60, 05E15, 11B75.
Key words and phrases: product-free sets, locally maximal, maximal, groups.
“adm-n2” — 2021/7/19 — 10:26 — page 168 — #4
168 Groups containing LMPFS of size 4
S of G such that given any other product-free set T in G with S ⊆ T ,
then S = T . Since every product-free set in a finite group G is contained
in a locally maximal product-free set in G, we can gain information about
product-free sets in a group by studying its locally maximal product-free
sets. In connection with Group Ramsey Theory, Street and Whitehead
[11] noted that every partition of a finite group G (or in fact, of G∗ =
G \ {1}) into product-free sets can be embedded into a covering by locally
maximal product-free sets, and hence to find such partitions, it is useful to
understand locally maximal product-free sets. Among other results, they
calculated locally maximal product-free sets in groups of small orders, up
to 16 in [11,12] as well as a few higher sizes. Giudici and Hart [9] started the
classification of finite groups containing small locally maximal product-free
sets. They classified finite groups containing locally maximal product-free
sets of sizes 1 and 2, as well as some of size 3. The size 3 problem was
resolved in [5]. The reader may see [1, 4, 6, 11] for a concept ‘filled groups’
studied for locally maximal product-free sets. A locally maximal product-
free set in a group G can be characterised as a product-free subset S of G
satisfying
G = S ∪ SS ∪ S−1S ∪ SS−1 ∪
√
S, (1)
where SS = {xy | x, y ∈ S}, S−1S = {x−1y | x, y ∈ S}, SS−1 = {xy−1 |
x, y ∈ S} and
√
S = {x ∈ G | x2 ∈ S} (see [9, Lemma 3.1]). Each
(locally maximal product-free) set S in a finite group of odd order satisfies
|
√
S| = |S|. No such result is known for finite groups of even order in
general. To better understand locally maximal product-free sets (LMPFS
for short), Bertram [7, p. 41] asked the question: does every locally maximal
product-free set S in a finite abelian group satisfy |
√
S| 6 2|S|? This
question was answered in the negation in [3, pp. 2–3]. The answer shows
that we can’t rely on |
√
S| 6 2|S| to obtain a reasonable bound on the
order of an arbitrary finite abelian group of even order containing a locally
maximal product-free set S. The set S = {x2} consisting of the unique
involution in the Quaternion group Q8 is locally maximal product-free
in Q8 and satisfies |
√
S| = 6|S| = 3
4 |Q8|. This shows that relying on
|
√
S| 6 2|S| to obtain a bound on all non-abelian finite groups containing
locally maximal product-free sets will be disastrous too. We note that
√
S
cannot always be removable from equation 1 as seen in Remark 3.4(a) of
this paper, with the locally maximal product-free subset S = {x, x6} of
G = 〈x | x7 = 1〉 ∼= C7. Though not every locally maximal product-free
set is a maximal (by cardinality) product-free set, the subset S of C7 given
here is clearly a maximal by cardinality product-free set. In particular,
“adm-n2” — 2021/7/19 — 10:26 — page 169 — #5
C. S. Anabanti 169
|S ∪ SS ∪ SS−1 ∪ S−1S| = |S ∪ SS| = 5 and |
√
S| = 2. Unfortunately,
this example shows that the proof of Theorem 3 of [10] is not correct, as
the author assumed that every element of a finite group G which is not an
element of a maximal product-free subset S of G is either an element of
SS, SS−1 or S−1S. The sizes of SS, SS−1 and S−1S were optimised in
[10, Theorem 3] that it is difficult to get a counter example of the theorem
itself, even though the absence of
√
S made the proof wrong. We devote
this paper to obtaining structures and sizes of finite groups in terms of
their locally maximal product-free sets S, without necessarily relying on
|
√
S| 6 2|S|. Throughout this discussion, G is an arbitrary finite group,
except where otherwise stated.
2. Preliminaries
Let S and V be subsets of G. We define the following:
SV = {sv | s ∈ S, v ∈ V }; S−1 = {s−1 | s ∈ S};
T (S) = S ∪ SS ∪ SS−1 ∪ S−1S;
√
S = {x ∈ G | x2 ∈ S}.
Lemma 2.1 ([9, Lemma 3.1]). Let S be a product-free set in a group G.
Then S is locally maximal product-free if and only if G = T (S) ∪
√
S.
Proposition 2.2 ([9, Proposition 3.2]). Let S be a LMPFS in G. Then
〈S〉 is a normal subgroup of G. Furthermore, G/〈S〉 is either trivial or an
elementary abelian 2-group.
Theorem 2.3 ([9, Theorem 3.4]). If S is a LMPFS in G, then |G| 6
2|T (S)| · |〈S〉|.
Notation 2.4. Let S ⊆ G. We define Ŝ := {s ∈ S |
√
{s} 6⊂ 〈S〉}.
Proposition 2.5 ([9, Proposition 3.6]). Suppose S is a LMPFS in G and
that 〈S〉 is not an elementary abelian 2-group. If |Ŝ| = 1, then |G| = 2|〈S〉|.
Proposition 2.6 ([9, Proposition 3.7]). Suppose S is locally maximal
product-free in G. Then every element s of Ŝ has even order. Moreover
all odd powers of s lie in S.
Proposition 2.7 ([9, Proposition 3.8]). Let S be a LMPFS in G. If there
exists s ∈ S and integers m1, . . . ,mt such that Ŝ = {s, sm1 , . . . , smt}, then
|G| divides 4|〈S〉|.
“adm-n2” — 2021/7/19 — 10:26 — page 170 — #6
170 Groups containing LMPFS of size 4
Lemma 2.8 ([9, Lemma 3.9]). Suppose S is a locally maximal product-free
set in a group G. If S ∩ S−1 = ∅, then G = T (S) ∪ T (S)−1.
Corollary 2.9 ([9, Corollary 3.10]). If S is a LMPFS in G such that
S ∩ S−1 = ∅, then |G| 6 4|S|2 + 1.
We write D2n = 〈x, y | xn = y2 = (xy)2 = 1〉 for the finite dihedral
group of order 2n.
Lemma 2.10 ([6, Lemma 3.10]). There is no locally maximal product-free
set of size 4 consisting of at most one involution in a finite dihedral group.
Theorem 2.11 ([6, Theorem 3.11]). Suppose S is a LMPFS of size 4 in
a finite dihedral group G. Then up to automorphisms of G, the possible
choices are given as follows:
|G| S
8 {y, xy, x2y, x3y}, {x, x3, y, x2y}
10 {x2, x3, y, x4y}
12 {x3, y, xy, x2y}, {x2, x3, y, x5y}, {x, x5, y, x3y}, {x, x4, y, x3y}
14 {x2, x3, y, x6y}
16 {x2, x3, y, x7y}, {x, x6, y, x4y}
18 {x2, x5, y, x8y}
20 {x, x8, y, x5y}
3. Main results
Let S be a locally maximal product-free set in a finite group G. If
the exponent of G is 2, then S−1S = SS = SS−1 and
√
S = ∅; so
G = S ∪ SS. If the exponent of G is 3, then for x ∈
√
S, we have that
x2 ∈ S, so x4 = x ∈ SS, and we conclude that
√
S ⊆ SS. In the light of
Equation (1) therefore G = S ∪ SS ∪ S−1S ∪ SS−1; whence
|G| = |S ∪ SS ∪ SS−1 ∪ S−1S| 6 3|S|2 − |S|+ 1
since |SS| 6 |S|2 and |SS−1∪S−1S| 6 2|S|2−2|S|+1. Now, suppose the
exponent of G is 4. If S ∩ S−1 = ∅, then S consists of elements of order 4
only, and as the square roots of elements of order 4 have order 8, we have
that
√
S = ∅; thus G = S∪SS∪S−1S∪SS−1. Again, |G| 6 3|S|2−|S|+1.
We begin this study by examining locally maximal product-free sets S
with the property that S ∩ S−1 = ∅. Lemma 2.8 is that if S is a locally
maximal product-free set in a finite group G such that S ∩ S−1 = ∅,
then G = S ∪ SS ∪ S−1S ∪ SS−1 ∪ S−1 ∪ (SS)−1. Corollary 2.9 is that
“adm-n2” — 2021/7/19 — 10:26 — page 171 — #7
C. S. Anabanti 171
if S is a locally maximal product-free set in a finite group G such that
S ∩ S−1 = ∅, then |G| 6 4|S|2 + 1. The first result here (Theorem 3.2
below) improves Lemma 2.8 and Corollary 2.9. For finite groups of odd
order, Proposition 3.3 below gives a much tighter and broader result for
Corollary 2.9.
Notation 3.1. For a subset S of a group G, we write I(S) for the set of
all involutions in S.
Theorem 3.2. Suppose S is a LMPFS in a finite group G such that
S ∩ S−1 = ∅. Then G = SS ∪ S−1S ∪ SS−1 ∪ (SS)−1. Moreover, |G| 6
4|S|2 − 2|S| − |I(G)|+ 1.
Proof. First, S−1 ∩ SS−1 = ∅; for if x−1 = yz−1 where x, y, z ∈ S, then
z = xy, a contradiction. Similarly,S−1∩S−1S = ∅. In the light of Equation
(1) therefore S−1 ⊆ SS ∪
√
S. Let x−1 ∈ S−1 be arbitrary. Suppose
x−1 ∈
√
S. Then x−2 ∈ S. As x ∈ S, we have that (x)(x−2) = x−1 ∈ SS.
So, S−1 ⊆ SS; whence S ⊆ (SS)−1, and G = SS∪S−1S∪SS−1∪ (SS)−1
follows from Lemma 2.8. Clearly, |G| 6 4|S|2 − 2|S|+ 1. However, each
involution used in obtaining the bound of |G| given as “ |G| 6 4|S|2 −
2|S|+1" is counted at least twice. Now, x ∈ SS if and only if x ∈ (SS)−1;
so all such involutions are counted twice. If x ∈ S−1S, then x = y−1z for
some y, z ∈ S. Thus, x = z−1y as well. So x is counted at least twice in
S−1S. The same applies to x ∈ SS−1. By removing the second count on
involutions of G, we obtain that |G| 6 4|S|2 − 2|S| − |I(G)|+ 1.
Proposition 3.3. If S is a locally maximal product-free set of size k in
a finite group G of odd order and 0 6 |S ∩ S−1| = l 6 k, then |G| 6
3k2 − l(2k − 1) + 1.
Proof. Suppose S is a LMPFS of size k in a finite group G of odd order
and 0 6 |S ∩ S−1| = l 6 k. As each element of S has a unique square
root, we have that |
√
S| = k. If l = 0, then |SS−1 ∪ S−1S| 6 2k2 − 2k+ 1
and |SS| 6 k2; so the result follows from Equation (1). Now, suppose
1 6 |S ∩ S−1| = l 6 k. Let S = {x1, · · · , xl, xl+1, · · · , xk} and S−1 =
{x1, · · · , xl, x−1
l+1, · · · , x−1
k }. Then |SS| 6 k2 − l + 1.
(SS ∪ SS−1) \ SS ⊆ x1{x−1
l+1, · · · , x−1
k } ∪ x2{x−1
l+1, · · · , x−1
k }
∪ · · · ∪ xl{x−1
l+1, · · · , x−1
k } ∪ xl+1{x−1
l+2, x
−1
l+3, · · · , x−1
k }
∪ xl+2{x−1
l+1, x
−1
l+3, x
−1
l+4, · · · , x−1
k }
∪ · · · ∪ xk{x−1
l+1, x
−1
l+2, · · · , x−1
k−1}.
“adm-n2” — 2021/7/19 — 10:26 — page 172 — #8
172 Groups containing LMPFS of size 4
Therefore
|(SS ∪ SS−1) \ SS| 6 l(k − l) + (k − l)(k − l − 1) = (k − l)(k − 1). (2)
Similarly,
|(SS ∪ S−1S) \ SS| 6 (k − l)(k − 1).
By Equation (1) therefore |G| 6 3k2 − l(2k − 1) + 1.
Remark 3.4. (a) The bound of |G| in Proposition 3.3 is tight as it is
attained with (k, l) = (2, 2) as S = {x, x−1} ⊂ G = 〈x | x7 = 1〉 ∼= C7
meets it. (b) Theorem 3.2 and Proposition 3.3 point at the need for a
bound on the sizes of finite groups of even order containing locally maximal
product-free sets S satisfying S∩S−1 6= ∅. Such universal bound is hard to
obtain (see Question 3.29 at the end of the paper) when G is an arbitrary
finite group of even order due to the difficulty in bounding |
√
S| as we saw
an example of a locally maximal product-free subset S of Q8 satisfying
|
√
S| = 3
4 |Q8|; an interested reader may see [2, Proposition 2.1] for a
family of groups G for which there exists a locally maximal product-free
set T such that |
√
T | is comparable to |G|. However, in the finite abelian
case, progress is possible.
In the light of Equation (1), a finite abelian group G containing a locally
maximal product-free set S can be characterised as:
G = S ∪ SS ∪ SS−1 ∪
√
S. (3)
If |G| is odd, then as each element of G has exactly one square root,
|
√
S| = |S|. Using
|SS| 6 |S|(|S|+ 1)
2
and |SS−1| 6 |S|2 − |S|+ 1 (4)
in equation (3), we obtain that
|G| 6 3|S|2 + 3|S|+ 2
2
. (5)
On the other hand, if |G| is even, then using (4) together with |
√
S| 6 |G|
2
yield
|G| 6 3|S|2 + |S|+ 2. (6)
We note here that |
√
S| 6 |G|
2 follows from the fact that
√
S is product-
free in a finite abelian group G whenever S is product-free, and that a
product-free set in G has size at most |G|
2 .
“adm-n2” — 2021/7/19 — 10:26 — page 173 — #9
C. S. Anabanti 173
Observation 3.5. Let S be a set of size k in a finite abelian group G
such that 1 6 |S ∩ S−1| = l 6 k. We define A(l) to be a non-negative
integer which is less than or equal to the maximal number of identity 1’s
in {x1x1, · · · , x1xk} ∪ {x2x2, · · · , x2xk} ∪ · · · ∪ {xk−1xk−1, xk−1xk} ∪
{xkxk}. So
|SS| 6 k(k + 1)
2
−A(l) + 1. (7)
Suppose |G| is even. Then
A(l) :=
{
l+1
2 if l is odd;
l
2 if l is even.
Whence
|SS| 6 k(k + 1)− (l − 1)
2
or
k(k + 1)− (l − 2)
2
(8)
according to whether l is odd or even. Now, suppose |G| is odd. Each
element of G has a unique inverse; so l > 2 and even. Thus, A(l) = l
2 ,
and we conclude from inequality (7) that
|SS| 6 k(k + 1)− l + 2
2
.
Inequality (7) can be used to obtain a better bound for |G| when S∩S−1 6= ∅.
For instance, if S is a locally maximal product-free set of size k in a
finite abelian group G of odd order such that 1 6 |S ∩ S−1| = l 6 k,
then |G| 6 3k2−l(2k−1)+(3k+2)
2 , and this result improves inequality (5) for
S ∩ S−1 6= ∅. Lemma 3.6 below gives a bound on |G| when S ∩ S−1 = ∅.
Lemma 3.6. Suppose S is a locally maximal product-free set in a finite
abelian group G such that S ∩ S−1 = ∅. Then G = SS ∪ SS−1 ∪ (SS)−1.
Moreover, |G| 6 2|S|2 − |I(G)|+ 1. Also, if I(G) ⊆ I(SS−1), then |G| 6
2|S|2 − 2|I(G)|+ 1.
Proof. As SS−1 = S−1S, the part “G = SS∪SS−1∪(SS)−1" follows from
G = SS ∪ SS−1 ∪ S−1S ∪ (SS)−1 in Theorem 3.2. As |SS| = |(SS)−1| 6
|S|(|S|+1)
2 and |SS−1| 6 |S|2 − |S|+ 1, we conclude that |G| 6 2|S|2 + 1.
However, each involution that gives rise to the bound on |G| is counted at
least twice. Suppose x is an involution of G. Now, x ∈ SS if and only if
x ∈ (SS)−1; so all such involutions are counted twice. If x ∈ SS−1, then
x = yz−1 for some y, z ∈ S. Thus, x = zy−1 as well. So x is counted at
least twice in SS−1. By removing the second count on involutions of G,
“adm-n2” — 2021/7/19 — 10:26 — page 174 — #10
174 Groups containing LMPFS of size 4
we obtain that |G| 6 2|S|2 − |I(G)|+ 1. Now, suppose I(G) ⊆ I(SS−1).
Let x ∈ SS−1 be an involution. Then there exist y, z ∈ S (for y 6= z) such
that x = yz−1. Now, 1 = x2 = y2z−2; so y2 = z2. This shows that given
each involution yz−1 ∈ SS−1, we can find two elements y2, z2 ∈ SS such
that y2 = z2. Thus, we can find a pair of repeated elements (y2, z2) for
y2, z2 ∈ SS such that y2 = z2. By discarding one element in each pair of
the repeated elements of SS, we obtain that |G| 6 2|S|2−2|I(G)|+1.
The first bound on |G| in Lemma 3.6 is tight as S = {x, y3, x3y} ⊂
G = 〈x, y | x4 = 1 = y4, xy = yx〉 ∼= C4 × C4 meets it.
Remark 3.7. An observation of [3, p. 2] is that if a finite abelian group
of order less than or equal to 52 contains a LMPFS S of size 6 or less, then
|
√
S| 6 2|S|. This observation, together with Equation (3), Inequalities
(2), (5) and (8) and Lemma 3.6 imply that if a finite abelian group G
contains a locally maximal product-free set of size 4, then |G| 6 32. The
locally maximal product-free sets of size 4 in abelian groups of order up to
32 were checked in GAP [8]. We also used the SmallGroup library in [8]
to restrict our search to the abelian groups. If at least two LMPFS of size
4 are found in a certain abelian group G, we check whether there is an
automorphism of G that takes one to another, and if there is, we display
only one such set. In other words, we display only one locally maximal
product-free set in each orbit of the action of the automorphism groups
of G. Our computational results are summarised in Table 1 below. For
notations in Table 1, n4 is the number of locally maximal product-free
sets of size 4 in G while M4 shows the corresponding sizes of each orbit
of the displayed locally maximal product-free sets under the action of
automorphism groups of G. We take this opportunity to correct a mistake
of [11, Table 1], where the authors mentioned that every locally maximal
product-free set of size 4 in C14 is mapped by an automorphism of C14 to
either S1 = {x2, x5, x9, x13} or S2 = {x4, x5, x6, x7}. This is not true as
Table 1 shows that there are five LMPFS up to automorphisms of C14. In
particular, as Aut(Cn) is isomorphic to the unit group C×
n whose order is
φ(n) (where φ(n) is Euler’s totient function), the automorphisms of C14
are maps Φi : x 7→ xi for x ∈ C14 and i ∈ {1, 3, 5, 9, 11, 13}. Therefore,
the LMPFS {x, x3, x8, x13} and {x, x4, x7, x12} are mapped by Φ9 and Φ5
respectively into S1 and S2, and the other three product-free sets in Table 1
(viz. {x, x3, x8, x10}, {x, x4, x6, x13} and {x, x6, x8, x13}) are mapped to
neither S1 nor S2.
“adm-n2” — 2021/7/19 — 10:26 — page 175 — #11
C. S. Anabanti 175
Table 1. Finite abelian groups containing LMPFS of size 4.
G n4 LMPFS S of size 4 in G M4
C8 = 〈x | x8 = 1〉 1 {x, x3, x5, x7} 1
C4 × C2 = 〈x1, x2 | x4
1 = 1 = x2
2,
x1x2 = x2x1〉 3
{x1, x
3
1, x2, x
2
1x2},
{x1, x
3
1, x1x2, x
3
1x2}
2,
1
C3
2 = 〈x1, x2, x3 | x2
i = 1,
xixj = xjxi for 1 6 i, j 6 3〉 7 {x1, x2, x3, x1x2x3} 7
C10 = 〈x | x10 = 1〉 2 {x, x4, x6, x9} 2
C11 = 〈x | x11 = 1〉 5 {x, x3, x8, x10} 5
C12 = 〈x | x12 = 1〉 9
{x, x4, x6, x11},
{x, x4, x7, x10},
{x2, x3, x8, x9},
{x2, x3, x9, x10}
4,
2,
2,
1
C6 × C2 = 〈x1, x2 | x6
1 = 1 = x2
2,
x1x2 = x2x1〉 11
{x1, x
5
1, x2, x
3
1x2},
{x1, x2, x
4
1, x
3
1x2},
{x1, x1x2, x
4
1, x
4
1x2}
3,
6,
2
C13 = 〈x | x13 = 1〉 21
{x, x3, x5, x12},
{x, x3, x10, x12},
{x, x5, x8, x12}
12,
6,
3
C14 = 〈x | x14 = 1〉 27
{x, x3, x8, x10},
{x, x3, x8, x13},
{x, x4, x6, x13},
{x, x4, x7, x12},
{x, x6, x8, x13}
6,
6,
6,
6,
3
C15 = 〈x | x15 = 1〉 16
{x, x3, x5, x7},
{x, x3, x7, x12}
8,
8
C16 = 〈x | x16 = 1〉 37
{x, x3, x10, x12},
{x, x4, x6, x9},
{x, x4, x6, x15},
{x, x4, x9, x14},
{x, x6, x9, x14},
{x, x6, x10, x14},
{x2, x6, x10, x14}
8,
4,
8,
4,
4,
8,
1
C4 × C4 = 〈x1, x2 | x4
1 = 1 = x4
2,
x1x2 = x2x1〉
6 {x1, x
3
1, x
2
2, x
2
1x
2
2} 6
“adm-n2” — 2021/7/19 — 10:26 — page 176 — #12
176 Groups containing LMPFS of size 4
Table 1, continued
G n4 LMPFS S of size 4 in G M4
C8 × C2 = 〈x1, x2 | x8
1 = 1 = x2
2,
x1x2 = x2x1〉
42
{x1, x2, x
6
1, x
5
1x2},
{x1, x2, x
6
1, x
4
1x2},
{x1, x2, x
2
1x2, x
5
1x2},
{x1, x1x2, x
6
1, x
6
1x2},
{x1, x
6
1, x
2
1x2, x
6
1x2},
{x2, x
2
1, x
6
1, x
4
1x2},
{x2
1, x
6
1, x
2
1x2, x
6
1x2}
8,
8,
8,
8,
8,
1,
1
C4 × C2 × C2 = 〈x1, x2, x3 |
x4
1 = 1 = x2
2, x
2
3 = 1,
xixj = xjxi for 1 6 i, j 6 3〉
4 {x2, x3, x
2
1, x
2
1x2x3} 4
C17 = 〈x | x17 = 1〉 48
{x, x3, x8, x13},
{x, x3, x8, x14},
{x, x3, x11, x13}
16,
16,
16
C18 = 〈x | x18 = 1〉 54
{x, x3, x5, x12},
{x, x3, x8, x14},
{x, x3, x9, x14},
{x, x3, x12, x14},
{x, x4, x9, x16},
{x, x4, x10, x17},
{x, x5, x8, x12},
{x, x5, x8, x17},
{x, x6, x9, x16}
6,
6,
6,
6,
6,
6,
6,
6,
6
C6 × C3 = 〈x1, x2 | x6
1 = 1 = x3
2,
x1x2 = x2x1〉
48
{x1, x
5
1, x2, x
3
1x2},
{x1, x2, x
5
1x
2
2, x
3
1}
24,
24
C19 = 〈x | x19 = 1〉 36
{x, x3, x5, x13},
{x, x4, x6, x9}
18,
18
C10×C2 = 〈x1, x2 | x10
1 = 1 = x2
2,
x1x2 = x2x1〉
28
{x1, x2, x
5
1x2, x
8
1},
{x1, x1x2, x
4
1, x
8
1x2},
{x1, x1x2, x
8
1, x
6
1x2}
12,
12,
4
C20 = 〈x | x20 = 1〉 36
{x, x3, x10, x16},
{x, x3, x14, x16},
{x, x4, x11, x18},
{x, x5, x14, x18},
{x, x6, x8, x11},
{x2, x5, x15, x16}
8,
8,
4,
8,
4,
4
“adm-n2” — 2021/7/19 — 10:26 — page 177 — #13
C. S. Anabanti 177
Table 1, continued
G n4 LMPFS S of size 4 in G M4
C21 = 〈x | x21 = 1〉 34
{x, x3, x5, x15},
{x, x4, x10, x17},
{x, x4, x14, x16},
{x, x8, x12, x18}
12,
12,
4,
6
C22 = 〈x | x22 = 1〉 10 {x, x4, x10, x17} 10
C24 = 〈x | x24 = 1〉 4 {x, x6, x17, x21} 4
C9 × C3 = 〈x1, x2 | x9
1 = 1 = x2
3,
x1x2 = x2x1〉
36
{x1, x1x2, x
3
1, x
7
1x
2
2},
{x1, x1x2, x
6
1, x
7
1x
2
2}
18,
18
C3
3 = 〈x1, x2, x3 | x3
i = 1,
xixj = xjxi for 1 6 i, j 6 3〉 468 {x1, x2, x3, x
2
1x
2
2x
2
3} 468
Suppose a finite nonabelian group G contains a locally maximal
product-free set S of size 4. If |G| is odd, then Proposition 3.3 tells
us that |G| 6 49. Now, suppose |G| is even. If S ∩ S−1 = ∅, then The-
orem 3.2 tells us that |G| 6 56. So the only case left is to bound the
size of a finite group G of even order which contains a locally maximal
product-free set of size 4 such that |S ∩ S−1| > 1. We use GAP [8] to
check all locally maximal product-free sets of size 4 in nonabelian groups
of order at most 56. The result shows that 45 nonabelian groups contains
locally maximal product-free sets of size 4, and over 15% of them are
dihedral groups. More importantly, the largest size of such group is 40.
We shall study a special case for the generating set of the locally maximal
product-free sets as we aim to prove the following:
Theorem 3.8. If G is a finite group containing a locally maximal product-
free set of size 4 such that every 2-element subset of S generates 〈S〉, then
|G| 6 40.
We develop preliminary results that we shall put together to prove
Theorem 3.8. In particular, we aim to prove Theorem 3.8 by considering
each of the following three cases: (a) S contains at least two involutions;
(b) S contains no involution; (c) S contains only one involution.
Before we proceed, we state the following result (Lemma 3.9 below)
for a finite group G which we shall employ whenever necessary, without
necessarily quoting the result.
Lemma 3.9. If S is a LMPFS in a group G, then S is locally maximal
product-free in 〈S〉.
“adm-n2” — 2021/7/19 — 10:26 — page 178 — #14
178 Groups containing LMPFS of size 4
Proof. Suppose S is a LMPFS in a finite group G. To show that S is
locally maximal product-free in 〈S〉 suffices to show that S is product-free
in 〈S〉 and T (S) ∪ {g ∈ 〈S〉 : g2 ∈ S} = 〈S〉. The first is clear since
〈S〉 ⊆ G and S is product-free in G. For the latter, we first recall that
T (S) ⊆ 〈S〉. Let g ∈ 〈S〉 \ T (S) be arbitrary. As g ∈ G and S is locally
maximal product-free in G, we have that g2 ∈ S. Therefore for all g ∈ 〈S〉,
either g ∈ T (S) or g2 ∈ S; whence S is locally maximal product-free in
〈S〉.
As a consequence to Theorem 2.11, we give the following result:
Corollary 3.10. No finite group contains a LMPFS S of size 4 such that
every two element subset of S generates 〈S〉, and S contains at least two
involutions.
Proof. Suppose a finite group G contains a LMPFS S of size 4 such that
every two element subset of S generates 〈S〉, and S contains at least two
involutions. Then 〈S〉 is dihedral. In the light of Lemma 3.9 and Theorem
2.11, 〈S〉 is one of D8, D10, D12, D14, D16, D18 or D20. Suppose 〈S〉 = D8.
Then 〈S〉 = 〈y, x2y〉 ∼= C2 × C2; a contradiction. Suppose 〈S〉 is one of
D10, D14, D16, D18 or D20. Then S contains two rotations; so the group
generated by S is also the group generated by such two rotations, which
is abelian (in particular, not dihedral); a contradiction. Finally, suppose
〈S〉 = D12. If S = {x3, y, xy, x2y}, then 〈S〉 = 〈x3, y〉 ∼= C2 × C2; a
contradiction. If S is any of the other three locally maximal product-free
sets in D12, then the group generated by any two rotations in such S
is not dihedral; a contradiction. Therefore, no such G (respectively S)
exists.
Before we proceed, we give the following result.
Table 2. Nonabelian groups (of order up to 40) that contain a LMPFS of
size 4.
G n4 LMPFS S of size 4 in G M4
D8 = 〈x, y | x4 = 1 = y2,
xy = yx−1〉
3 {y, xy, x2y, x3y},
{x, x3, y, x2y}
1,
2
Q8 = 〈x, y | x4 = 1, x2 = y2,
xy = yx−1〉
3 {x, x3, y, x2y} 3
D10 = 〈x, y | x5 = 1 = y2,
xy = yx−1〉
10 {x2, x3, y, x4y} 10
Q12 = 〈x, y | x6 = 1, x3 = y2,
xy = yx−1〉
9 {x, x5, y, x3y},
{x, x4, y, x3y}
3,
6
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C. S. Anabanti 179
Table 2, continued
G n4 LMPFS S of size 4 in G M4
A4 =
〈x, y | x3 = y2 = (xy)3 = 1〉
2 {x, yx, x2yx, xy} 2
D12 = 〈x, y | x6 = 1 = y2, xy =
yx−1〉
27 {x3, y, xy, x5y},
{x2, x3, y, x5y},
{x, x5, y, x3y},
{x, x4, y, x3y}
6,
12,
3,
6
D14 = 〈x, y | x7 = 1 = y2, xy =
yx−1〉
42 {x2, x3, y, x6y} 42
(C4 × C2)⋊α C2 = 〈x, y | x4 =
y2 = (xyx)2 = (yx−1)4 =
(yxyx−1)2 = 1〉
2 {x2, y, (xy)2, x3yx} 2
C4 ⋊ C4 = 〈x, y | x4 = y4 =
x−1yxy = x2y−1x2y =
(y−1x−2y−1)2 = 1〉
2 {x2, y, x3yx, x2y2} 2
M16 = C8 ⋊ C2 = 〈x, y | x8 =
1 = y2, xy = yx5〉
26 {x, x4, x7, y},
{x, x6, y, x4y},
{x2, x6, y, x4y},
{x, x6, x2y, x6y},
{x2, x6, x2y, x6y}
8,
8,
1,
8,
1
D16 = 〈x, y | x8 = y2 = (xy)2 =
1〉
48 {x2, x3, y, x7y},
{x, x6, y, x4y}
32,
16
QD16 = 〈x, y | x8 = 1 =
y2, xy = yx3〉
16 {x, x6, y, x4y},
{x, x6, x3y, x7y}
8,
8
Q16 = 〈x, y | x8 = 1, x4 =
y2, xy = yx−1〉
16 {x, x6, y, x4y} 16
D8 ∗ C4 = 〈x, y, z |
x2 = y2 = z4 = 1 =
z−1xzx = z−1yzy = yz2xyx〉
8 {x, y, xyz, (xy)2} 8
D18 = 〈x, y | x9 = 1 = y2, xy =
yx−1〉
54 {x2, x5, y, x8y} 54
C3 × S3 = 〈x, y, z | x2 = y3 =
z3 = (xz)2 = y−1xyx =
z−1y−1zy = 1〉
72 {x, y, xz, yzx},
{x, y, z, xyz},
{x, y, z, yzx},
{x, y, z, xy2zx},
{x, y, xyz, yzx},
{x, z, xyz, xy2z},
{x, z, yzx, y2zx},
{y, xy, z, xy2zx}
6,
12,
12,
12,
6,
6,
6,
12
(C3 × C3)⋊ C2 = 〈x, y, z |
x2 = y3 = z3 = (xy)2 = (xz)2 =
z−1y−1zy = 1〉
144 {y, x, xz, xyxzx},
{y, x, xz, yzx}
72,
72
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180 Groups containing LMPFS of size 4
Table 2, continued
G n4 LMPFS S of size 4 in G M4
Q20 = 〈x, y | x10 = 1, x5 = y2,
xy = yx−1〉
20 {x, x8, y, x5y} 20
Suz(2) = 〈x, y | x5 = 1 =
y4, xy = yx2〉 (the only
non-simple Suzuki group).
40 {x, y2, x2y, xy3},
{x3, y2, x2y, xy3}
20,
20
D20 = 〈x, y | x10 = 1 = y2, xy =
yx−1〉
20 {x, x8, y, x5y} 20
C7 ⋊ C3 = 〈x, y | x3 = y7 =
y−1x−1yxy−1 = 1〉
280 {x, y, (xy)2x, xyx},
{x, y, (xy)2x, x2yx2},
{x, y, (yx)2, x(xy)2},
{x, y, xyx, x(xy)2},
{x, y, x2yx2, (xy)2},
{x, y, yx2y, (xy)2},
{x, x2y, x(xy)2x, yx},
{x, x2y, xyx, yx2},
{x2, y, (xy)2x, xyx}
42,
21,
42,
42,
42,
42,
14,
14,
21
C3 ⋊ C8 = 〈a, x | x8 = a3 =
x−1axa = 1〉
39 {x, a, x6, x4a},
{x, xa, x6, x4ax},
{a, x2, x6, x4a},
{x2, x6, x4a, x3ax}
24,
12,
2,
1
SL(2, 3) = 〈x, y | x3 = y4
= 1 = y−1xyxy−1x =
x−1y−1(x−1y)2 = (x−1y−1)3〉
72 {x, x2y, y3, yx2},
{x, x2y, y3, x(xy)2y},
{x, y, x2y3, x2yx},
{x2y, y, x2yx, xy2}
24,
24,
12,
12
Q24 = 〈x, y | x12 = 1, x6 = y2,
xy = yx−1〉
4 {x, x6, x8, x11} 4
Q12 × C2 = 〈x, y, a | x4 = y2 =
a3 = x−1axa = yx−1yx =
a−1yay = 1〉
6 {y, a, x2, x2ya},
{y, x2, x2ya, xyax}
4,
2
B(2, 3) = 〈x, y | x3 = y3 = 1
= (x−1y−1)3 = (y−1x)3〉
252 {y, x, (xy)2, x2y2xy},
{y, x, (xy)2y, x2y2x}
144,
108
C9 ⋊ C3 = 〈x, y | x9 = 1 = y3,
xy = yx4〉
144 {x6, x8, y, x4y2},
{x3, y2, x4y2, x8y2},
{x, x3, xy, x4y2},
{x, x6, xy, x4y2}
54,
54,
18,
18
C7 ⋊ C4 = 〈x, z | x4 = z7 =
x−1zxz = (x−1z)2x−2 = 1〉
6 {z, x2, x3z2x, x2z3} 6
(C4 × C2)⋊ C4 = 〈x, y | x4 =
y4 = y2x−1y−2x = xyx2y−1x =
y2x−1y−2x = (yx)2(y−1x)2 =
(yxy−1x−1)2 = (xyx−1y)2 = 1〉
1 {x2, y2, (xy)2, x2(xy)2y2} 1
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C. S. Anabanti 181
Table 2, continued
G n4 LMPFS S of size 4 in G M4
C4 ⋊ C8 = 〈x, y | x8 = y4 =
x−1yxy = y−1x−1y2xy−1 = 1〉
1 {x6, x2, x4y2, y2} 1
C8 ⋊Quasidih type C4 = 〈x, y|
x8 = 1 = y4, xy = yx3〉
9 {x, x6, y2, x4y2},
{x2, x6, y2, x4y2}
8,
1
C8 ⋊Dih type C4 = 〈x, y | x8 =
1 = y4, xy = yx−1〉
17 {x, x6, y2, x5y2},
{x, x6, y2, x4y2},
{x, y2, x2y2, x5y2},
{x2, x6, y2, x4y2}
4,
8,
4,
1
C4 ⋊D8 = 〈x, y | xy−2x−1y−2 =
x(xy)2x = x2(xy−1)2 =
y−1x−1y3x = 1〉
9 {y, x2, x4y2, x6},
{x5yx, x2, x4y2, x6},
{x2, y2, x4y2, x6}
4,
4,
1
(C4 ⋊ C4)× C2 = 〈x, y, z | z2 =
y4 = x4 = x−1yxy = zx−1zx =
1 = zy−1zy = x2y−1x2y =
(y−1x−2y−1)2〉
2 {x2y2z, z, x2, y2} 2
C4 ×Q8 = 〈x, y, z | x4 = y4 =
z4 = 1 = x−1yxy = z−1x−1zx
= z−1y−1zy = y2zx−2z =
x−1zx−1y−2z = x2y−1x2y〉
2 {x2, y2, z, x2y2z} 2
(C2 ×Q8)⋊ C2 = 〈x, y, z | x4 =
z2 = xyxy−1 = y2x2 =
yxyx−1 = zy−1zy = (xzx)2
= (zx−1)4 = (zxzx−1)2 = 1〉
2 {x2, z, (xz)2, x3zx} 2
C2
2 ⋊ C3
2 = 〈x, y, z | x4 = y4 =
z2x2 = y−1x−1yx−1 =
z−1y−1zy = y2x−1y2x =
zxy2zx = (x−1y−2x−1)2 = 1〉
2 {x2, y2, yz, x2y3z} 2
C4 ⋊Q8 = 〈x, y, z | y4 = z4 =
x2y2 = yxyx−1 = x−1zxz =
z−1y−1zy = 1〉
4 {x2, z, x3zx, x2z2} 4
C2 × C2 ×Q8 = 〈x, y, z, a | x4 =
a2 = z2 = y2x2 = yxyx−1 =
ax−1ax = ay−1ay = zx−1zx =
(az)2 = zy−1zy = 1〉
4 {x2, z, a, x2za} 4
C9 ⋊ C4 = 〈x, z | x4 = z9 =
x−1zxz = 1〉
18 {x2, z, x2z2, xz3x},
{x2, z, x3z2x, x2z4},
{x2, z, x3z3x, x2z4}
6,
6,
6
(C3 × C3)⋊ C4 = 〈x, z, a | x4 =
z3 = a3 = x−1zxz = x−1axa =
a−1z−1az = (x−1z)2x−2 =
z−1ax−1az−1x = 1〉
24 {z, x2, x2a, xzax} 24
“adm-n2” — 2021/7/19 — 10:26 — page 182 — #18
182 Groups containing LMPFS of size 4
Table 2, continued
G n4 LMPFS S of size 4 in G M4
C3 ×A4 = 〈x, y, z | x3 = y3 =
z2 = 1 = y−1x−1yx = zy−1zy =
(zx)3 = (x−1z)3〉
144 {x2, yzx, x2y2, x2zx2} 144
C5 ⋊inverse map C8 = 〈x, y |
x8 = 1 = y5, xyx−1 = y−1〉
4 {x2, x−2, y, x4y2} 4
Q40 = 〈x, y | x20 = 1, x10 =
y2, xy = yx−1〉
8 {x10, x11, x12, x17} 8
Q20 × C2 = 〈x, y, z | x10 = 1 =
z2, x5 = y2, xy = yx−1, yz =
zy, xz = zx〉
8 {x−2, y2, z, xz} 8
For notation in Table 2, n4 is the number of locally maximal product-
free sets of size 4 in G while M4 shows the corresponding sizes of each
orbit of the displayed locally maximal product-free sets under the action
of automorphism groups of G.
Theorem 3.11. Let S be a locally maximal product-free set of size 4 in a
finite group G such that |G| 6 57. Then the possibilities for S and G are
given in Tables 1 and 2.
Proof. We checked for groups of order from 8 up to 57 that contains locally
maximal product-free sets of size 4 in GAP [8]. Then listed all such locally
maximal product-free sets S of size 4 up to automorphisms of each such
group G in Tables 1 and 2.
Corollary 3.12. If S is a LMPFS of size 4 in a finite group G of odd
order, then both S and G are contained in Tables 1 and 2.
Proof. Follows from Proposition 3.3 and Theorem 3.11.
Lemma 3.13. Let S be a LMPFS of size 4 in a finite group G such that
every 2-element subset of S generates 〈S〉. If S contains no involution,
then either |G| 6 40 or 〈S〉 is cyclic.
Proof. If S∩S−1 = ∅, then by Theorem 3.2, |G| 6 57. Theorem 3.11 tells
us that (G,S) is one of the possibilities in Tables 1 and 2. In particular,
|G| 6 40. Suppose S ∩ S−1 6= ∅. Then S contains two elements a and b
such that b = a−1. As every 2-element subset of S generates 〈S〉, we have
that 〈S〉 = 〈a, a−1〉 = 〈a〉. So 〈S〉 is cyclic.
Proposition 3.14. Suppose S is a LMPFS of size 4 in a group G. If 〈S〉
is cyclic, then |G| 6 40.
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C. S. Anabanti 183
Proof. As 〈S〉 is cyclic, in the light of Lemma 3.9 and Remark 3.7,
|〈S〉| 6 24. Table 1 shows various possibilities for G and S. Proposition
2.6 tells us that each element s of Ŝ has even order; whence if 〈S〉 is any of
the cyclic groups of odd order, then Ŝ = ∅ and we conclude that G = 〈S〉.
In the light of Table 1 therefore |G| 6 21 < 40. Suppose 〈S〉 is cyclic of
even order. Consider 〈S〉 = C8 = 〈x | x8 = 1〉 and S = {x, x3, x5, x7}. If
any of x, x3, x5 or x7 is contained in Ŝ, then Ŝ consists of power of a single
element; by Proposition 2.7, |G| divides 32. If none of x, x3, x5 or x7 is con-
tained in Ŝ, then Ŝ = ∅; so G = 〈S〉. Consider 〈S〉 = C10 = 〈x | x10 = 1〉
and S = {x, x4, x6, x9}. As x4 and x6 have odd order, by Proposition 2.6,
x4, x6 6∈ Ŝ. Clearly, x, x9 6∈ Ŝ since x3, (x9)3 6∈ S. Therefore Ŝ = ∅; so
G = 〈S〉. Consider 〈S〉 = C12 = 〈x | x12 = 1〉. By Table 1, there are four
such LMPFS up to automorphisms of C12. By Proposition 2.6, x4 6∈ Ŝ
because it has odd order. Suppose S = {x, x4, x6, x11}. By Proposition 2.6,
x, x11 6∈ Ŝ because x3, (x11)3 6∈ S. So Ŝ 6 1 and we conclude by Propo-
sition 3.6 that |G| 6 24. If S = {x, x4, x7, x10}, then by Proposition 2.6,
Ŝ = ∅ since (x)3, (x7)3, (x10)3 6∈ S; so G = 〈S〉. Now, suppose S is any of
{x2, x3, x8, x9} or {x2, x3, x9, x10}. In the light of Proposition 2.6, in the
first case, x2, x8 6∈ Ŝ, and in the latter case, x2, x10 6∈ Ŝ. If none of x3 or x9
is an element of Ŝ, then Ŝ = ∅, and we conclude that G = 〈S〉. If any of x3
or x9 is contained in Ŝ, then both are contained in Ŝ; by Proposition 2.7
therefore |G| divides 48. Suppose |G| = 48. As
√
S has only elements of or-
der at least 3, we note that the number of elements of order at least 3 in G is
46. Among all the 47 nonabelian groups of order 48, only the groups whose
GAP ID are [48, 1], [48, 8], [48, 18], [48, 27] and [48, 28] have 46 elements of
order at least 3. We checked each of them for a LMPFS of size 4, and could
not find such. Therefore, if S is any of {x2, x3, x8, x9} or {x2, x3, x9, x10},
then |G| 6 24. Now, consider 〈S〉 = C14 = 〈x | x14 = 1〉. Up to au-
tomorphisms of C14, the LMPFS of size 4 in C14 are {x, x3, x8, x10},
{x, x3, x8, x13}, {x, x4, x6, x13}, {x, x6, x8, x13} and {x, x4, x7, x12}. In the
light of Proposition 2.6, all elements of order 7 and 14 in the respective
sets S do not lie in Ŝ. This means that in the first four cases, G = 〈S〉.
For the latter case, only the involution is a possible element of Ŝ; thus,
|Ŝ| 6 1, and we conclude by Proposition 2.5 that |G| 6 28. Consider
〈S〉 = C16 = 〈x | x16 = 1〉. Up to automorphisms of C16, the LMPFS S of
size 4 are {x, x3, x10, x12}, {x, x4, x6, x9}, {x, x4, x6, x15}, {x, x4, x9, x14},
{x, x6, x9, x14}, {x, x6, x10, x14} and {x2, x6, x10, x14}. For the first six
cases, Ŝ = ∅; so G = 〈S〉. For the last case, S = S−1; so 〈S〉 ∼= C8; a
contradiction as 〈S〉 ∼= C16. Consider 〈S〉 = C18 = 〈x | x18 = 1〉. By
Table 1, there are 9 such LMPFS up to automorphisms of C18. In the
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184 Groups containing LMPFS of size 4
light of Proposition 2.6, any of the 9 locally maximal product-free sets
S which does not contain the unique involution x9 gives rise to Ŝ = ∅;
so G = 〈S〉. For the LMPFS which contains the unique involution, we
have that |Ŝ| 6 1; whence by Proposition 2.5 therefore |G| 6 36. Consider
〈S〉 = C20 = 〈x | x20 = 1〉. Here, there are six such LMPFS up to automor-
phisms of C20. They are {x, x3, x14, x16}, {x, x4, x11, x18}, {x, x5, x14, x18},
{x, x6, x8, x11}, {x, x3, x10, x16} and {x2, x5, x15, x16}. The first four cases
can be handled with Proposition 2.6 to give that Ŝ = ∅; so G = 〈S〉.
For S = {x, x3, x10, x16}, we have that |Ŝ| 6 1; so |G| 6 40. Finally, let
S = {x2, x5, x15, x16}. By Proposition 2.6, the only possible element of
Ŝ are x5 and x15. If none of them are in Ŝ, then Ŝ = ∅ and G = 〈S〉.
Suppose at least one of them is in Ŝ, then as all odd powers of such element
lies in S, both elements must belong to Ŝ, and Proposition 2.7 tells us that
|G| divides 80. Suppose |G| = 80. As
√
S has only elements of order at
least 4, we note that the number of elements of order at least 4 in G is 78.
Among all the 47 nonabelian groups of order 80, only the groups whose
GAP ID are [80, 1], [80, 3], [80, 8], [80, 18] and [80, 27] contain 78 elements
of order at least 4. We checked each of them for a LMPFS of size 4, and
couldn’t find such. Therefore, if S = {x2, x5, x15, x16}, then |G| 6 40. For
〈S〉 = C22 or C24, there is only one such LMPFS up to automorphisms
of the respective groups and a direct check using Proposition 2.6 tells us
that Ŝ = ∅; so G = 〈S〉.
Corollary 3.15. If S is a LMPFS of size 4 in a finite group G such that
every two element subset of S generates 〈S〉 and S contains no involution,
then |G| 6 40.
Proof. Follows from Lemma 3.13 and Proposition 3.14.
Proposition 3.16. Suppose S is a LMPFS of size 4 in a finite group G
such that every two element subset of S generates 〈S〉 and S contains only
one involution. Then either |G| 6 40 or S = {a, b, c, d}, where c is the
unique involution in S and either a, b and d have order 3, or that a has
order greater than 3 together with a−1 = bd and none of b and d is an
involution.
Proof. Suppose S = {a, b, c, d}, where c is an involution, and each of a, b
and d has order at least 3. Consider a−1. Recall that G = T (S) ∪
√
S.
Suppose a−1 ∈
√
S. Then a−2 ∈ S. This implies that either a has order
3 (by a−2 = a) or 〈S〉 is cyclic
(
by a−2 ∈ {b, c, d}; for instance a−2 = b
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C. S. Anabanti 185
implies that 〈a, b〉 = 〈a〉
)
. In the latter case, Proposition 3.14 tells us that
|G| 6 40. Suppose a−1 ∈ T (S). Then
T (S) ⊆
1, a, b, c, d, a2, b2, d2, ab, ba, ac, ca, ad, da, bc, cb, bd, db, cd, dc,
ab−1, ba−1, ca−1, a−1c, cb−1, b−1c, a−1b, b−1a, ad−1, d−1a,
da−1, a−1d, bd−1, d−1b, cd−1, d−1c, db−1, b−1d
(9)
If
a−1 ∈ {b, b2, d, d2, ab, ba, ad, da, ab−1, ba−1, ad−1, da−1, a−1b,
b−1a, a−1d, d−1a},
then 〈S〉 is cyclic, generated by either a, b or d. In the light of Proposition
3.14 therefore |G| 6 40. If a−1 ∈ {c, ac, ca, a−1c, ca−1}, then 〈S〉 is cyclic;
again |G| 6 40. Since a has order at least 3, we have that a−1 6∈ {1, a}. Note
that a−1 6∈ {bc, cb, dc, cd, b−1c, cb−1, d−1c, cd−1, b−1d, db−1, d−1b, bd−1};
otherwise S is not product-free. The only remaining possibilities is that
a−1 ∈ {a2, bd, db}. We also perform a similiar analysis with b−1 and d−1.
Our conclusion is that either 〈S〉 is cyclic (which by Proposition 3.14
implies that |G| 6 40) or either a has order 3 with either a−1 = bd or
a−1 = db with similar statement for b and d. In the latter case, we can
assume without loss of generality that either all of a, b and d have order
3, or that a has order greater than 3 together with a−1 = bd and none of
b and d is an involution.
In the latter part of Proposition 3.16, our goal is to show that |G| 6 40.
We resolve the first case of the latter part of Proposition 3.16 in Lemma
3.20 below, and the second case in Corollary 3.24, Lemma 3.25 and Remark
3.26. We will be considering the following special case: S is a LMPFS of
size 4 in a group G such that every two-element subset of S generates 〈S〉.
Furthermore S = {a, b, c, d} where c is an involution but none of a, b and
d is an involution. We shall impose an additional condition that ‘〈S〉 is
neither abelian nor dihedral’ (see Assumption 3.18 below). To do so, we
first clear the air with the following:
Lemma 3.17. Let S be a LMPFS of size 4 in a finite group G such
that S contains exactly one involution and every two element subset of S
generates 〈S〉. Suppose 〈S〉 is either abelian or dihedral. Then 〈S〉 must
be abelian and |G| 6 40.
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186 Groups containing LMPFS of size 4
Proof. In the light of Lemmas 3.9 and 2.10, 〈S〉 cannot be dihedral. So, it
must be that 〈S〉 is abelian. If 〈S〉 is cyclic, then Proposition 3.14 tells us
that |G| 6 40. Now, suppose 〈S〉 is a non-cyclic abelian group. By Remark
3.7 and Table 1, 〈S〉 is one of C4 × C2, C
3
2 , C6 × C2, C4 × C4, C8 × C2,
C4×C2×C2, C6×C3 and C10×C2. The LMPFS in the groups C4×C2, C
3
2 ,
C6×C2, C4×C4, C4×C2×C2 and C10×C2 do not meet the requirement
of our defined S in terms of the orders of its elements. In fact, the only
possibilities (up to automorphisms of respective group) that satisfy the
condition that S has only one element of order 2 and other elements have
order at least 3 is that (〈S〉, S) ∈ {(C8 × C2, {x1, x2, x61, x51x2}), (C8 ×
C2, {x1, x2, x21x2, x51x2}), (C6 × C3, {x1, x2, x51x22, x31})}. Proposition 2.6
tells us that elements of Ŝ have even order, and if s ∈ Ŝ, then all odd pow-
ers of s lies in S. In the listed representatives of S, we see immediately that
if a non-involution x ∈ S ⊂ C8 ×C2, then x3 /∈ S, and if a non-involution
y ∈ S ⊂ C6×C3, then y5 /∈ S. So in all cases |Ŝ| 6 1. In the light of Propo-
sition 2.5 therefore |G| 6 2|S| in each of the possibilities, from where we de-
duce that |G| 6 32 or 36 according as 〈S〉 = C8×C2 or 〈S〉 = C6×C3. How-
ever, the only possibility is
(
〈S〉, S, |G|
)
=
(
C8×C2, {x1, x2, x51x22, x31}, 32
)
since 〈S〉 is cyclic in each of the other two cases; for instance if G = C6×C3,
then 〈S〉 = 〈x1, x31〉 = 〈x1 | x61 = 1〉 ∼= C6.
Assumption 3.18. Suppose S is a locally maximal product-free set of
size 4 in a finite group G such that every two-element subset of S generates
〈S〉. Furthermore S = {a, b, c, d} where c is an involution but none of a, b
or d is an involution, and 〈S〉 is neither abelian nor dihedral.
Lemma 3.19. Suppose Assumption 3.18 holds. Then
I(G) ⊆ {c, a2, b2, d2, ab, ba, ad, da, bd, db, ab−1, b−1a,
ad−1, d−1a, bd−1, d−1b}.
Proof. The set I(G) consisting of all involutions in G is a subset of
T (S), because G = T (S) ∪
√
S and no element of
√
S can be an involu-
tion. Clearly 1, a, b, d are not involutions. Suppose x ∈ {a±1, b±1, d±1},
and cx is an involution. Now 〈S〉 = 〈c, x〉 because 〈S〉 is generated
by every 2-element subset of S. But 〈c, x〉 = 〈c, cx〉, which is dihe-
dral; a contradiction. The case where xc is an involution is the same.
So we can eliminate ac, bc, dc, a−1c, b−1c, d−1c, ca, cb, cd, ca−1, cb−1
and cd−1. This means I(G) ⊆ {c, a2, b2, d2, ab, ba, ad, da, bd, db, ab−1, ba−1,
a−1b, b−1a, ad−1, da−1, a−1d, d−1a, bd−1, db−1, b−1d, d−1b}. If an element
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C. S. Anabanti 187
g is an involution, then g−1 = g, so we only need to include one repre-
sentative from {g, g−1} in the list of possible involutions. This means we
need only include one of ab−1 and ba−1, for example. There are six such
pairs, allowing us to remove ba−1, a−1b, da−1, a−1d, db−1 and b−1d from
the list as if they are involutions then they will equal an element that is on
the list. Thus I(G) ⊆ {c, a2, b2, d2, ab, ba, ad, da, bd, db, ab−1, b−1a, ad−1,
d−1a, bd−1, d−1b}.
Lemma 3.20. Suppose Assumption 3.18 holds. If a, b and d all have
order 3, then 〈S〉 ∼= A4 and |G| 6 24.
Proof. By Lemma 3.19, and the fact that a2, b2 and d2 are not involutions,
we have
I(G) ⊆ {c, ab, ba, ad, da, bd, db, ab−1, b−1a, ad−1, d−1a, bd−1, d−1b}.
If c is the only involution in G, then c is central, and hence commutes with a.
But 〈S〉 = 〈a, c〉, which implies 〈S〉 is abelian, contrary to Assumption 3.18.
Therefore there are elements x, y ∈ {a±1, b±1, d±1} with y 6= x±1, such
that xy is an involution. This implies 〈S〉 = 〈x, y : x3 = y3 = (xy)2 = 1〉,
which is a presentation of A4. By Proposition 2.6, any element of Ŝ must
have even order; so Ŝ ⊆ {c}. In the light of Proposition 2.5 therefore
|G| 6 24.
Lemma 3.21. Suppose S is a LMPFS of size 4 in a finite group G such
that every two-element subset of S generates 〈S〉. Furthermore suppose
S = {a, b, c, d} where c is an involution, a−1 = bd has order at least 4
and none of b or d is an involution. Then G has either 1, 3, 5, 7 or 9
involutions, and I(G) ⊆ {c, a2, b2, d2, ab−1, b−1a, ad−1, d−1a, bd−1, d−1b}.
Proof. Since a−1 = bd, we get b−1 = da and d−1 = ab. None of these
elements can be involutions. Now o(bd) = o(db) and o(da) = o(ad) and
o(ab) = o(ba). Hence these elements can’t be involutions either. The result
now follows immediately from Lemma 3.19 and the fact that any group
containing involutions has an odd number of them.
Proposition 3.22. Suppose Assumption 3.18 holds. Let K be the cen-
tralizer CG(c) of c in G, and let J = K ∩ T (S). Then |K| 6 4|J |, and
hence |G| 6 4|J | · |cG|.
Proof. Let K be the centralizer in G of c. If x ∈ K and x2 = a, then
that would imply 〈S〉 is abelian, because 〈S〉 = 〈a, c〉 = 〈x2, c〉 and x
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188 Groups containing LMPFS of size 4
commutes with c. Similarly x cannot be a square root of b or d. Hence
K ⊆ √
c ∪ J . Suppose there exist x, y ∈ √
c ∩ K with xy /∈ J . Let
z ∈ √
c ∩ K. Now xy, xz and yz are elements of K because K is a
subgroup. Suppose xz /∈ J and yz /∈ J . Then we have (yz)2 = c, which
implies zy = cyz. Similarly yx = cxy and zx = cxz. But now (xyz)2 =
xyzxyz = xycxzyz = cxyxzyz = cxcxyzyz = c2x2(yz)2 = 1. Therefore
xyz /∈ √
c. Thus xyz ∈ J . So either xz ∈ J , yz ∈ J or xyz ∈ J . Hence√
c∩K ⊆ x−1J ∪ y−1J ∪ (xy)−1J . Remembering that K = J ∪ (
√
c∩K)
we immediately derive |K| 6 4|J |. The remaining possibility is that there
do not exist x, y ∈ √
c∩K with xy /∈ J . This means either K = J (because√
c ∩K = ∅), or that there is some x ∈ √
c ∩K, but for all y ∈ √
c ∩K
we have xy ∈ J . Hence
√
c ∩K ⊆ x−1J . Either way, |K| 6 2|J |. Hence
|K| 6 4|J | and so |G| 6 4|J | · |cG|.
Lemma 3.23. Suppose Assumption 3.18 holds. Let K be the centralizer
CG(c) of c in G, and let J = K ∩ T (S). Then
J ⊆
{
1, c, a2, b2, d2, ba, ad, db, ab−1, ba−1, a−1b, b−1a,
ad−1, da−1, a−1d, d−1a, bd−1, db−1, b−1d, d−1b
}
.
In particular, |J | 6 20.
Proof. Since 〈S〉 is not abelian, J doesn’t contain a, b or d. Similarly no ele-
ment of the form xc or cx can be contained in J , where x ∈ {a±1, b±1, d±1}
as this would imply the presence in J of either a, b or d. Hence we remove
these elements from our original list for T (S). The other observation is
that since a−1 = bd, b−1 = da and d−1 = ab, these three elements cannot
be contained in J , because again this would imply the presence in J of
a, b or d.
Corollary 3.24. Suppose S is a LMPFS of size 4 in a finite group G
such that every two-element subset of S generates 〈S〉. In particular, take
S = {a, b, c, d} where c is an involution, a−1 = bd has order at least 4 and
none of b or d is an involution. Furthermore, suppose 〈S〉 is not dihedral
or abelian, then |G| 6 720. Moreover, G has 1, 3, 5, 7 or 9 involutions
and I(G) ⊆ {c, a2, b2, d2, ab−1, b−1a, ad−1, d−1a, bd−1, d−1b}.
Proof. As a−1 = bd, by Lemma 3.21 we know |cG| 6 9 and by Lemma
3.23 we have |J | 6 20. Hence by Proposition 3.22, |G| 6 4× 20× 9 = 720.
The latter fact
(
that G has 1, 3, 5, 7 or 9 involutions and possible elements
of I(G)
)
follows from Lemma 3.21.
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C. S. Anabanti 189
Lemma 3.25. In Corollary 3.24, no LMPFS S exists if G contains either
1 or 9 involutions.
Proof. Let G and S be as defined in Corollary 3.24. If G contains only
one involution, then c is central; so 〈S〉 = 〈a, c〉 is abelian, contradicting
the hypothesis. Suppose G contains exactly 9 involutions. So nine out of
the ten likely elements of I(G) listed in Corollary 3.24 are involutions.
Note that for any g, h ∈ G, o(gh) = o(hg). So ab−1 is an involution if
and only if b−1a is, and so on. Since only one of the above list of ten
things is not an involution, it must be the case that all of ab−1, b−1a,
ad−1, d−1a, bd−1 and d−1b are involutions, as is c, and exactly two of
a2, b2 and d2 are involutions. Without loss of generality, suppose a2 and
b2 are involutions. Using a−1 = bd, we get ad−1 = a2b and so on; so the
nine involutions of G as c, a2, b2, ab−1, b−1a, a2b, aba, bab and ab2. From
(b−1a)2 = (aba)2, we obtain b−1a = aba2b2; whence ab−1a = a2ba2b2 = b.
Now ab−1a−1 = ba−2 = ba2; a contradiction since o(b) = 4, however
(ba2)2 = 1 = (a2b)2. Thus no such locally maximal product-free set S
exists.
Remark 3.26. The aim of this remark is to assert that in Corollary
3.24, |G| < 40 if G contains either 3, 5 or 7 involutions. If G has ex-
actly 7 involutions, then at least two of the three pairs ab−1, b−1a, ad−1,
d−1a, bd−1, d−1b are involutions; so it means we can remove at least 4
elements from the count of J . Hence |J | 6 16 and |cG| 6 7. By Propo-
sition 3.22 therefore |G| 6 4 × 16 × 7 = 448. We show that this case
is not easily resolvable like the case where the number of involutions
in G is 9 as seen in Lemma 3.25. For instance, if the two pairs ab−1,
b−1a, bd−1 and d−1b are involutions, together three other involutions c,
(ab)2 and a2, then |〈S〉| yields different values. As d−1 = ab, we write
〈S〉 = 〈a, b | (ab−1)2 = (ab2)2 = (bab)2 = a4 = (ab)4 = bn = 1〉, where n
is the order of b; for example if n = 3, 4, 5 or 6, then 〈S〉 ∼= S4, D8, D10 or
C2 × S4 respectively. So we can’t make further deductions from 〈S〉, with-
out knowing at least a dihedral subgroup of 〈S〉. Now, suppose G contains
exactly 5 involutions. Then we can remove at least 2 elements from the
count of J . So |J | 6 18 and |cG| 6 5. Hence |G| 6 4×18×5 = 360. Finally,
if G contains exactly 3 involutions, then we can’t remove anything from
J necessarily; so |J | 6 20 and |cG| 6 3. Therefore |G| 6 4× 20× 3 = 240.
In each of these cases, S = {c, b, d, (bd)−1}, where b and d are arbitrary
elements of order at least three. We checked in GAP for existence of a
LMPFS S in the respective groups of even orders. Our result is summarised
below.
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190 Groups containing LMPFS of size 4
|I(G)| |NAG|I(G)|| GAP I.D of groups G contain-
ing required S
8 6 n 6 40 3, 5, 7 47 [20, 3] and [32, 14]
42 6 n 6 240 3, 5, 7 1665 ————————
242 6 n 6 360 5, 7 4525 ————————
362 6 n 6 448 7 3036 ————————
In the table above, n gives the range of orders of the nonabelian
groups of even order tested. We performed four checks in GAP. The first
line of result is the outcome of our first check. Our first check was for
nonabelian groups of orders from 8 up to 40 that contain either 3, 5
or 7 involutions. Only 47 nonabelian groups of even order n ∈ [8, 40]
contain either 3, 5 or 7 involutions. Among these groups, only in the
groups whose GAP IDs are [20, 3] and [32, 14] that we found our required
LMPFS. The group of order 20 mentioned is mainly referred to as the
only non-simple Suzuki group; it is denoted by Suz(2) = 〈x, y | x5 = 1 =
y4, xy = yx2〉. There are 40 LMPFS of size 4 in this group
(
examples
are {x, y2, x2y, xy3} and {x3, y2, x2y, xy3}
)
; each of the 40 LMPFS is of
our required form, and under automorphisms of the group, is one of the
two mentioned. On the other hand, the group of order 32 mentioned is
called the semidihedral group of C8 and C4 of dihedral type. It has a
presentation as C8 ⋊Dih type C4 = 〈x, y | x8 = 1 = y4, xy = yx−1〉. This
group has 17 LMPFS of size 4; only 8 of them are of our required form,
and up to automorphisms of the group, any LMPFS of our kind is either
{x, x6, y2, x5y2} or {x, y2, x2y2, x5y2} (4 belonging to each class). Our
second check was for nonabelian groups of even orders from order 42 up
till 240 that contain either 3, 5 or 7 involutions. Only 1665 nonabelian
groups in that range contain either 3, 5 or 7 involutions, and our search
shows that no such group contains our desired LMPFS of size 4. The
table is now easily understood for the third and fourth check, which are
precisely the last two rows of the table.
We now turn back to give a proof of Theorem 3.8.
Proof of Theorem 3.8. Suppose G is a finite group containing a LMPFS of
size 4 such that every 2-element subset of S generates 〈S〉. Corollary 3.10
tells us that no such S exists if S were to contain at least two involutions.
If S contains no involution, then Corollary 3.15 tells us that |G| 6 40.
Finally, if S contains exactly one involution, then Proposition 3.16, Lemma
3.20, Lemma 3.17, Corollary 3.24, Lemma 3.25 and Remark 3.26 yield
|G| 6 40.
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C. S. Anabanti 191
A deduction from the classification of finite groups containing locally
maximal product-free sets of size 4 studied in this paper is the following:
Corollary 3.27. If a finite group G contains a LMPFS S of size 4, then
either |G| 6 40 or G is nonabelian, |G| is even, S ∩ S−1 6= ∅ and not
every 2-element subset of S generates 〈S〉.
Proof. Follows from Theorems 3.2, 3.8, 3.11, Remark 3.7 and Corollary
3.12 .
We close the discussion on finite groups containing LMPFS of size 4
with the following:
Conjecture 3.28. If a finite group G contains a LMPFS of size 4, then
|G| 6 40.
In the light of Theorem 3.2 and Proposition 3.3 as well as some
computational investigations, we ask the following question:
Question 3.29. Does there exists a finite group G containing a locally
maximal product-free set of size k (for k > 4) such that |G| > 2k(2k− 1)?
We write nk (resp. mk) for the maximal size of a finite abelian group of
even (resp. odd) order containing a LMPFS of size k. Some experimental
results on finite abelian groups G of even order nk (resp. odd order mk)
containing LMPFS of size k are reported below.
k nk G
1 4 C4
2 8 C8 and C4 × C2
3 16 C4 × C4
4 24 C24
5 36 C36, C18 × C2, C12 × C3, C6 × C6
6 48 C48, C12 × C4, C12 × C2
2 , C6 × C3
2
7 64 C3
4 , C8 × C4 × C2
k mk G
1 3 C3
2 7 C7
3 15 C15
4 27 C3
3 , C9 × C3
5 35 C35
6 45 C45, C15 × C3
7 61 C61
One may be moved by the above result to conjecture that ‘if a finite abelian
group G of even order contains a LMPFS of size k, then |G| 6 (k + 1)2’.
However, such conjectural statement cannot hold in general as C84 contains
some locally maximal product-free sets of size 8. On another remark,
let C2n = 〈x | x2n = 1〉 be the finite cyclic group of order 2n, and
suppose S is a locally maximal product-free set in a finite cyclic group
of even order containing the unique involution in the group. Then S is
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192 Groups containing LMPFS of size 4
also a locally maximal product-free subset of the finite dicyclic group
Q4n = 〈x, y | x2n = 1, xn = y2, xy = yx−1〉 of order 4n; the reason is
because {xiy | 0 6 i 6 2n− 1} ⊂
√
xn ⊆
√
S, and already we know that
{xi | 0 6 i 6 2n−1} ⊆ S∪SS∪SS−1∪
√
S. Thus, one may obtain a lower
bound on the maximal size of a finite group containing a locally maximal
product-free set of size k by using the bound from the dicyclic group
counterpart. For instance, the maximal size of a finite cyclic group of even
order containing a locally maximal product-free set of size k which contains
the unique involution is 4, 6, 12, 20, 30 and 40 for k = 1, 2, 3, 4, 5 and 6
respectively. An example of a locally maximal product-free set containing
the unique involution in C4, C6, C12, C20, C30 and C40 is given respectively
as {x2}, {x2, x3}, {x2, x6, x10}, {x4, x7, x9, x10}, {x2, x6, x15, x22, x27} and
{x3, x8, x20, x29, x33, x39}. This tells us that there are locally maximal
product-free set(s) of sizes 1, 2, 3, 4, 5 and 6 in Q8, Q12, Q24, Q40, Q60
and Q80 respectively. So if a finite group G contains a locally maximal
product-free set of size 1, 2, 3, 4, 5 or 6, then |G| > 8, 12, 24, 40, 60 or 80
respectively. Experimental results as well as results of [5, 9] suggest that
the largest size of a finite group containing a LMPFS of size 1, 2, 3, 4, 5
or 6 is 8, 16, 24, 40, 64 or 96 respectively. We take this opportunity to list
all finite groups G of expected highest possible size which contain locally
maximal product-free sets of size k for k ∈ {1, 2, 3, 4, 5}.
k G
1 G8 := 〈x, y | x4 = 1, x2 = y2, xy = yx−1〉 ∼= Q8
2 G16A := 〈x, y | x4 = 1 = y4, xy = y−1x〉
G16B := 〈x, y, z | x4 = 1 = z2, x2 = y2, xy = yx−1, yz = zy, xz =
zx〉 ∼= Q8 × C2
3 G24A := 〈x, y | x12 = 1, x6 = y2, xy = yx−1〉 ∼= Q24
G24B := 〈x, y, z | x4 = 1 = z3, x2 = y2, xy = yx−1, yz = zy, xz =
zx〉 ∼= Q8 × C3
4 G40A := 〈x, y | x8 = 1 = y5, xy = y−1x〉
G40B := 〈x, y | x20 = 1, x10 = y2, xy = yx−1〉 ∼= Q40
G40C := 〈x, y, z | x10 = 1 = z2, x5 = y2, xy = yx−1, yz = zy, xz =
zx〉 ∼= Q20 × C2
5 G64A := 〈a, b | a8 = b4 = b−2a−1b−2a = ab−1a2ba =
a−1b−1ab−1a−1b−1ab = 1〉
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C. S. Anabanti 193
G64B := 〈a, b | a4 = b8 = a2b−1a2b = a−1b2ab2 = (aba−1b)2 = 1〉
G64C := 〈a, b, c | a4 = b4 = c2 = cb−1cb = ca−1ca = a2b−1a−2b =
ba−1b2ab = b−1a−1bab−1aba−1 = (a−1b−2a−1)2 =
a−1(ba)2ba−1b = 1〉
G64D := 〈a, b, c | a4 = b4 = c2 = cb−1cb = a−1bab = bab−2a−1b =
(aca)2 = a2b−1a2b = (ca−1)4 = (b−1a−2b−1)2 = (caca−1)2 = 1〉
G64E := 〈a, b, c | a4 = b4 = c2 = ba−1ba = b2cb−2c = a2ba2b−1 =
ca−1b−2ca−1 = (cb−1)2(cb)2 = (a−1b2a−1)2 = (cbacba−1)2 = 1〉
G64F := 〈a, b, c | a4 = b8 = c2 = cbcb−1 = ca−1ca = a−1bab =
a2b−1a2b = 1〉
G64G := 〈x, y, z | x8 = 1 = z4, x4 = y2, xy = yx−1, yz = zy, xz =
zx〉 ∼= Q16 × C4
G64H := 〈a, b, c | a4 = c−1b−1cb = c4a2 = c2ac2a−1 = b4a2 =
b2ab2a−1 = b−1ac2ba−1 = c−1a−1b−1cb−1a−1 = 1〉
G64I := 〈a, b, c | a4 = c4 = a−1bab = a−1cac = c−1b−1cb =
a−1b−4a−1 = a2c−1a2c = (c−1a−2c−1)2 = 1〉
G64J := 〈a, b, c | b4 = c8 = ab−1a−1b−1 = c−1b−1cb = a−1cac =
b−1a−1ba−1 = 1〉
G64K := 〈a, b, c, d | a4 = b4 = c2 = d2 = da−1da = a−1bab =
ca−1ca = db−1db = (dc)2 = cb−1cb = a2b−1a2b = (b−1a−2b−1)2 =
1〉 ∼= G16A × C2
2
G64L := 〈a, b, c, d | a4 = c2 = d2 = aba−1b = db−1db = (dc)2 =
cb−1cb = ba2b = da−1da = b−1a−1ba−1 = a2ca−2c = (ca)4 =
cacdaca−1cda−1 = 1〉
G64M := 〈a, b, c, d | a4 = b2 = d2 = c2a2 = caca−1 = c2a−2 =
(db)2 = dc−1dc = (aba)2 = a−1ba−1cbc = (a2d)2 = da−1bdab =
(ba−1)4 = c−1bcbaba−1b = 1〉
G64N := 〈a, b, c, d, g | a4 = c2 = d2 = g2 = b2a2 = baba−1 =
gb−1gb = ca−1ca = ga−1ga = (gc)2 = cb−1cb = da−1da =
(gd)2 = (dc)2 = db−1db = 1〉 ∼= Q8 × C3
2 .
Acknowledgements
The author is grateful to Professor Sarah Hart for useful discussion
during the writing of this paper. He is also thankful to Birkbeck college
for the financial support provided.
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194 Groups containing LMPFS of size 4
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Contact information
Chimere
Stanley
Anabanti
Department of Mathematics and Applied Mathematics,
University of Pretoria, South Africa
E-Mail(s): chimere.anabanti@up.ac.za
Web-page(s): https://www.up.ac.za/
mathematics-and-applied-mathematics/
article/2953786/
dr-cs-chimere-anabanti
Received by the editors: 05.03.2019.
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