Extremal product-one free sequences in Dihedral and Dicyclic groups

TL;DR

This paper characterizes all extremal sequences in dihedral and dicyclic groups that are just below the Davenport constant and contain no subsequence with product 1, extending understanding of zero-sum problems in these groups.

Contribution

It provides explicit descriptions of sequences of length one less than the Davenport constant that are product-one free in dihedral and dicyclic groups.

Findings

Characterizations of extremal product-one free sequences in dihedral groups

Characterizations of extremal product-one free sequences in dicyclic groups

Extension of Davenport constant results to explicit sequence structures

Abstract

Let $G$ be a finite group, written multiplicatively. The Davenport constant of $G$ is the smallest positive integer $D(G)$ such that every sequence of $G$ with $D(G)$ elements has a non-empty subsequence with product $1$. Let $D_{2n}$ be the Dihedral Group of order $2n$ and $Q_{4n}$ be the Dicyclic Group of order $4n$. J. J. Zhuang and W. Gao (European J. Combin. 26 (2005), 1053-1059) showed that $D(D_{2n}) = n+1$ and J. Bass (J. Number Theory 126 (2007), 217-236) showed that $D(Q_{4n}) = 2n+1$. In this paper, we give explicit characterizations of all sequences $S$ of $G$ such that $|S| = D(G) - 1$ and $S$ is free of subsequences whose product is $1$, where $G$ is equal to $D_{2n}$ or $Q_{4n}$ for some $n$.

Equations

$$S=(\underset{n-1\text{ times}}{\underbrace{g,\dots ,g}}),\text{ where g is a generator of Cn.}$$

$$S=(\underset{\exp (G)-1\text{ times}}{\underbrace{T,T,\dots ,T}}),$$

$$C_q{⋊}_s{C}_m=⟨x,y|x^m=1,y^q=1,yx=x{y}^s,{\mathrm{ord}}_q(s)=m,q\text{ prime}⟩.$$

$$S=(\underset{q-1\text{ times}}{\underbrace{y^t,y^t,\dots ,y^t}},x^i{y}^{{\nu }_1},x^i{y}^{{\nu }_2},\dots ,x^i{y}^{{\nu }_{m-1}}).$$

$$S=(y^t,y^t,x{y}^{\nu })\text{ for }t\in \{2,3\}\text{ and }\nu \in \{0,1,2\}\text{ or }S=(x,xy,x{y}^2).$$

$$x^2=y^n=1,yx=x{y}^{-1}.$$

$$x^2=y^n,y^{2n}=1,yx=x{y}^{-1}.$$

$$Q_{4n}/\{1,y^n\}\simeq D_{2n},$$

$$S=(\underset{n-1\text{ times}}{\underbrace{y^t,y^t,\dots ,y^t}},x{y}^s).$$

$$S=(y^t,y^t,x{y}^{\nu })\text{ for }t\in \{2,3\}\text{ and }\nu \in \{0,1,2\}\text{ or }S=(x,xy,x{y}^2).$$

$$S=(\underset{2n-1\text{ times}}{\underbrace{y^t,y^t,\dots ,y^t}},x{y}^s).$$

$$(y^r,y^r,y^r,x{y}^s),(y^r,x{y}^s,x{y}^s,x{y}^s)\text{ or }(x{y}^s,x{y}^s,x{y}^s,x{y}^{r+s}).$$

$$⟨e,i,j,k|i^2=j^2=k^2=ijk=e,e^2=1⟩.$$

$$\pm (i,i,i,\pm j),\pm (i,i,i,\pm k),\pm (j,j,j,\pm i),\pm (j,j,j,\pm k),\pm (k,k,k,\pm i)\text{ or }\pm (k,k,k,\pm j).$$

$$g_{\sigma (n_1)}{g}_{\sigma (n_2)}\dots g_{\sigma (n_k)}=1$$

$$x{y}^{\alpha }\cdot x{y}^{\beta }=y^{\beta -\alpha }.$$

$$x{y}^{\alpha }\cdot x{y}^{\beta }=y^{\beta -\alpha +n}.$$

$$\sum _{j\in J}{\epsilon }_j{y}_j\equiv 0(\mathrm{mod}n).$$

$$(g{)}^{n-3},(g{)}^{n-4}(g^2),(g{)}^{n-4}(g^3),(g{)}^{n-5}(g^2{)}^2.$$

$$x{y}^u\cdot \underset{k\text{ times}}{\underbrace{y^t\cdot y^t\dots y^t}}\cdot x{y}^v=y^{v-u-kt}.$$

$$x{y}^v\cdot y^t\cdot x{y}^u=y^{u-v-t}=1.$$

$$x{y}^u\cdot \underset{k\text{ times}}{\underbrace{y^t\cdot y^t\dots y^t}}\cdot x{y}^v=y^{v-u-kt}.$$

$$S\cap N_D=(x{y}^{\alpha },x{y}^{\beta },x{y}^{\gamma }).$$

$$\alpha -\beta \in \{0,1,2,\dots ,\lfloor n/3\rfloor \}(\mathrm{mod}n).$$

$$x{y}^{\beta }\cdot y^r\cdot x{y}^{\alpha }=y^{\alpha -\beta -r}=1$$

$$\{\begin{array}{l}S=(x{y}^{\alpha },x{y}^{\beta },x{y}^{\gamma },y,y,y);\\ S=(x{y}^{\alpha },x{y}^{\beta },x{y}^{\gamma },y,y,y^2);\\ S=(x{y}^{\alpha },x{y}^{\beta },x{y}^{\gamma },y,y,y^3);\text{ or }\\ S=(x{y}^{\alpha },x{y}^{\beta },x{y}^{\gamma },y,y^2,y^2).\end{array}$$

$$x{y}^{\alpha }\cdot x{y}^{\beta }\cdot y=y^{\beta +1-\alpha }=1.$$

$$x{y}^{\alpha }\cdot x{y}^{\beta }\cdot y^2=y^{\beta +2-\alpha }=1.$$

$$S\cap H_D$$

$$S\cap N_D$$

$$\lfloor k/2\rfloor >\lfloor {\log }_{2}n\rfloor$$

Full text

1. Introduction

Given a finite group $G$ written multiplicatively, the Zero-Sum Problems study conditions that guarantee a non-empty subsequence in $G$ has a non-empty subsequence with certain prescribed properties such that the product of its elements, in some order, is equal to the identity $1\in G$. Interesting properties include length, repetitions, and weights.

One of the first problem of this type was considered by Erdös, Ginzburg, and Ziv [6]. They proved that given $2n-1$ integers, it is possible to select $n$ of them, such that their sum is divisible by $n$. In the language of group theory, every sequence $S$ consisting of at least $2n-1$ elements in a finite cyclic group of order $n$ has a subsequence $T$ of length $n$ such that the product of the elements in $T$ in some order is equal to the identity. Further, they proved that the number $2n-1$ is the smallest positive integer with this property.

These problems have been studied extensively for abelian groups; see the surveys by Y. Caro [5] and W. Gao - A. Geroldinger [9].

An important problem of zero-sum type is to determine the so-called Davenport constant of a finite group $G$ (written multiplicatively). This constant, denoted by $D(G)$, is the smallest positive integer $d$ such that every sequence $S$ with $d$ elements in $G$ (repetition allowed) contains some subsequence $T$ such that the product of the elements in $T$ in some order is $1$.

For $n\in\mathbb{N}$, let $C_{n}\simeq\mathbb{Z}_{n}$ denote the cyclic group of order $n$ written multiplicatively. The Davenport constant is known for the following groups:

• •

$D(C_{n})=n$;

• •

$D(C_{m}\times C_{n})=m+n-1$ if $m|n$ (J. Olson, [14]);

• •

$D(C_{p^{e_{1}}}\times\dots\times C_{p^{e_{r}}})=1+\sum_{i=1}^{r}(p^{e_{i}}-1)$ (J. Olson, [13]);

• •

$D(D_{2n})=n+1$ where $D_{2n}$ is the Dihedral Group of order $2n$ (see [19]);

• •

$D(C_{q}\rtimes_{s}C_{m})=m+q-1$ where $q\geq 3$ is a prime number and $\mathop{\rm ord}\nolimits_{q}(s)=m\geq 2$ (J. Bass, [2]).

For most finite groups, the Davenport constant is not known.

By the definition of the Davenport constant, for a given finite group $G$ there exist sequences $S$ with elements in $G$ such that $|S|=D(G)-1$ and such that $S$ is free of product-1 subsequences. That is, there exists $S=(g_{1},\cdots,g_{D(G)-1})$ of $G$ such that $g_{i_{1}}\cdots g_{i_{k}}\neq 1$ for every non-empty subset $\{i_{1},\cdots,i_{k}\}$ of $\{1,\cdots,D(G)-1\}$. The Inverse Zero-Sum Problems study the structure of these extremal-length sequences which are free of product-$1$ subsequences with some prescribed property. For an overview, see articles by W. Gao, A. Geroldinger, D. J. Grynkiewicz and W. A. Schmid [11], [16] and [10].

The inverse problems associated with Davenport constant are solved for few abelian groups. For example, the following theorem is direct consequence of Theorem 2.2 and gives a complete characterization of sequences free of product-$1$ subsequences in the finite cyclic group $C_{n}$.

Theorem 1.1**.**

Let $S$ be a sequence in $C_{n}$ free of product-$1$ subsequences with $n-1$ elements, where $n\geq 2$. Then

[TABLE]

Observe that in the cyclic case, sequences free of product-$1$ subsequences contain an element repeated many times. It is natural to ask if this is true in general. Specifically, let us say that a given finite abelian group $G$ has Property C if every maximal sequence $S$ of $G$ free of product-$1$ subsequences with at most $\exp(G)$, the exponent of the group $G$, elements has the form

[TABLE]

for some subsequence $T$ of $S$. The above theorem states that $C_{n}$ has Property C. It then follows from a result of C. Reiher [15] that $C_{p}^{2}$ possesses Property C (see also [11] and [8]). In [10], W. Gao, A. Geroldinger and D. J. Grynkiewicz showed that this result is multiplicative, extending our conclusion to the groups $C_{n}^{2}$ for a composite number $n$. In [17], W. A. Schmid discusses the case $C_{n}\times C_{m}$, where $n|m$. Not much is known about groups of rank exceeding $2$, only few specific cases (see, for example, [17]).

A minimal zero sequence $S$ in a finite abelian group $G$ is a sequence such that the product of its elements is $1$, but each proper subsequence is free of product-$1$ subsequences. In [7, Theorem 6.4], W. Gao and A. Geroldinger showed that if $|S|=D(G)$ then $S$ contains some element $g\in G$ with order $\mathop{\rm ord}\nolimits(g)=\exp(G)$ for certain groups such as $p$-groups, cyclic groups, groups with rank two and groups that are the sum of two elementary $p$-groups. They also conjectured that the same conclusion holds for every finite abelian group.

For non-abelian groups, nothing was known, until the paper [4] where we solved the inverse problem associated to the metacyclic group

[TABLE]

Specifically, we proved the following result:

Theorem 1.2** ([4, Theorem 1.2]).**

Let $q$ be a prime number, $m\geq 2$ be a divisor of $q-1$ and $s\in\mathbb{Z}_{q}^{*}$ such that $\mathop{\rm ord}\nolimits_{q}(s)=m$.

Let $S$ be a sequence in the metacyclic group $C_{q}\rtimes_{s}C_{m}$ with $m+q-2$ elements.

1. (1)

If $(m,q)\neq(2,3)$ then the following statements are equivalent:

1. (i)

$S$* is free of product-$1$ subsequences;* 2. (ii)

For some $1\leq t\leq q-1$, $1\leq i\leq m-1$ such that $\gcd(i,m)=1$ and $0\leq\nu_{1},\dots,\nu_{m-1}\leq q-1$,

[TABLE] 2. (2)

If $(m,q)=(2,3)$ then $S$ is free of product-$1$* subsequences if and only if*

[TABLE]

In this article, we characterize the maximal sequences which are free of product-$1$ subsequences for the Dihedral Groups and the Dicyclic Groups. In particular, we show that these sequences have a property similar to Property C.

Let $n\geq 2$. Denote by $D_{2n}\simeq C_{n}\rtimes_{-1}C_{2}$ the Dihedral Group of order $2n$, i.e., the group generated by $x$ and $y$ with relations:

[TABLE]

Let $n\geq 2$. Denote by $Q_{4n}$ the Dicyclic Group, i.e., the group generated by $x$ and $y$ with relations:

[TABLE]

We have $Z(Q_{4n})=\{1,y^{n}\}$, where $Z(G)$ denotes the center of a group $G$. In addition:

[TABLE]

where $D_{2n}$ is the Dihedral Group of order $2n$.

Specifically, we prove the following result:

Theorem 1.3**.**

Let $S$ be a sequence in the Dihedral Group $D_{2n}$ with $n$ elements, where $n\geq 3$.

1. (1)

If $n\geq 4$ then the following statements are equivalent:

1. (i)

$S$* is free of product-$1$ subsequences;* 2. (ii)

For some $1\leq t\leq n-1$ with $\gcd(t,n)=1$ and $0\leq s\leq n-1$,

[TABLE] 2. (2)

If $n=3$ then $S$ is free of product-$1$* subsequences if and only if*

[TABLE]

Notice that this theorem reduces to Theorem 1.2 in the case when $n$ is prime and when $m=2$, since we have $s\equiv-1\pmod{n}$. For $n\geq 4$, it is easy to check that $(ii)\Longrightarrow(i)$. Observe that the case $n=3$ is exactly the item $(2)$ of Theorem 1.2, since $D_{6}\simeq C_{3}\rtimes_{-1}C_{2}$. For $n=2$, we have $D_{4}\simeq\mathbb{Z}_{2}^{2}$ and it is easy to check that $S$ is free of product-$1$ subsequences if and only if $S=(x,y)$, $S=(xy,y)$ or $S=(x,xy)$.

As a consequence of previous theorem, we obtain:

Theorem 1.4**.**

Let $S$ be a sequence in the Dicyclic Group $Q_{4n}$ with $2n$ elements, where $n\geq 2$.

1. (1)

If $n\geq 3$ and $|S|=2n$ then the following statements are equivalent:

1. (i)

$S$* is free of product-$1$ subsequences;* 2. (ii)

For some $1\leq t\leq n-1$ with $\gcd(t,2n)=1$ and $0\leq s\leq 2n-1$,

[TABLE] 2. (2)

If $n=2$ then $S$ is free of product-$1$* subsequences if and only if, for some $r\in\mathbb{Z}_{4}^{*}$ and $s\in\mathbb{Z}_{4}$, $S$ has one of the forms*

[TABLE]

Again, if $n\geq 3$ then it is easy to check that $(ii)\Longrightarrow(i)$, therefore we just need to show that $(i)\Longrightarrow(ii)$. If $n=2$ then it is easy to check that the sequences of the form $(y^{r},y^{r},y^{r},xy^{s})$, $(y^{r},xy^{s},xy^{s},xy^{s})$, $(xy^{s},xy^{s},xy^{s},xy^{r+s})$ are free of product-$1$ subsequences, therefore we just need to show that all other sequences with $2n$ elements have subsequences with product $1$.

Also note that if $n=2$ then $Q_{8}$ is isomorphic to the Quaternion Group, i.e. the group defined by

[TABLE]

This isomorphism may be described, for example, by $x\mapsto i,y\mapsto j$ (or its natural permutations, by the simmetry of $i,j,k$). The above theorem says that, in terms of Quarternion Group, extremal sequences free of product-$1$ subsequence are of the forms

[TABLE]

The main technical difficulty in our present proofs lies on the fact that $n$ may not be prime, therefore we cannot use Cauchy-Davenport inequality (see [12, p. 44-45]) and Vosper’s Theorem (see [18]), as used in the proof of Theorem 1.2 (see [4]). Instead, we now exhibit sequences with product-$1$ in cases not covered by those given forms. In section 3 we prove Theorem 1.3, solving the extremal inverse zero-sum problem associated to Davenport constant for dihedral groups. The proof of Theorem 1.4 is split up into sections 4, 5 and 6, where we solve the extremal inverse zero-sum problem associated to Davenport constant for dicyclic groups of orders $4n$, where $n\geq 4$, $n=3$ and $n=2$, respectively. The case $n\geq 4$ is a direct consequence of Theorem 1.3 and the case $n=3$ follows from Theorem 1.3, but in the special case $(2)$. The proof in the case $n=2$ is done manually, using only Theorem 2.2 to reduce the number of cases, without using Theorem 1.3.

2. Notation and Auxiliary Results

In this section we present the notation and auxiliary lemmas and theorems that we use throughout the paper.

Let $G$ be a finite group written multiplicatively and $S=(g_{1},g_{2},\dots,g_{l})$ be a sequence of elements of $G$. We denote by $|S|=l$ the length of $S$.

For each subsequence $T$ of $S$, i.e. $T=(g_{n_{1}},g_{n_{2}},\dots,g_{n_{k}})$, where $\{n_{1},n_{2},\dots,n_{k}\}$ is a subset of $\{1,2,\dots,l\}$, we say that $T$ is a product-$1$ subsequence when

[TABLE]

for some permutation $\sigma$ of $\{n_{1},\dots,n_{k}\}$, and if there are no such product-$1$ subsequences then we say that $S$ is free of product-$1$ subsequences.

If $S_{1}=(g_{i_{1}},\dots,g_{i_{u}})$ and $S_{2}=(g_{j_{1}},\dots,g_{j_{v}})$ are subsequences of $S$, then

• •

$SS_{1}^{-1}$ denotes the subsequence formed by the elements of $S$ without the elements of $S_{1}$;

• •

$S_{1}S_{2}=(g_{i_{1}},\dots,g_{i_{u}},g_{j_{1}},\dots,g_{j_{v}})$ denotes the concatenation of $S_{1}$ and $S_{2}$;

• •

$S^{k}=SS\dots S$ denotes the concatenation of $k$ identical copies of $S$’s.

For the group $G=D_{2n}=\langle x,y|x^{2}=1,\ y^{n}=1,\ yx=xy^{-1}\rangle$, let

• •

$H_{D}$ be the normal cyclic subgroup of order $n$ generated by $y$;

• •

$N_{D}=D_{2n}\setminus H_{D}=x\cdot H_{D}$.

The product of any even number of elements in $N_{D}$ is in $H_{D}$, since

[TABLE]

For the group $G=Q_{4n}=\langle x,y|x^{2}=y^{n},\ y^{2n}=1,\ yx=xy^{-1}\rangle$, let

• •

$H_{Q}$ be the normal cyclic subgroup of order $2n$ generated by $y$;

• •

$N_{Q}=Q_{4n}\setminus H_{Q}=x\cdot H_{Q}$.

The product of any even number of elements in $N_{Q}$ is in $H_{Q}$, since

[TABLE]

The following theorem is known as “Davenport constant of $\mathbb{Z}_{n}$ with weights $\{\pm 1\}$”, and will be used in the proof of Theorem 1.3.

Lemma 2.1** ([1, Lemma 2.1]).**

Let $n\in\mathbb{N}$ and $(y_{1},\dots,y_{s})$ be a sequence of integers with $s>\log_{2}n$. Then there exist a nonempty $J\subset\{1,2,3,\dots,s\}$ and $\varepsilon_{j}\in\{\pm 1\}$ for each $j\in J$ such that

[TABLE]

As a generalization of Theorem 1.1, we will use the following:

Theorem 2.2** ([11, Theorem 2.1], see also [3]).**

Let $G$ be a cyclic group of order $n\geq 3$ and $S$ be a zero-sum free sequence in $G$ of length $|S|\geq(n+1)/2$. Then there exists some $g\in S$ that appear at least $2|S|-n+1$ times in $S$. In particular, $D(G)=n$ and the following statements hold:

1. (1)

If $|S|=n-1$, then $S=(g)^{n-1}$. 2. (2)

If $|S|=n-2$, then either $S=(g)^{n-2}$ or $S=(g)^{n-3}(g^{2})$. 3. (3)

If $|S|=n-3$, then $S$ has one of the following forms:

[TABLE]

3. Proof of Theorem 1.3

We just need to show that $(i)\Longrightarrow(ii)$. Let $S$ be a sequence in $D_{2n}$ with $n$ elements that is free of product-$1$ subsequences. If $S\cap N_{D}$ contains two identical elements then $S$ is not free of product-$1$ subsequences by Equation 2.1. Hence, we assume that the elements of $S\cap N_{D}$ are all distinct.

From now on, we consider some cases, depending on the cardinality of $S\cap H_{D}$:

1. (a)

Case $|S\cap H_{D}|=n$: In this case, $S$ is contained in the cyclic subgroup of order $n$. Since $D(\mathbb{Z}_{n})=n$, $S$ contains some non-empty subsequence with product $1$. 2. (b)

Case $|S\cap H_{D}|=n-1$: In this case, by Theorem 2.2, the elements of $S\cap H_{D}$ must all be equal, say, $S\cap H_{D}=(y^{t})^{n-1}$ where $\gcd(t,n)=1$, and so $S=(y^{t})^{n-1}(xy^{s})$. 3. (c)

Case $|S\cap H_{D}|=n-2$: In this case, by Theorem 2.2, $S$ must have one of these forms:

1. (c-1)

Subcase $S=(y^{t})^{n-2}(xy^{u},xy^{v})$: Notice that we can obtain the products

[TABLE]

Since $1\leq k\leq n-2$, it’s enough to take $k\equiv(v-u)t^{-1}\pmod{n}$. The only problem occurs when $k=n-1$, that is, when $v-u+t\equiv 0\pmod{n}$, but in this case we switch $u$ and $v$ and so

[TABLE] 2. (c-2)

Subcase $S=(y^{t})^{n-3}(y^{2t},xy^{u},xy^{v})$: Notice that we can obtain the products

[TABLE]

Since $1\leq k\leq n-1$, it’s enough to take $k\equiv(v-u)t^{-1}\pmod{n}$. 4. (d)

Case $|S\cap H_{D}|=n-3$: In this case, by Theorem 2.2, $S$ must contain at least $n-5$ copies of some $y^{t}$, where $\gcd(t,n)=1$. Also, suppose that

[TABLE]

By renaming $z=y^{t}$, we may assume without loss of generality that $t=1$. By Pigeonhole Principle it follows that there exist two exponents of $y$ with difference in $\{1,2,\dots,\lfloor n/3\rfloor\}\pmod{n}$, say,

[TABLE]

We may ensure that

[TABLE]

for some $1\leq r\leq n/3$ provided there are enough $y$’s. But if $n\geq 8$ then $r\leq n/3\leq n-5$, so there are enough $y$’s and the theorem follows in these cases. Since the theorem is already proved for $n$ prime, it only remains to prove for $n\in\{4,6\}$.

For $n=4$, we have $S=(xy^{\alpha},xy^{\beta},xy^{\gamma},y)$ and some of the $\alpha,\beta,\gamma$ are consecutives modulo $4$. Without loss of generality, suppose that $\alpha-\beta\equiv 1\pmod{4}$, so $xy^{\beta}\cdot y\cdot xy^{\alpha}=1$.

For $n=6$, we have the cases

[TABLE]

If the set $\{\alpha,\beta,\gamma\}\pmod{6}$ contains two consecutive elements, say $\alpha\equiv\beta+1\pmod{6}$, then

[TABLE]

Otherwise, the only possibilities are $\{\alpha,\beta,\gamma\}\pmod{6}=\{0,2,4\}$ or $\{\alpha,\beta,\gamma\}\pmod{6}=\{1,3,5\}$. Suppose that $\alpha\equiv\beta+2\pmod{6}$ and notice that in any option for $S$ it is possible to take a product $y^{2}$ coming from $S\cap H_{D}$. Therefore

[TABLE] 5. (e)

Case $|S\cap H_{D}|=n-k,4\leq k\leq n$: Suppose that

[TABLE]

It follows from Theorem 2.1 that if

[TABLE]

then there exist a linear combination of a subset of

[TABLE]

with coefficients $\pm 1$ summing [math]. Suppose without loss of generality that, in this combination,

[TABLE]

appear with signal $-1$ and

[TABLE]

appear with signal $+1$. Then

[TABLE]

Thus the theorem is true for $k>2\lfloor\log_{2}n\rfloor$ + 1.

Hence, we may assume $4\leq k\leq 2\lfloor\log_{2}n\rfloor+1$. Theorem 2.2 implies without loss of generality that $t_{i}=t$ for $1\leq i\leq n-2k+1$. By renaming $z=y^{t}$, we may assume without loss of generality that $t=1$. Since $k\geq 4$, Pigeonhole Principle implies that there exist $\alpha_{i},\alpha_{j}$ such that

[TABLE]

Notice that if

[TABLE]

then

[TABLE]

for some $0\leq r\leq\lfloor n/4\rfloor$. But if $n\geq 8$ then $3n\geq 8\log_{2}n\geq 4(k-1)$, therefore Equation 3.2 holds in this case. Since the theorem is already proved for $n$ prime, it only remains to prove for $n\in\{4,6\}$.

For $n=4$, the only possibility is $S=(x,xy,xy^{2},xy^{3})$. Thus, $x\cdot xy\cdot xy^{3}\cdot xy^{2}=1$.

For $n=6$, there are three subcases to consider:

• •

Subcase $k=4$: Let $S=(y^{t_{1}},y^{t_{2}},xy^{\alpha_{1}},xy^{\alpha_{2}},xy^{\alpha_{3}},xy^{\alpha_{4}})$. Then either there are two pairs of consecutive $\alpha_{i}$’s modulo $6$ or there are three consecutive $\alpha_{i}$’s modulo $6$.

• –

If there are two pairs of consecutive $\alpha_{i}$’s, say $\alpha_{1}+1\equiv\alpha_{2}$ and $\alpha_{3}+1\equiv\alpha_{4}\pmod{6}$, then

[TABLE]

• –

If there are three consecutive $\alpha_{i}$’s, say $\alpha_{1}+2\equiv\alpha_{2}+1\equiv\alpha_{3}\pmod{6}$, then $\alpha_{4}$ can be any element in the set $\{\alpha_{3}+1,\alpha_{3}+2,\alpha_{3}+3\}$. If $\alpha_{4}\equiv\alpha_{3}+1$ or $\alpha_{4}\equiv\alpha_{3}+3\pmod{6}$ then we return to the previous item. Therefore we may assume $\alpha_{4}\equiv\alpha_{3}+2\pmod{6}$. Taking the products $xy^{\alpha_{i}}\cdot xy^{\alpha_{j}}=y^{\alpha_{j}-\alpha_{i}}$, we can get any element in $\{y,y^{2},y^{3},y^{4},y^{5}\}$. For example,

[TABLE]

Let $i,j$ such that $\alpha_{i}-\alpha_{j}\equiv t_{1}\pmod{6}$. Then

[TABLE]

• •

Subcase $k=5$: Let $S=(y^{t},xy^{\alpha_{1}},xy^{\alpha_{2}},\dots,xy^{\alpha_{5}})$. In this case, there are four consecutive $\alpha_{i}$’s modulo $6$, say $\alpha_{1}+3\equiv\alpha_{2}+2\equiv\alpha_{3}+1\equiv\alpha_{4}\pmod{6}$. Then

[TABLE]

• •

Subcase $k=6$: The only possibility is $S=(x,xy,xy^{2},xy^{3},xy^{4},xy^{5})$. So

[TABLE]

$\square$

4. Proof of Theorem 1.4 in the case $n\geq 4$

We just need to prove that $(i)\Longrightarrow(ii)$. Let $S$ be a sequence in $Q_{4n}$ with $2n$ elements that is free of product-$1$ subsequences. We consider some cases depending on the cardinality of $S\cap H_{Q}$:

1. (5.1)

Case $|S\cap H_{Q}|=2n$: In this case, $S$ is contained in the cyclic subgroup of order $2n$. Since $D(H_{Q})=D(\mathbb{Z}_{2n})=2n$, $S$ contains some non-empty subsequence with product $1$. 2. (5.2)

Case $|S\cap H_{Q}|=2n-1$: In this case, by Theorem 2.2, the elements of $S\cap H_{Q}$ must all be equal, say, $S\cap H_{Q}=(y^{t})^{2n-1}$ where $\gcd(t,2n)=1$, and so $S=(y^{t})^{2n-1}(xy^{s})$. 3. (5.3)

Case $|S\cap H_{Q}|=2n-2$: In this case, let $S_{1}$ be a subsequence of $S$ such that $|S_{1}\cap H_{Q}|=n-2$ and let $S_{2}=SS_{1}^{-1}$. Then $S_{2}$ is a sequence in $H_{Q}$ with $n$ elements. Since

[TABLE]

Theorem 1.3 tells us that $S_{1}$ and $S_{2}$ must contain subsequences $T_{1}$ and $T_{2}$, respectively, with products in $\{1,y^{n}\}$. If some of these products is $1$ then we are done. Otherwise, both products are $y^{n}$, therefore

[TABLE]

thus $S$ is not free of product-$1$ subsequences. 4. (5.4)

Case $|S\cap H_{Q}|=2n-3$: In this case, let $S_{1}$ be a subsequence of $S$ such that $|S_{1}\cap H_{Q}|=n-3$ and let $S_{2}=SS_{1}^{-1}$. Then $S_{2}$ is a sequence in $H_{Q}$ with $n$ elements. The argument is similar to the above case, thus $S$ is not free of product-$1$ subsequence. 5. (5.5)

Case $|S\cap H_{Q}|=2n-k,4\leq k\leq 2n$: In this case, let $S_{1}$ be a subsequence of $S$ such that $|S_{1}\cap H_{Q}|\leq n-2$ and $S_{2}=SS_{1}^{-1}$ is such that $|S_{2}\cap H_{Q}|\leq n-2$. The argument is similar to the above cases, thus $S$ is not free of product-$1$ subsequence.

Therefore, the proof for $n\geq 4$ is complete. $\square$

5. Proof of Theorem 1.4 in the case $n=3$

We have $Q_{12}=\langle x,y|\;x^{2}=y^{3},y^{6}=1,yx=xy^{5}\rangle$ and we just need to prove that $(i)\Longrightarrow(ii)$. Let $S$ be a sequence in $Q_{12}$ with $6$ elements that is free of product-$1$ subsequences. We consider some cases depending on the cardinality of $S\cap H_{Q}$:

1. (6.1)

Case $|S\cap H_{Q}|=6$: In this case, $S$ must contain a product-$1$ subsequence, since $D(H_{Q})=D(\mathbb{Z}_{6})=6$. 2. (6.2)

Case $|S\cap H_{Q}|=5$: In this case, Theorem 2.2 says that $S\cap H_{Q}=(y^{r})^{5}$ where $r\in\{1,5\}$, therefore $S$ is of the form $(y^{r})^{5}(xy^{s})$. 3. (6.3)

Case $|S\cap H_{Q}|=4$: In this case, we decompose $S=S_{1}S_{2}$ where $|S_{i}|=3$ for $i\in\{1,2\}$, $|S_{1}\cap H_{Q}|=1$ and $|S_{2}\cap H_{Q}|=3$, and use the same argument than item $(5.3)$, therefore $S$ is not free of product-$1$ subsequences. 4. (6.4)

Case $|S\cap H_{Q}|=3$: In this case, we decompose $S=S_{1}S_{2}$ where $|S_{i}|=3$ for $i\in\{1,2\}$, $|S_{1}\cap H_{Q}|=1$ and $|S_{2}\cap H_{Q}|=2$, therefore

[TABLE]

Notice that $S_{1}$ contains a subsequence with product in $\{1,y^{3}\}$. Observe that if $r\neq t$ then we could also decompose $S=S_{1}^{\prime}S_{2}^{\prime}$ where

[TABLE]

So, $S_{1}^{\prime}$ and $S_{2}^{\prime}$ have subsequences with products in $\{1,y^{3}\}$ and we can use the same argument than item $(5.3)$. Therefore, $r=t$ and we can decompose $S=S_{1}^{\prime\prime}S_{2}^{\prime\prime}$ such that

[TABLE]

Notice that $S_{1}^{\prime\prime}$ contains a subsequence with product in $\{1,y^{3}\}$ and $S_{2}^{\prime\prime}$ does not contains a subsequence with product in $\{1,y^{3}\}$ if and only if $S_{2}^{\prime\prime}\!\pmod{\{1,y^{3}\}}=(x,xy,xy^{2})$. Hence, the only possibility for $S\!\pmod{\{1,y^{3}\}}$ is:

[TABLE]

and so the possibilities for $S\cap H_{Q}$ are

[TABLE]

Notice that the second, third, fifth and seventh possibilities contain subsequences with product $1$, therefore it only remains

[TABLE]

Observe that in every case, it is possible to find a subsequence with product either $y$ or $y^{5}$. We claim that $S\cap N_{Q}$ contains subsequences with product $y$ and $y^{5}$, which we can join with those $y$ or $y^{5}$ coming from $S\cap H_{Q}$ to get a product-$1$ subsequence. For this, a sufficient condition is the existence of two elements in $S\cap N_{Q}$ such that the exponents of $y$ have difference $2$, since $xy^{\alpha}\cdot xy^{\alpha+2}=y$ and $xy^{\alpha+2}\cdot xy^{\alpha}=y^{5}$. In fact, if $xy^{\beta}\in S\cap N_{Q}$ and $xy^{\beta-2},xy^{\beta+2}\not\in S\cap N_{Q}$ then $xy^{\beta+1},xy^{\beta-1}\in S\cap N_{Q}$, and so $(\beta+1)-(\beta-1)=2$. 5. (6.5)

Case $|S\cap H_{Q}|=2$: In this case, we decompose $S=S_{1}S_{2}$ where $|S_{i}|=3$ and $|S_{i}\cap H_{Q}|=1$, and use the same argument than item $(5.3)$, therefore $S$ is not free of product-$1$ subsequences. 6. (6.6)

Case $|S\cap H_{Q}|=1$: In this case, we decompose $S=S_{1}S_{2}$ where $|S_{i}|=3$, $|S_{1}\cap H_{Q}|=1$ and $|S_{2}\cap H_{Q}|=0$. Notice that $S_{2}$ contains a subsequence with product in $\{1,y^{3}\}$. The same argument than item $(5.3)$ does not apply if and only if $S_{2}\pmod{\{1,y^{3}\}}=(x,xy,xy^{2})$. Observe that we could also decompose $S=S_{1}^{\prime}S_{2}^{\prime}$ in such way that $|S_{i}^{\prime}|=3$ for $i\in\{1,2\}$, $|S_{1}^{\prime}\cap H_{Q}|=1$, $|S_{2}^{\prime}\cap H_{Q}|=0$ and $S_{2}^{\prime}\pmod{\{1,y^{3}\}}$ contains two elements with the same exponent in $y$. Therefore, the same argument than item $(5.3)$ applies for $S_{1}^{\prime}$ and $S_{2}^{\prime}$. 7. (6.7)

Case $|S\cap H_{Q}|=0$: In this case, we decompose $S=S_{1}S_{2}$ where $|S_{i}|=3$ and $|S_{i}\cap H_{Q}|=0$ for $i\in\{1,2\}$. Notice that $S_{i}$ contains a subsequence with product in $\{1,y^{3}\}$ if and only if $S_{i}\pmod{\{1,y^{3}\}}$ is not of the form $(x,xy,xy^{2})$. Hence, the same argument than item $(5.3)$ does not apply if and only if $S_{i}\pmod{\{1,y^{3}\}}=(x,xy,xy^{2})$ for $i\in\{1,2\}$. Observe that we could also decompose $S=S_{1}^{\prime}S_{2}^{\prime}$ in such way that $S_{1}^{\prime}\pmod{\{1,y^{3}\}}=(x,x,xy)$ and $S_{2}^{\prime}\pmod{\{1,y^{3}\}}=(xy,xy^{2},xy^{2})$. Therefore, the same argument than item $(5.3)$ applies for $S_{1}^{\prime}$ and $S_{2}^{\prime}$.

Therefore, the proof for $n=3$ is complete. $\square$

6. Proof of Theorem 1.4 in the case $n=2$

We have $Q_{8}=\langle x,y|\;x^{2}=y^{2},y^{4}=1,yx=xy^{3}\rangle$. Suppose that $S$ is a sequence in $Q_{8}$ with $4$ elements that is free of product-$1$ subsequences. We want to show that $S$ has some of those forms given in item $(2)$. For this, we consider some cases depending on the cardinality of $S\cap H_{Q}$:

1. (7.1)

Case $|S\cap H_{Q}|=4$: In this case, $S$ must contain a product-$1$ subsequence, since $D(H_{Q})=D(\mathbb{Z}_{4})=4$. 2. (7.2)

Case $|S\cap H_{Q}|=3$: In this case, Theorem 2.2 says that $S\cap H_{Q}=(y^{r})^{3}$ where $r\in\{1,3\}$, therefore $S$ is of the form $(y^{r})^{3}(xy^{s})$. 3. (7.3)

Case $|S\cap H_{Q}|=2$: In this case, the only possibilities for $S\cap H_{Q}$ making $S$ be free of product-$1$ subsequences are

[TABLE]

and in all these possibilities $S\cap H_{Q}$ possesses a subsequence with product $y^{2}$, namely $y\cdot y$, $y^{3}\cdot y^{3}$ or $y^{2}$ itself. On the other hand, the possibilities for $S\cap N_{Q}$ are

[TABLE]

for $s\in\mathbb{Z}_{4}$.

The cases $(xy^{s},xy^{s})$ can be eliminated, since $xy^{s}\cdot xy^{s}\cdot y^{2}=1$.

The cases $(x,xy^{2})$ and $(xy,xy^{3})$ can also be eliminated, since $xy^{s}\cdot xy^{s+2}=1$.

The other cases can be eliminated by the following table:

[TABLE] 4. (7.4)

Case $|S\cap H_{Q}|=1$: In this case, the possibilities for $S\cap H_{Q}$ are $(y)$, $(y^{2})$ and $(y^{3})$. On the other hand, $S\cap N_{Q}$ has three elements and, by Equation 2.2, we may assume that $x\in S$. Thus, the possibilities for $S\cap N_{Q}$ are

$(x,x,x)$, $(x,x,xy)$, $(x,x,xy^{2})$, $(x,x,xy^{3})$, $(x,xy,xy)$, $(x,xy,xy^{2})$, $(x,xy,xy^{3}),$

$(x,xy^{2},xy^{2})$, $(x,xy^{2},xy^{3})$ and $(x,xy^{3},xy^{3}).$

If $S$ contains $(x,xy^{2})$ or $(xy,xy^{3})$ then $S$ is not free of product-$1$ subsequences, since

[TABLE]

Therefore, the remainder possibilities are $(x,x,x)$, $(x,x,xy)$, $(x,x,xy^{3})$, $(x,xy,xy)$ and $(x,xy^{3},xy^{3})$. Notice that these last four possibilities contains two identical terms and two terms such that the exponents of $y$ have difference $1$ modulo $4$. Since

[TABLE]

we may discard these four cases. Therefore, the only remainder possibility is $(x,x,x)$, and so $S=(y)(xy^{s})^{3}$ or $S=(y^{3})(xy^{s})^{3}$. 5. (7.5)

Case $|S\cap H_{Q}|=0$: In this case, we also may assume $x\in S$. Therefore, the possibilities for $S$ are

$(x,x,x,x)$, $(x,x,x,xy)$, $(x,x,x,xy^{2})$, $(x,x,x,xy^{3})$, $(x,x,xy,xy)$, $(x,x,xy,xy^{2}),$

$(x,x,xy,xy^{3})$, $(x,x,xy^{2},xy^{2})$, $(x,x,xy^{2},xy^{3})$, $(x,x,xy^{3},xy^{3})$, $(x,xy,xy,xy),$

$(x,xy,xy,xy^{2})$, $(x,xy,xy,xy^{3})$, $(x,xy,xy^{2},xy^{2})$, $(x,xy,xy^{2},xy^{3}),$

$(x,xy^{2},xy^{2},xy^{2})$, $(x,xy^{2},xy^{2},xy^{3})$, $(x,xy^{2},xy^{3},xy^{3})$ and $(x,xy^{3},xy^{3},xy^{3}).$

If $S$ contains two pairs of identical elements, say $(x,x,x^{\alpha},x^{\alpha})$, then

[TABLE]

so we remove $(x,x,x,x)$, $(x,x,xy,xy)$, $(x,x,xy^{2},xy^{2})$ and $(x,x,xy^{3},xy^{3})$.

If $S$ contains some of the pairs $(x,xy^{2})$ or $(xy,xy^{3})$ then, by equations in 6.1, we may remove other $11$ possibilities, thus it only remains $(x,x,x,xy)$, $(x,x,x,xy^{3})$, $(x,xy,xy,xy)$ and $(x,xy^{3},xy^{3},xy^{3})$. Therefore, $S=(xy^{s})(xy^{r+s})^{3}$ for $r\in\mathbb{Z}_{4}^{*}$ and $s\in\mathbb{Z}_{4}$.

$\square$

Acknowledgements**.**

The second author would like to thank CAPES/Brazil for the PhD student fellowship.