On Erdős-Ginzburg-Ziv inverse theorems for Dihedral and Dicyclic groups
aa r X i v : . [ m a t h . C O ] A ug ON ERD ˝OS-GINZBURG-ZIV INVERSE THEOREMS FORDIHEDRAL AND DICYCLIC GROUPS
JUN SEOK OH AND QINGHAI ZHONG
Abstract.
Let G be a finite group and exp( G ) = lcm { ord( g ) | g ∈ G } . A finite unordered sequence ofterms from G , where repetition is allowed, is a product-one sequence if its terms can be ordered such thattheir product equals the identity element of G . We denote by s ( G ) (or E ( G ) respectively) the smallestinteger ℓ such that every sequence of length at least ℓ has a product-one subsequence of length exp( G )(or | G | respectively). In this paper, we provide the exact values of s ( G ) and E ( G ) for Dihedral andDicyclic groups and we provide explicit characterizations of all sequences of length s ( G ) − E ( G ) − G ) (or | G | respectively). Introduction
Let G be a finite group and exp( G ) = lcm { ord( g ) | g ∈ G } . By a sequence S over G , we mean a finitesequence of terms from G which is unordered, repetition of terms allowed. We say that S is a product-onesequence if its terms can be ordered so that their product equals the identity element of G . The smallDavenport constant d ( G ) is the maximal integer ℓ such that there is a sequence of length ℓ which hasno non-trivial product-one subsequence. We denote by s ( G ) (or E ( G ) respectively) the smallest integer ℓ such that every sequence of length at least ℓ has a product-one subsequence of length exp( G ) (or | G | respectively). When G is cyclic, we have that exp( G ) = | G | and s ( G ) = E ( G ) = d ( G ) + | G | , which is dueto Erd˝os, Ginzburg, and Ziv in 1961, whence s ( G ) is called the Erd˝os-Ginzburg-Ziv constant of G .Both invariants, s ( G ) and E ( G ), found wide attention for finite abelian groups (see [13, 18, 16, 21] forsurveys). Caro ([3]) and Gao ([11]) proved independently that E ( G ) = d ( G ) + | G | (for a short proof see[18, Proposition 5.7.9]). This result has seen far reaching generalizations ([21, Chapter 16]). Much less isknown for s ( G ). If G ∼ = C n ⊕ C n with 1 ≤ n | n , then s ( G ) = 2 n + 2 n − s ( G ) is known only in very special cases ([9, 10, 19, 20]).The study of sequences and associated invariants for non-abelian groups dates back to the 1970s (e.g.,[28]), and fresh impetus came from applications in factorization theory and invariant theory ([6, 17, 22,4, 5, 8, 25, 26]). Among others, the formula E ( G ) = d ( G ) + | G | turned out to hold true for variousnon-abelian groups (see [14, 23, 24]). Gao and Lu ([15]) verified such formula for the dihedral groups oforder 2 n , where n ≥
23, and Bass ([1]) improved the result to all dihedral and dicyclic groups.
Theorem A. If G is a dihedral group of order n for n ≥ or a dicyclic group of order n for n ≥ respectively ) , then E ( G ) = d ( G ) + | G | = 3 n ( or E ( G ) = d ( G ) + | G | = 6 n respectively ) . In some earlier papers ([7, 24]), authors also studied an invariant s ′ ( G ) defined as the smallest in-teger ℓ such that every sequence over G of length at least ℓ has a product-one subsequence of lengthmax { ord( g ) | g ∈ G } . If G is nilpotent (in particular, if G is abelian), then exp( G ) = max { ord( g ) | g ∈ G } ,whence s ( G ) = s ′ ( G ). Similarly, if G is a dihedral group of order 2 n or a dicyclic group of order 4 n for Mathematics Subject Classification.
Key words and phrases. product-one sequences, Erd˝os-Ginzburg-Ziv constant, Dihedral groups, Dicyclic groups.This work was supported by the Austrian Science Fund FWF, W1230 Doctoral Program “Discrete Mathematics” andProject No. P28864–N35. even n , then we also have exp( G ) = max { ord( g ) | g ∈ G } . If G is a dihedral group of order 2 n with n odd, then there are arbitrarily long sequences over G having no product-one subsequence of lengthmax { ord( g ) | g ∈ G } (see the discussion after Definition 2.1). In the present paper, we provide the precisevalue of s ( G ) for the dihedral and dicyclic groups. Theorem 1.1.
Let n ∈ N . If G is a dihedral group of order n for n ≥ , then s ( G ) = (cid:26) n if n is odd n if n is even . If G is a dicyclic group of order n for n ≥ , then s ( G ) = (cid:26) n if n is odd n if n is even . After discussing the direct problem, which asks for the precise value of group invariants, we considerthe associated inverse problem, which asks for the structure of extremal sequences. Structural resultscharacterizing which sequences achieve equality are rare. Even in the abelian case, little is known preciselyoutside of groups of rank at most 2. Among others, we refer to [2, 27] for very recent works associatedthe Davenport constants for specific non-abelian groups. In the present paper, we focus on the inverseproblem associated to the Erd˝os-Ginzburg-Ziv constant, and we prove the following results.
Theorem 1.2.
Let n ∈ N ≥ , let G be a dihedral group of order n , and let S ∈ F ( G ) be a sequence. The following statements are equivalent :(a) n ≥ is even, | S | = s ( G ) − , and S has no product-one subsequence of length exp( G ) = n . (b) There exist α, τ ∈ G such that G = h α, τ | α n = τ = 1 G and τ α = α − τ i and S = ( α r ) [ n − · ( α r ) [ n − · α r τ , where r , r , r ∈ [0 , n − with gcd( r − r , n ) = 1 . The following statements are equivalent :(a) | S | = E ( G ) − and S has no product-one subsequence of length | G | . (b) There exist α, τ ∈ G such that G = h α, τ | α n = τ = 1 G and τ α = α − τ i and S has one ofthe following forms :(1) n ≥ and S = ( α r ) [2 n − · ( α r ) [ n − · α r τ , where r , r , r ∈ [0 , n − with gcd( r − r , n ) = 1 . (2) n = 3 , and either S = 1 [5] G · τ · ατ · α τ or S = ( α t ) [5] · ( α t ) [2] · α t τ , where t , t , t ∈ [0 , with gcd( r − r ,
3) = 1 . Theorem 1.3.
Let n ∈ N ≥ , let G be a dicyclic group of order n , and let S ∈ F ( G ) be a sequence. The following statements are equivalent :(a) n ≥ is even, | S | = s ( G ) − , and S has no product-one subsequence of length exp( G ) = 2 n . (b) There exist α, τ ∈ G such that G = h α, τ | α n = 1 G , τ = α n , and τ α = α − τ i and S has oneof the following forms :(1) n ≥ and S = ( α r ) [2 n − · ( α r ) [2 n − · α r τ , where r , r , r ∈ [0 , n − with gcd( r − r , n ) = 1 . (2) n = 2 and S ∈ (cid:8) ( α t ) [3] · ( α t ) [3] · α t τ, ( α t ) [3] · ( α t τ ) [3] · α t , ( α t ) [3] · ( α t τ ) [3] · α t τ (cid:9) ,where t , t , t , t ∈ [0 , such that t is even, t is odd, and t t (mod 2) . The following statements are equivalent :(a) | S | = E ( G ) − and S has no product-one subsequence of length | G | . (b) There exist α, τ ∈ G such that G = h α, τ | α n = 1 G , τ = α n , and τ α = α − τ i and S has oneof the following forms :(1) n ≥ and S = ( α r ) [4 n − · ( α r ) [2 n − · α r τ , where r , r , r ∈ [0 , n − with gcd( r − r , n ) = 1 . (2) n = 2 and S ∈ n(cid:0) ( α t ) [3] · ( α t ) [3] · α t τ (cid:1) · S , (cid:0) ( α t ) [3] · ( α t τ ) [3] · α t (cid:1) · S , (cid:0) ( α t ) [3] · ( α t τ ) [3] · α t τ (cid:1) · S o , where S ∈ (cid:8) ( α t ) [4] , ( α t ) [4] (cid:9) , S ∈ (cid:8) ( α t ) [4] , ( α t τ ) [4] (cid:9) , and t , t , t , t ∈ [0 , such that t is even, t is odd, and t t (mod 2) . RD ˝OS-GINZBURG-ZIV CONSTANT 3 Preliminaries
Much of the following notation can be found in [27] and is repeated here for the convenience of the reader.We denote by N the set of positive integers and we set N = N ∪{ } . For each k ∈ N , we also denote by N ≥ k the set of positive integers greater than or equal to k . For integers a, b ∈ Z , [ a, b ] = { x ∈ Z | a ≤ x ≤ b } is the discrete interval. Sequences over groups.
Let G be a multiplicatively written finite group with identity element 1 G and let G ⊂ G be a subset. For an element g ∈ G , we denote by ord( g ) ∈ N the order of g , byexp( G ) = lcm { ord( g ) | g ∈ G } the exponent of G , and by h G i ⊂ G the subgroup generated by G .The elements of the free abelian monoid F ( G ) will be called sequences over G . This terminology goesback to Combinatorial Number Theory. Indeed, a sequence over G can be viewed as a finite unorderedsequence of terms from G , where the repetition of elements is allowed. We briefly discuss our notationwhich follows the monograph [21, Chapter 10.1]. In order to avoid confusion between multiplication in G and multiplication in F ( G ), we denote multiplication in F ( G ) by the boldsymbol · and we use bracketsfor all exponentiation in F ( G ). In particular, a sequence S ∈ F ( G ) has the form(2.1) S = g · . . . · g ℓ = Y • i ∈ [1 ,ℓ ] g i ∈ F ( G ) , where g , . . . , g ℓ ∈ G are the terms of S . For g ∈ G , • v g ( S ) = |{ i ∈ [1 , ℓ ] | g i = g }| denotes the multiplicity of g in S , • supp( S ) = { g ∈ G | v g ( S ) > } denotes the support of S , and • h ( S ) = max { v g ( S ) | g ∈ G } denotes the maximal multiplicity of S .A subsequence T of S is a divisor of S in F ( G ) and we write T | S . For a subset H ⊂ G , we denote by S H the subsequence of S consisting of all terms from H . Furthermore, T | S if and only if v g ( T ) ≤ v g ( S ) forall g ∈ G , and in such case, S · T [ − denotes the subsequence of S obtained by removing the terms of T from S so that v g (cid:0) S · T [ − (cid:1) = v g ( S ) − v g ( T ) for all g ∈ G . On the other hand, we set S − = g − · . . . · g − ℓ to be the sequence obtained by taking elementwise inverse from S .Moreover, if S , S ∈ F ( G ) and g , g ∈ G , then S · S ∈ F ( G ) has length | S | + | S | , S · g ∈ F ( G )has length | S | + 1, g g ∈ G is an element of G , but g · g ∈ F ( G ) is a sequence of length 2. If g ∈ G , T ∈ F ( G ), and k ∈ N , then g [ k ] = g · . . . · g | {z } k ∈ F ( G ) and T [ k ] = T · . . . · T | {z } k ∈ F ( G ) . Let S ∈ F ( G ) be a sequence as in (2.1). Then we denote by π ( S ) = { g τ (1) . . . g τ ( ℓ ) ∈ G | τ a permutation of [1 , ℓ ] } ⊂ G and Π n ( S ) = [ T | S | T | = n π ( T ) ⊂ G , the set of products and n -products of S , and more generally, the subsequence products of S is denoted byΠ( S ) = [ n ≥ Π n ( S ) ⊂ G .
It can easily be seen that π ( S ) is contained in a G ′ -coset, where G ′ is the commutator subgroup of G .Note that | S | = 0 if and only if S = 1 F ( G ) , and in that case we use the convention that π ( S ) = { G } .The sequence S is called • a product-one sequence if 1 G ∈ π ( S ), • product-one free if 1 G / ∈ Π( S ), and • square-free if h ( S ) ≤ JUN SEOK OH AND QINGHAI ZHONG If S = g · . . . · g ℓ ∈ B ( G ) is a product-one sequence with 1 G = g . . . g ℓ , then 1 G = g i . . . g ℓ g . . . g i − forevery i ∈ [1 , ℓ ]. Every map of groups θ : G → H extends to a monoid homomorphism θ : F ( G ) → F ( H ),where θ ( S ) = θ ( g ) · . . . · θ ( g ℓ ). If θ is a group homomorphism, then θ ( S ) is a product-one sequence if andonly if π ( S ) ∩ ker( θ ) = ∅ . We denote by B ( G ) = (cid:8) S ∈ F ( G ) | G ∈ π ( S ) (cid:9) the set of all product-one sequences over G , and clearly B ( G ) ⊂ F ( G ) is a submonoid. We denote by A ( G ) the set of irreducible elements of B ( G ) which, in other words, is the set of minimal product-onesequences over G . Moreover, D ( G ) = sup (cid:8) | S | | S ∈ A ( G ) (cid:9) ∈ N ∪ {∞} is the large Davenport constant of G , and d ( G ) = sup (cid:8) | S | | S ∈ F ( G ) is product-one free (cid:9) ∈ N ∪ {∞} is the small Davenport constant of G . Ordered sequences over groups.
These are an important tool used to study (unordered) sequencesover non-abelian groups. Indeed, it is quite useful to have related notation for sequences in which the orderof terms matters. Thus, for a subset G ⊂ G , we denote by F ∗ ( G ) = (cid:0) F ∗ ( G ) , · (cid:1) the free (non-abelian)monoid with basis G , whose elements will be called the ordered sequences over G .Taking an ordered sequence in F ∗ ( G ) and considering all possible permutations of its terms gives riseto a natural equivalence class in F ∗ ( G ), yielding a natural map[ · ] : F ∗ ( G ) → F ( G )given by abelianizing the sequence product in F ∗ ( G ). For any sequence S ∈ F ( G ), we say that anordered sequence S ∗ ∈ F ∗ ( G ) with [ S ∗ ] = S is an ordering of the sequence S ∈ F ( G ).All notation and conventions for sequences extend naturally to ordered sequences. We sometimesassociate an (unordered) sequence S with a fixed (ordered) sequence having the same terms, also denotedby S . While somewhat informal, this does not give rise to confusion, and will improve the readability ofsome of the arguments.For an ordered sequence S = g · . . . · g ℓ ∈ F ∗ ( G ), we denote by π ∗ : F ∗ ( G ) → G the uniquehomomorphism that maps an ordered sequence onto its product in G , so π ∗ ( S ) = g . . . g ℓ ∈ G . If G is a multiplicatively written abelian group, then for every sequence S ∈ F ( G ), we always use π ∗ ( S ) ∈ G to be the unique product, and Π( S ) = S (cid:8) π ∗ ( T ) | T divides S and | T | ≥ (cid:9) ⊂ G . Definition 2.1.
We denote by • s ( G ) the smallest integer ℓ such that every sequence of length at least ℓ has a product-one subse-quence of length exp( G ), and • E ( G ) the smallest integer ℓ such that every sequence of length at least ℓ has a product-one subse-quence of length | G | .Note that max { ord( g ) | g ∈ G } ≤ exp( G ), and equality holds for nilpotent groups G . If G is a dihedralgroup of order 2 n with n even, then the equality holds true, whence our definition for s ( G ) coincides withthe one defined in the papers [7, 24]. However, when n is odd, the concepts are different. To see this, let n ∈ N ≥ be odd and let G = h α, τ | α n = τ = 1 G and τ α = α − τ } be a dihedral group of order 2 n . Thenmax { ord( g ) | g ∈ G } = n is odd, and it follows by the relation between α and τ that every sequence over G \ h α i cannot have a product-one subsequence of any odd length, in particular of length n . Lemma 2.2.
Let G be a finite group. Then s ( G ) ≥ d ( G ) + exp( G ) and E ( G ) ≥ d ( G ) + | G | . RD ˝OS-GINZBURG-ZIV CONSTANT 5
Proof.
We need to find a sequence of length d ( G ) + exp( G ) − d ( G ) + | G | − G ) (or | G | respectively). Take a product-one free sequence S over G of length | S | = d ( G ). Then S · [exp( G ) − G (or S · [ | G |− G respectively) is the desired sequence. (cid:3) For the rest of this section, we list the following preliminary results.
Lemma 2.3. [15, Lemma 7]
Let G be a finite abelian group of order n , let r ∈ N ≥ , and let S ∈ F ( G ) bea sequence of length | S | = n + r − . If S has no product-one subsequence of length n , then | Π n − ( S ) | = | Π r ( S ) | ≥ r − . The following lemma can be found in [12, Theorem 1] directly, and also in [16, Theorem 5.1.16 andProposition 5.1.14] as a consequence of the characterization of long sequences having required property.
Lemma 2.4.
Let G be a cyclic group of order n ≥ and let S ∈ F ( G ) be a sequence of length | S | =2 n − k ≥ n − for k ≥ . If S has no product-one subsequence of length n , then there exist g, h ∈ G with ord( gh − ) = n such that g [ u ] · h [ v ] | S , where u, v ≥ n − k + 3 and u + v ≥ n − k + 2 . Finally, we conclude the following direct result from the previous lemma.
Lemma 2.5.
Let G be a cyclic group of order n ≥ , and let S ∈ F ( G ) be a sequence of length | S | = 3 n − .Then the following statements are equivalent :(a) S has no product-one subsequence of length n . (b) S = g [2 n − · h [ n − , where g, h ∈ G with ord( gh − ) = n .Proof. (b) ⇒ (a) Since any sequence of the form g [ n − · h [ n − , where g, h ∈ G with ord( gh − ) = n , hasno product-one subsequence of length n , the sequence S = g [2 n − · h [ n − has no product-one subsequenceof length 2 n .(a) ⇒ (b) Let S ∈ F ( G ) be a sequence of length | S | = 3 n −
2. Suppose that S has no product-onesubsequence of length 2 n . Then it follows by s ( G ) = 2 n − S has a product-one subsequence S of length | S | = n , whence | S · S [ − | = 2 n −
2. Then Lemma 2.4 ensures that there exist g, h ∈ G withord( gh − ) = n such that S = S · g [ n − · h [ n − . If n = 2, then we clearly obtain that S = g [2] or S = h [2] .Thus we suppose that n ≥
3, and assume to the contrary that there exists x ∈ supp( S ) such that x = g, h .Then it follows again by Lemma 2.4 that x · g [ n − · h [ n − has a product-one subsequence of length n , say x · g [ r ] · h [ r ] , where r , r ∈ [1 , n −
3] such that r + r = n −
1. Then ( S · x [ − ) · g [ n − − r ] · h [ n − − r ] is a sequence of length 2 n −
2. Since S has no product-one subsequence of length 2 n , it follows again byLemma 2.4 that S · x [ − = g [ r ] · h [ r ] , whence g [ n ] · h [ n ] is a product-one subsequence of S of length2 n , a contradiction. Thus every element in supp( S ) is either g or h . If g, h ∈ supp( S ), then g [ n ] · h [ n ] is a product-one subsequence of S of length 2 n , a contradiction. Therefore we obtain either S = g [ n ] or S = h [ n ] . (cid:3) On Dihedral groups
In this section, let n ∈ N ≥ and let G be a dihedral group of order 2 n . Then we denote by H the cyclicsubgroup of G of index 2, and by G = G \ H . Note that d ( G ) = n and exp( G ) = lcm(2 , n ).We prove the technical lemmas by improving the argument from [1], and then drive our main results. Lemma 3.1.
Let n ∈ N ≥ be even and let S ∈ F ( G ) be a sequence of length | S | = 2 n − . If S has noproduct-one subsequence of length n , then | S G | = 1 . JUN SEOK OH AND QINGHAI ZHONG
Proof.
Let α, τ ∈ G such that G = h α, τ | α n = τ = 1 G and τ α = α − τ i . Then H = h α i . Let S ∈ F ( G )be a sequence of length | S | = 2 n − S has no product-one subsequence of length n . Note that | S G | ≥
1. Assume to the contrary that | S G | ≥ H = h α i , H = H \ H , G = (cid:8) α k τ | k ∈ [0 , n − (cid:9) , and G = G \ G . Suppose that S H = a · . . . · a r , S H = b · . . . · b s , S G = c · . . . · c t , and S G = d · . . . · d ℓ , where r, s, t, ℓ ∈ N . Then r + s + t + ℓ = 2 n − T = ( a a ) · . . . · ( a ⌊ r ⌋− a ⌊ r ⌋ ) · ( b b ) · . . . · ( b ⌊ s ⌋− b ⌊ s ⌋ ) · ( c c ) · . . . · ( c ⌊ t ⌋− c ⌊ t ⌋ ) · ( d d ) · . . . · ( d ⌊ ℓ ⌋− d ⌊ ℓ ⌋ ) . Then T ∈ F ( H ) of length | T | = (cid:4) r (cid:5) + (cid:4) s (cid:5) + (cid:4) t (cid:5) + (cid:4) ℓ (cid:5) ≥ n − (cid:0) n (cid:1) −
2. Since S has no product-onesubsequence of length n , we have that T has no product-one subsequence of length n . It follows by s ( H ) = 2 (cid:0) n (cid:1) − | T | = n − { r, s, t, ℓ } are odd. By Lemma 2.4, there exist g, h ∈ H with ord( gh − ) = n such that T = g [ n − · h [ n − . CASE 1. n = 4.If { r, s, t, ℓ } ⊂ N , then a b , c d ∈ H . Since | H | = 2, we obtain that either a b = c d or a b = ( c d ) − = d c . It follows that either c · a · b · d or a · b · c · d is a product-one subsequenceof S of length 4, a contradiction. Thus one element of { r, s, t, ℓ } must be zero.Suppose that r = 0, and the case when s = 0 follows by the same argument. Then all s, t, ℓ must beodd with s + t + ℓ = 7. Since | H | = | H | = | G | = | G | = 2, we obtain that there exist subsequences W , W of S such that W = x [2] and W = y [2] for some x, y ∈ G . If x, y ∈ G , then W · W is aproduct-one subsequence of S of length 4, a contradiction. If x ∈ H and y ∈ G (or else x ∈ G and y ∈ H respectively), then xyxy = 1 G (or yxyx = 1 G respectively), which implies that W · W is aproduct-one subsequence of S of length 4, a contradiction. If x, y ∈ H , then x = y or x = y − , whichimplies that W · W is a product-one subsequence of S of length 4, a contradiction.Suppose that t = 0, and the case when ℓ = 0 follows by the same argument. Then all r, s, ℓ must beodd with ℓ ≥ r + s + ℓ = 7. Since | H | = | H | = | G | = | G | = 2, we obtain that there existsubsequences W , W of S such that W = x [2] and W = y [2] , where x ∈ G and y ∈ G . If x ∈ G , then W · W is a product-one subsequence of S of length 4, a contradiction. If x ∈ H , then xyxy = 1 G , whichimplies that W · W is a product-one subsequence of S of length 4, a contradiction. CASE 2. n ≥ t + ℓ ≥ t ≥ ℓ ≥
2. If there exist distinct t , t ∈ [1 , t ] (or ℓ , ℓ ∈ [1 , ℓ ] respectively) such that c t = c t (or d ℓ = d ℓ respectively), then after renumbering if necessary, we may assume that c = c (or d = d respectively). By symmetry, we may suppose that c c = g (or d d = g respectively). Then h [ n − ] · c · h [ n − ] · c (cid:0) or h [ n − ] · d · h [ n − ] · d respectively (cid:1) is a product-one sequence, and by splittingelements equal to h to be subsequence of S of length 2, we infer that S has a product-one subsequence oflength n , a contradiction. Therefore h ( S G ) = 1 , and hence t + ℓ ≤ n . We only prove the case when t ≥
2, and the case when ℓ ≥ t ≥
2, and without loss of generality that c c = g .Suppose that n is even. Then n − ≥
3, and h [ n − ] · c · h [ n − ] · g · c is a product-one sequence.It follows by splitting the elements equals h and g to be subsequences of S length 2 that there exists aproduct-one subsequence of S of length n , a contradiction.Suppose that n is odd, and both t and ℓ are odd. Then we set T = T · ( c c ) [ − · ( c c t ) which stillhas no product-one subsequence of length n . Comparing with T , we have c c = c c t , and hence c = c t ,contradicting that h ( S G ) = 1. RD ˝OS-GINZBURG-ZIV CONSTANT 7
Suppose that n is odd, and both r and s are odd. Then we set T = T · ( a a ) [ − · ( a a r ) which stillhas no product-one subsequence of length n . Comparing with T , we have a a = a a r , and we obtain byproceeding the same argument that S H = a [ r ]1 and S H = b [ s ]1 . Since r + s = 2 n − − ( t + ℓ ) ≥ n − ≥ x, y ∈ H such that x [2] · y [2] | S H . If n ≥
10, then h [ n − ] · c · h [ n − ] · g · x · c · x is a product-onesequence, and by splitting the elements equal to h and g to be subsequences of S of length 2, we infer that S has a product-one subsequence of length n , a contradiction. If n = 6, then x = y , and hence either c · x [2] · y · c · y or c · y [2] · x · c · x is a product-one subsequence of S of length 6, a contradiction. CASE 3. n ≥ t + ℓ = 2.Then | S H | = 2 n −
3. Since t or ℓ must be odd, we have t = ℓ = 1. Suppose that S G = α i τ and S G = α j τ , where i, j ∈ [0 , n − S H has no product-one subsequence of length n , it follows bylemma 2.4 that there exist r , r ∈ [0 , n −
1] such that r is odd, r is even, and gcd( r − r , n ) = 1 suchthat ( α r ) [ u ] · ( α r ) [ u ] | S H for some u , u ≥ n − ≥ n with u + u ≥ n − ≥ n + 2. Then there exist v, w ∈ [0 , n −
1] such that i ≡ v ( r − r ) (mod n ) and j ≡ w ( r − r ) (mod n ) , and we set x to be | w − v | if | w − v | ≤ n , and n − | w − v | if | w − v | > n . Then x ∈ [1 , n ] is odd, and V = ( α r ) [ x ] · ( α r ) [ x ] · α i τ · α j τ is a product-one subsequence of S having even length. If x ≤ n −
1, thenby adding even terms of either α r or α r to V , we obtain a product-one subsequence of S of length n , acontradiction. Therefore x = n is odd and j ≡ n + i (mod n ).Suppose that gcd( r , n ) >
1. Since r is odd, we have gcd( r , n ) | n , and hence there exists x ∈ [1 , n − r x ≡ n (mod n ). Since n is odd, we obtain that x is odd. Hence n − x − ≥ α r ) [ x + n − x − ] · ( α i τ ) · ( α r ) [ n − x − ] · ( α j τ ) · ( α r ) [ n ] is a product-one subsequence of S of length n , a contradiction.Suppose that gcd( r , n ) = 1. If r = 0, then since u , u ≥ n , it follows that ( α r ) [ n − · ( α r ) [ n ] · α i τ · α j τ is a product-one subsequence of S of length n , a contradiction. Thus we assume that r = 0, and thenthere exists an odd x ∈ [1 , n ] \ (cid:8) n (cid:9) such that r x ≡ n + r (mod n ). Thus | n − x | ≥
2. If x ∈ [1 , n − n − x − ≥ α r ) [ x + n − x − ] · ( α i τ ) · ( α r ) [ n − x − ] · ( α r ) [1+ n − ] · ( α j τ ) · ( α r ) [ n − ] is a product-one subsequence of S of length n , a contradiction. If x ∈ [ n + 1 , n ], then 2 x − n − ≥ α r ) [ n − x + x − n − ] · ( α i τ ) · ( α r ) [ x − n − ] · ( α r ) [1+ n − ] · ( α j τ ) · ( α r ) [ n − ] is a product-one subsequence of S of length n , a contradiction. (cid:3) Proof of Theorem 1.1.1. If n ≥ s ( G ) = E ( G ) and the assertion follows from [1]. Supposethat n ≥ s ( G ) = 2 n . By Lemma 2.2, it suffices to show that every sequenceof length 2 n has a product-one subsequence of length n . Let S ∈ F ( G ) be a sequence of length 2 n , andassume to the contrary that S has no product-one subsequence of length n .Let T | S be a subsequence of length 2 n −
1. Then T has no product-one subsequence of length n .By Lemma 3.1, we have | T G | = 1, and hence | S · T [ − G | = 2 n −
1. Again by Lemma 3.1, we have (cid:12)(cid:12) ( S · T [ − G ) G (cid:12)(cid:12) = 1, which implies that | S G | = 2. Let W | S be a subsequence of length 2 n − S G | W . Then Lemma 3.1 ensures that W , and hence S , has a product-one subsequence of length n , acontradiction. (cid:3) Lemma 3.2.
Let n ∈ N ≥ and let S ∈ F ( G ) be a sequence of length | S | = 3 n − . If S has no product-one subsequence of length n , then either | S G | = 1 , or that n = 3 and S = 1 [5] G · g · g · g , where { g , g , g } = G \ H . JUN SEOK OH AND QINGHAI ZHONG
Proof.
Let α, τ ∈ G such that G = h α, τ | α n = τ = 1 G and τ α = α − τ i . Then H = h α i , and if n = 3,then G \ H = { τ, ατ, α τ } . Let S ∈ F ( G ) be a sequence of length | S | = 3 n − S has noproduct-one subsequence of length 2 n . Note that | S G | ≥
1. Assume to the contrary that | S G | ≥ S = 1 [5] G · τ · ατ · α τ when n = 3.1. We first prove the case when n ≥ g , g ∈ G be such that g · g | S G . Then | S · ( g · g ) [ − | = 3 n − ≥ n , and by Theorem 1.1.1, there exists a product-one subsequence T | S · ( g · g ) [ − of length n . Then | S · T [ − | = 2 n − | ( S · T [ − ) G | ≥ | g · g | = 2, whence Lemma 3.1 ensuresthat there exists a product-one subsequence T | S · T [ − of length n . Therefore T · T is a product-onesubsequence of S of length 2 n , a contradiction.2. Now we prove the case when n ≥ S G = (cid:0) a [2]1 · . . . · a [2] r (cid:1) · ( c · . . . · c u ) , where c i = c j for all distinct i, j ∈ [1 , u ]. Then u ≤ n . Similarly we set S H = (cid:0) b [2]1 · . . . · b [2] t (cid:1) · ( e · e − ) · . . . · ( e s · e − s ) · ( d · . . . · d v ) , where d i / ∈ { d j , d − j } for all distinct i, j ∈ [1 , v ], and ord( b k ) = 2 for all k ∈ [1 , v ]. Then v ≤ n +12 . CASE 1. r ≥ W be the maximal product-one subsequence of c · . . . · c u · d · . . . · d v of even length. Then | W | ≤ n and W = c · . . . · c u · d · . . . · d v · W [ − has no product-one subsequence of even length. If | W | ≤ n −
1, then 2 t + 2 r + 2 s + | W | ≥ n , and hence there exist t ∈ [0 , t ], r ∈ [0 , r ], and s ∈ [0 , s ]with 2 t + 2 r + 2 s + | W | = 2 n such that W · ( e · e − ) · . . . · ( e s · e − s ) · ( b · . . . · b t ) · a · ( b · . . . · b t ) · a · a [2]2 · . . . · a [2] r is a product-one subsequence of S of length 2 n , a contradiction. Since n is odd, we have | W | ≥ n + 1.Suppose that | W H | ≥
2. Let Ω = (cid:8) x ∈ H | x ∈ Π ( W H ) or x − ∈ Π ( W H ) (cid:9) . Since Π ( W H ) ⊂ H \ { G } , we have | Ω | ≥ | W H | , and hence | Π ( W G ) | + | Ω | ≥ (cid:0) | W G | − (cid:1) + | W H | ≥ n = | H | . It followsby Π ( W G ) ∪ Ω ⊂ H \ { G } that there exist ordered sequences T | W G and T | W H with | T | = | T | = 2such that π ∗ ( T ) = π ∗ ( T ) ∈ Π ( W G ) ∩ Ω, which implies that T · T is a product-one subsequence of W of length 4, a contradiction to the maximality of W .Suppose that | W H | = 1. Then | W G | = n . If n ≥
5, then W G has a product-one subsequence oflength 4, a contradiction to the maximality of W . If n = 3, then it is easy to verify that there are g , g ∈ supp( W G ) such that T = a [2]1 · g · g is a product-one sequence. Note that 4 = 2 r + 2 t + 2 s + | W | .If W is non-trivial, then since r ≥
1, we must have | W | = 2, whence W · T is a product-one subsequenceof S of length 6, a contradiction. Thus W is a trivial sequence, and then r + t + s = 2. Hence we inferthat S has a product-one subsequence of length 6, a contradiction. CASE 2. r = 0.Then 2 ≤ | S G | = u ≤ n , and we proceed by the following assertion. A. For every non-trivial product-one subsequence W of S G , we have u − | W | ≥ n +12 .Proof of A . Assume to the contrary that S G = c · . . . · c u has a non-trivial product-one subsequence W such that u − | W | ≤ n − . Since | W | is even and (cid:12)(cid:12) (cid:0) b [2]1 · . . . · b [2] t (cid:1) · ( e · e − ) · . . . · ( e s · e − s ) · W (cid:12)(cid:12) ≥ (3 n − − n + 12 − n −
12 = 2 n − , it follows that (cid:0) b [2]1 · . . . · b [2] t (cid:1) · ( e · e − ) · . . . · ( e s · e − s ) · W has a product-one subsequence of length 2 n , acontradiction. (cid:3) RD ˝OS-GINZBURG-ZIV CONSTANT 9
SUBCASE 2.1. u = n .Since all c i are distinct, we may assume by renumbering if necessary that c i = α i τ for every i ∈ [1 , n − c u = c n = τ . If n = 3, then S G = τ · ατ · α τ and | S H | = 5. If α i ∈ Π ( S H ) for some i ∈ [1 , S has a product-one subsequence of length 6, a contradiction. Thus Π ( S H ) = { G } , which implies that S H = 1 [5] G and S = τ · ατ · α τ · [5] G , a contradiction to our assumption.Suppose that n ≥
5. If n − is even, then ατ · . . . · α n − τ is a product-one subsequence of S G , acontradiction to A . If n − is odd, then n ≥ τ · α τ · α τ · . . . · α n − τ is a product-one subsequenceof S G , again a contradiction to A . SUBCASE 2.2. u = 2.Let S G = α i τ · α j τ for distinct i, j ∈ [0 , n − n = 3, then | S H | = 6 and π ( S G ) = { α i − j , α j − i } = { α, α } . If α k ∈ Π ( S H ) for some k ∈ [1 , S has a product-one subsequence of length 6, acontradiction. Thus Π ( S H ) = { G } , and hence S H = 1 [6] G is a product-one subsequence of length 6, acontradiction.Suppose that n ≥
5. Since | S H | = 3 n −
3, it follows by s ( H ) = 2 n − T | S H of length n , whence | S H · T [ − | = 2 n −
3. Since S has no product-one subsequenceof length 2 n , it follows by Lemma 2.4 that there exist r , r ∈ [0 , n −
1] with gcd( r − r , n ) = 1 such that( α r ) [ γ ] · ( α r ) [ γ ] (cid:12)(cid:12) S H · T [ − , where γ , γ ≥ n − ≥ n −
12 and γ + γ ≥ n − ≥ n + 1 . Since gcd( r − r , n ) = 1, there exists x ∈ [0 , n −
1] such that ( r − r ) x ≡ j − i + r (mod n ). If x ≤ n − ,then V = ( α r ) [ x − · α i τ · ( α r ) [ x ] · α j τ is a product-one subsequence of S . Since γ + γ ≥ n + 1, byadding even terms of either α r or α r to V , we obtain that S · T [ − has a product-one subsequence oflength n , a contradiction. If x ≥ n +32 , then n − x + 1 ≤ n − and V = ( α r ) [ n − x +1] · α j τ · ( α r ) [ n − x ] · α i τ is a product-one subsequence of S . Since γ + γ ≥ n + 1, by adding even terms of either α r or α r to V , we obtain that S · T [ − has a product-one subsequence of length n , a contradiction. Thus x = n +12 ,and again by gcd( r − r , n ) = 1, there exists y ∈ [0 , n −
1] such that ( r − r ) y ≡ i − j + r (mod n ). Asimilar argument shows that y = n +12 = x , and hence i − j + r ≡ j − i + r (mod n ). Since n is odd, itfollows that i = j , a contradiction. SUBCASE 2.3. u ∈ [3 , n − | S H | = 3 n − u − ≥ s ( H ) = 2 n − S H has a product-one subsequence T of length n . Then S H · T [ − is a sequence over H of length 2 n − u − n , and thus Lemma 2.3 ensures that (cid:12)(cid:12) Π n − (cid:0) S H · T [ − (cid:1)(cid:12)(cid:12) ≥ n − u . Since all c i are distinct, we havethat all c c , c c , . . . , c c u are distinct, and hence (cid:12)(cid:12) Π ( c · . . . · c u ) (cid:12)(cid:12) ≥ u − SUBCASE 2.3.1 (cid:12)(cid:12) Π ( c · . . . · c u ) (cid:12)(cid:12) ≥ u .If Π n − (cid:0) S H · T [ − (cid:1) ∩ Π ( c · . . . · c u ) = ∅ , then there exists a product-one subsequence T | S · T [ − oflength n , a contradiction. Thus Π n − (cid:0) S H · T [ − (cid:1) ∩ Π ( c · . . . · c u ) = ∅ , implying that (cid:12)(cid:12) Π ( c · . . . · c u ) (cid:12)(cid:12) = u and (cid:12)(cid:12) Π n − (cid:0) S H · T [ − (cid:1)(cid:12)(cid:12) = n − u . Note that if h ∈ Π ( c · . . . · c u ), then h − ∈ Π ( c · . . . · c u ), for otherwise, S · T [ − must have a product-one subsequence of length n , again a contradiction. Since n is odd andΠ ( c · . . . · c u ) ⊂ H \ { G } , we must have that u is even. Since all c c , c c , . . . , c c u are distinct, afterrenumbering if necessary, we may assume that c c k = ( c c k +1 ) − for all k ∈ [1 , u − c i = α r i τ for all i ∈ [1 , u ]. Then 2 r ≡ r + r ≡ . . . ≡ r u − + r u − (mod n ), and hence for every k ∈ N such that (2 + 4 k ) + 3 ≤ u −
1, we have c k c (2+4 k )+2 c (2+4 k )+1 c (2+4 k )+3 = α r k + r (2+4 k )+1 − r (2+4 k )+2 − r (2+4 k )+3 = 1 G . Suppose that u ≥ | u −
2. Then W = c · . . . · c u − is a product-one sequence of length u − n ≥ u ≥
6, we have u − | W | = 2 ≤ n − , a contradiction to A . Suppose that u ≥ ∤ u −
2. Then 4 | u − W = c · . . . · c u − is a product-one sequence oflength u −
4. Since u ≥
8, we obtain n ≥
9, and hence u − | W | = 4 ≤ n − , again a contradiction to A .Suppose that u = 4. Then c c ∈ Π ( c · . . . · c ) = { c c , c c , c c , c c } . Since all c i are distinct,we have c c = c c . If c c = c c or c c = c c = c c , then c · . . . · c is a product-one sequence,a contradiction to A , whence c c = c c . Similarly we can prove that c c = c c , which implies that c c = c c , and thus c = c , a contradiction. SUBCASE 2.3.2 (cid:12)(cid:12) Π ( c · . . . · c u ) (cid:12)(cid:12) = u − i ∈ [1 , u ], we set c i = α r i τ . Since (cid:12)(cid:12) Π ( c · . . . · c u ) (cid:12)(cid:12) = u −
1, it follows thatΠ ( c · . . . · c u ) = { c c , . . . , c c u } = (cid:8) α r − r , α r − r , . . . , α r − r u (cid:9) = (cid:8) α r − r , α r − r , . . . , α r − r u (cid:9) . By multiplying all the elements of each set, we have the modulo equation( u − r − ( r + r + . . . + r u ) ≡ ( u − r − ( r + r + . . . + r u ) (mod n ) , which implies that ur ≡ ur (mod n ). Similarly, ur i ≡ ur j (mod n ) for all distinct i, j ∈ [1 , u ]. Thus allelements α r i − r j are in the subgroup of H having order gcd( u, n ). But u = (cid:12)(cid:12) { G , α r − r , . . . , α r − r u } (cid:12)(cid:12) ≤ gcd( u, n ) ≤ u , whence gcd( u, n ) = u and u | n . Since n is odd, u is also odd, and Π ( c · . . . · c u ) = (cid:8) α nu , . . . , α ( u − nu (cid:9) . By renumbering if necessary, we can assume that c = α r τ, c = α r + nu τ, . . . , c u = α r +( u − nu τ . Suppose that u ≥ | u −
1. Then c · c · c · c is a product-one sequence, and thus c · . . . · c u − is a product-one subsequence of S G of length u −
1, a contradiction to A .Suppose that u ≥ ∤ u −
1. Then u ≥
7, and c · c · c · c · c · c is a product-one sequence.Thus c · c · c · . . . · c u − is a product-one subsequence of S G of length u −
1, again a contradiction to A .Suppose that u = 3. Since u | n and u ≤ n −
1, we obtain n ≥
9, and since | S H | = 3 n − ≥ s ( H ) = 2 n − S H has a product-one subsequence T of length n . Then | S H · T [ − | = 2 n − S H · T [ − has no product-one subsequence of length n . It follows by Lemma 2.4 that there exist r , r ∈ [0 , n − r − r , n ) = 1 such that( α r ) [ γ ] · ( α r ) [ γ ] (cid:12)(cid:12) S H · T [ − for some γ , γ ≥ n − ≥ n −
12 and γ + γ ≥ n − ≥ n + 3 . Since gcd( r − r , n ) = 1, there exists x ∈ [0 , n −
1] such that ( r − r ) x ≡ n + r (mod n ). If x ≤ n − ,then V = ( α r ) [ x − · c · ( α r ) [ x ] · c is a product-one subsequence of S . Since γ + γ ≥ n + 3, by addingeven terms of either α r or α r to V , we obtain that S · T [ − has a product-one subsequence of length n , a contradiction. If x ≥ n +32 , then n − x + 1 ≤ n − and V = ( α r ) [ n − x +1] · c · ( α r ) [ n − x ] · c is aproduct-one subsequence of S . Since γ + γ ≥ n + 3, by adding even terms of either α r or α r to V ,we obtain that S · T [ − has a product-one subsequence of length n , a contradiction. Thus x = n +12 , andagain by gcd( r − r , n ) = 1, there exists y ∈ [0 , n −
1] such that ( r − r ) y ≡ n + r (mod n ). A similarargument shows that y = n +12 = x , and hence n + r ≡ n + r (mod n ), a contradiction. (cid:3) Proof of Theorem 1.2.
1. (b) ⇒ (a) Suppose that α, τ ∈ G are generators, H = h α i , and S = ( α r ) [ n − · ( α r ) [ n − · α r τ , where r , r , r ∈ [0 , n −
1] with gcd( r − r , n ) = 1. It follows that S H = g [ n − · h [ n − with g, h ∈ H and ord( gh − ) = n . Then S H has no product-one subsequence of length n , and thus theassertion follows.(a) ⇒ (b) Let n ≥ s ( G ) = 2 n by Theorem 1.1.1. Thus the assertion follows by Lemmas3.1 and 2.4.2. (b) ⇒ (a) Clearly, S = 1 [5] G · g · g · g , where n = 3 and { g , g , g } = G \ H , has no product-onesubsequence of length 6, and the remains follow from Lemma 2.5.(a) ⇒ (b) Let n ≥
3. Note that E ( G ) = 3 n . Then the assertion follows by Lemmas 3.2 and 2.5. (cid:3) RD ˝OS-GINZBURG-ZIV CONSTANT 11 On Dicyclic groups
In this section, let n ∈ N ≥ and let G be a dicyclic group of order 4 n . Then we denote by H a cyclicsubgroup of G of index 2, and by G = G \ H . Note that H is unique if n ≥
3, and we observe that d ( G ) = 2 n and exp( G ) = lcm(4 , n ). We denote by ψ : G → G = G/ Z ( G ) , where Z ( G ) is the center of G , the natural epimorphism. If n ≥
3, then G is a dihedral group of order 2 n ,and thus we also denote by H the cyclic subgroup of G of index 2, and by G = G \ H .Likewise, we prove the technical lemmas by using results from the dihedral case and by improving theargument from [1], and then derive our main results. Lemma 4.1.
Let n ∈ N ≥ be even and let S ∈ F ( G ) be a sequence of length | S | = 4 n − . If n ≥ and S has no product-one subsequence of length n , then | S G | = 1 . If n = 2 and S has no product-one subsequence of length , then for any α, τ ∈ G with G = h α, τ | α = 1 G , τ = α , and τ α = α − τ i , we have that S has one of the following forms : S ∈ (cid:8) ( α r ) [3] · ( α r ) [3] · α r τ , ( α r ) [3] · ( α r τ ) [3] · α r , ( α r ) [3] · ( α r τ ) [3] · α r τ (cid:9) , where r , r , r , r ∈ [0 , such that r is even, r is odd, and r r (mod 2) . In particular, | supp( S ) | = 3 .Proof. Let α, τ ∈ G such that G = h α, τ | α n = 1 G , τ = α n , and τ α = α − τ i and let H = h α i . Let S ∈ F ( G ) be a sequence of length | S | = 4 n − S has no product-one subsequence of length 2 n .Note that | S G | ≥ n ≥
4. Assume to the contrary that | S G | ≥
2. Then ψ ( S ) is a sequence of length 4 n − G and | ψ ( S ) G | ≥
2. Let g , g ∈ G be such that g · g | S G . Since | ψ ( S · ( g · g ) [ − ) | = 4 n − ≥ n ,it follows by Theorem 1.1.1 that there exists T | S · ( g · g ) [ − of length n such that ψ ( T ) is a product-one sequence. Since | ψ ( S · ( g · g · T ) [ − ) | = 3 n − ≥ n , it follows again by Theorem 1.1.1 thatthere exists T | S · ( g · g · T ) [ − of length n such that ψ ( T ) is a product-one sequence. Therefore | ψ ( S · ( T · T ) [ − ) | = 2 n − | ψ ( S · ( T · T ) [ − ) G | ≥ | ψ ( g · g ) | = 2. We deduce by Lemma3.1 that there exists T | S · ( T · T ) [ − of length n such that ψ ( T ) is a product-one sequence. Since π ( T i ) ∩ { G , α n } 6 = ∅ for all i ∈ [1 , i, j ∈ [1 ,
3] such that T i · T j is aproduct-one subsequence of S of length 2 n , a contradiction.2. Let n = 2. If | S G | = 1, then | S H | = 6, and hence Lemma 2.4 ensures that S has the desired structure.We assume that | S G | ≥
2. Let H = { G , α } , H = { α, α } , G = { τ, α τ } , and G = { ατ, α τ } .Suppose that S H = a · . . . · a r , S H = b · . . . · b s , S G = c · . . . · c t , and S G = d · . . . · d ℓ , where r, s, t, ℓ ∈ N . Then r + s + t + ℓ = 7. We set T = ( a a ) · . . . · ( a ⌊ r ⌋− a ⌊ r ⌋ ) · ( b b ) · . . . · ( b ⌊ s ⌋− b ⌊ s ⌋ ) · ( c c ) · . . . · ( c ⌊ t ⌋− c ⌊ t ⌋ ) · ( d d ) · . . . · ( d ⌊ ℓ ⌋− d ⌊ ℓ ⌋ ) . Then T ∈ F ( H ) of length | T | = (cid:4) r (cid:5) + (cid:4) s (cid:5) + (cid:4) t (cid:5) + (cid:4) ℓ (cid:5) ≥
2. Since S has no product-one subsequence oflength 4, we obtain that T has no product-one subsequence of length 2. Since | H | = 2, we obtain that | T | = 2, and hence three elements of { r, s, t, ℓ } are odd.If { r, s, t, ℓ } ⊂ N , then a b , c d ∈ H . Since | H | = 2, we obtain that either a b = c d or a b = ( c d ) − . It follows that either c · b · a · d or a · b · c · d is a product-one subsequence of S of length 4, a contradiction. Thus one element of { r, s, t, ℓ } must be zero, and we distinguish four cases.Suppose that r = 0. Then all s, t, ℓ must be odd with s + t + ℓ = 7. Since | H | = | H | = | G | = | G | = 2,we obtain that there exist subsequences W , W of S such that W = x [2] and W = y [2] for some x, y ∈ G .If x ∈ G or y ∈ G , then W · W is a product-one subsequence of S of length 4, a contradiction. If x, y ∈ H , then x = y or x = y − , which implies that W · W is a product-one subsequence of S of length4, a contradiction.Suppose that s = 0. Then all r, t, ℓ must be odd with r + t + ℓ = 7. Since | H | = | H | = | G | = | G | = 2,we obtain that there exist subsequences W , W of S such that W = x [2] and W = y [2] for some x, y ∈ G .If x, y ∈ H or x, y ∈ G , then W · W is a product-one subsequence of S of length 4, a contradiction. Thuswe may assume that x ∈ H and y ∈ G , whence r = 3 and t + ℓ = 4. Thus either S = (cid:0) a [2]1 · a (cid:1) · (cid:0) c [2]1 · c (cid:1) · d or S = (cid:0) a [2]1 · a (cid:1) · c · (cid:0) d [2]1 · d (cid:1) . In the former case, if a = a (or c = c respectively), then a · a · c [2]1 (or a [2]1 · c · c respectively) is a product-one subsequence of S of length 4, a contradiction, whence S = a [3]1 · c [3]1 · d has the desired structure. In the latter case, we similarly obtain S = a [3]1 · c · d [3]1 , whichhas the desired structure.Suppose that t = 0. Then all r, s, ℓ must be odd with ℓ ≥ r + s + ℓ = 7. Since | H | = | H | = | G | = | G | = 2, we obtain that there exist subsequences W , W of S such that W = x [2] and W = y [2] ,where x ∈ G and y ∈ G . If x ∈ G , then W · W is a product-one subsequence of S of length 4, acontradiction. Thus we must have ℓ = 3 and r + s = 4. If s = 3, then x ∈ H , and hence W · W is a product-one subsequence of S of length 4, a contradiction. Hence we must have r = 3 and s = 1,and thus after renumbering if necessary, we have S = (cid:0) a [2]1 · a (cid:1) · b · (cid:0) d [2]1 · d (cid:1) . If a = a (or d = d respectively), then a · a · d · d (or a · a · d · d respectively) is a product-one subsequence of S oflength 4, a contradiction. Therefore S = a [3]1 · b · d [3]1 has the desired structure.Suppose that ℓ = 0. A similar argument as used in the case t = 0 shows that S = a [3]1 · b · c [3]1 , whichhas the desired structure. (cid:3) Proof of Theorem 1.1.2. If n ≥ s ( G ) = E ( G ) and the assertion follows from [1]. Supposethat n ≥ s ( G ) = 4 n . By Lemma 2.2, it suffices to show that every sequenceof length 4 n has a product-one subsequence of length 2 n . Let S ∈ F ( G ) be a sequence of length 4 n , andassume to the contrary that S has no product-one subsequence of length 2 n . Let T | S be a subsequenceof length 4 n −
1. Then T has no product-one subsequence of length 2 n .Suppose that n ≥
4. Then Lemma 4.1.1 ensures that | T G | = 1, and in addition that (cid:12)(cid:12) ( S · T [ − G ) G (cid:12)(cid:12) = 1,implying | S G | = 2. Let W | S be a subsequence of length 4 n − S G | W . Again by Lemma 4.1.1,we obtain that W , and hence S , has a product-one subsequence of length 2 n , a contradiction.Suppose that n = 2. Then T must have the desired structure in Lemma 4.2.2. We fix α, τ ∈ G suchthat G = h α, τ | α = 1 G , τ = α , and τ α = α − τ i . If | T G | = 1, then S = x · ( α r ) [3] · ( α r ) [3] · α r τ forsome x ∈ G and r , r , r ∈ [0 ,
3] with gcd( r − r ,
4) = 1. Since S · ( α r ) [ − is a sequence of length 7having no product-one subsequence of length 4, Lemma 4.1.2 ensures that x = α r , whence ( α r ) [4] is aproduct-one subsequence of S of length 4, a contradiction. If | T G | ≥
2, then the same argument showsthat ( α r ) [4] is a product-one subsequence of S of length 4, a contradiction. (cid:3) Lemma 4.2.
Let n ∈ N ≥ and let S ∈ F ( G ) be a sequence of length | S | = 6 n − . If n ≥ and S has no product-one subsequence of length n , then | S G | = 1 . If n = 2 and S has no product-one subsequence of length , then for any α, τ ∈ G with G = h α, τ | α = 1 G , τ = α , and τ α = α − τ i , we have that S has one of the following forms : S ∈ n(cid:0) ( α r ) [3] · ( α r ) [3] · α r τ (cid:1) · S , (cid:0) ( α r ) [3] · ( α r τ ) [3] · α r (cid:1) · S , (cid:0) ( α r ) [3] · ( α r τ ) [3] · α r τ (cid:1) · S o , where S ∈ (cid:8) ( α r ) [4] , ( α r ) [4] (cid:9) , S ∈ (cid:8) ( α r ) [4] , ( α r τ ) [4] (cid:9) , and r , r , r , r ∈ [0 , such that r iseven, r is odd, and r r (mod 2) .Proof. Let α, τ ∈ G such that G = h α, τ | α n = 1 G , τ = α n , and τ α = α − τ i and let H = h α i . Let S ∈ F ( G ) be a sequence of length 6 n − S has no product-one subsequence of length 4 n . Notethat | S G | ≥ RD ˝OS-GINZBURG-ZIV CONSTANT 13
1. Let n ≥
3. Assume to the contrary that | S G | ≥
2. We first suppose that n ≥ g , g ∈ G be such that g · g | S G . Then | S · ( g · g ) [ − | = 6 n − ≥ n , and by Theorem 1.1.2,there exists a product-one subsequence T | S · ( g · g ) [ − of length 2 n . Then | S · T [ − | = 4 n − | ( S · T [ − ) G | ≥ | g · g | ≥
2. Thus Lemma 4.1.1 ensures that S · T [ − has a product-one subsequence T of length 2 n , whence T · T is a product-one subsequnce of S of length 4 n , a contradiction.Suppose now that n ≥ CASE 1. | S G | ≤ n .Then | S H | ≥ n −
1, and by mapping under ψ , we obtain a sequence ψ ( S H ) ∈ F ( H ) of length at least5 n −
1. Since 5 n − > n − s ( H ), we obtain that there exists T | S H of length n such that ψ ( T )is a product-one sequence over H , whence π ∗ ( T ) ∈ { G , α n } . Continuing this process, we obtain thatthere exist subsequences T , T , T of S H such that | T | = | T | = | T | = n and π ∗ ( T i ) ∈ { G , α n } for all i ∈ [2 , S has no product-one subsequence of length 4 n , we may assume that π ∗ ( T ) = π ∗ ( T ) = π ∗ ( T ) = π ∗ ( T ). Consider the sequence ψ (cid:0) S · ( T · T · T ) [ − (cid:1) ∈ F ( G ) of length 3 n − n ≥
5. Since | S G | ≥
2, it follows by Lemma 3.2 that there exists an ordered subsequence T | S · ( T · T · T ) [ − of length 2 n such that π ∗ ( T ) ∈ { G , α n } . It follows that either T · T · T or T · T · T is a product-one subsequence of S of length 4 n , a contradiction.Suppose that n = 3. Since | S G | ≥
2, Lemma 3.2 ensures that S = T · T · T · (cid:0) α r · α r · α r · α r · α r (cid:1) · (cid:16) α r τ · α r τ · α r τ (cid:17) , where r , . . . , r ∈ { , } , r ∈ { , } , and r ∈ { , } . Since there exist distinct i, j ∈ [6 ,
8] such that r i ≡ r j (mod 2), we obtain { α, α } = (cid:8) α r i τ α r j τ, α r j τ α r i τ (cid:9) ⊂ Π ( α r τ · α r τ · α r τ ), and assert that S · ( T · T ) [ − has a product-one subsequenc of length 6, leading a contradiction to the fact that S hasno product-one subsequence of length 12.If π ∗ ( T ) ∈ Π ( α r · . . . · α r ), then T · α r · . . . · α r has a product-one subsequence of length 6. Supposethat (cid:12)(cid:12) Π ( α r · . . . · α r ) (cid:12)(cid:12) = 1 and π ∗ ( T ) / ∈ Π ( α r · . . . · α r ). Then r = . . . = r and π ∗ ( T ) = α r , whence π ∗ ( T · α r ) = α . If { G , α } ⊂ Π ( T · α r ), then together with α r · . . . · α r , we obtain a product-onesubsequence of length 6. If { α, α } ⊂ Π ( T · α r ) or { α , α } ⊂ Π ( T · α r ), then ( T · α r ) · α r τ · α r τ · α r τ has a product-one subsequence of length 4. Together with α r · α r , we obtain a product-one subsequenceof length 6. CASE 2. | S G | ≥ n + 1.By renumbering if necessary, we can assume that S G = ( a · a − ) · . . . · ( a r · a − r ) · (cid:0) W · . . . W x (cid:1) · ( c · . . . · c u ) , where W i is a product-one sequence of length 4 for all i ∈ [1 , x ], and c · . . . · c u has no product-onesubsequence of length 2 and 4. Then h ( c · . . . · c u ) ≤
3, and thus u − ≤ | supp( c · . . . · c u ) | ≤ n , whence u ≤ n + 2. Similarly we set S H = ( e · e − ) · . . . · ( e s · e − s ) · (cid:0) b [2]1 · . . . · b [2] t (cid:1) · ( d · . . . · d v ) , where d i / ∈ { d j , d − j } for all distinct i, j ∈ [1 , v ], and ord( b k ) = 2 for all k ∈ [1 , t ]. Then d · . . . · d v issquare-free, and hence v ≤ n + 1. Let R = c · . . . · c u · d · . . . · d v . SUBCASE 2.1. r + s + t = 0.If h ( R G ) = 1, then u ≤ n , and hence 4 x ≥ (6 n − − u − v ≥ n −
2. It follows that x ≥ n ,whence W · . . . W n is a product-one subsequence of S of length 4 n , a contradiction. Thus we must have h ( R G ) ≥
2, and by renumbering if necessary, we may assume that c = c . If v ≥ n +12 + 1, then bymapping under ψ , we infer that there exist i, j ∈ [1 , v ] such that π ∗ ( d i d j ) = α n , whence c [2]1 · d i · d j is aproduct-one sequence of length 4. It follows by 4 x ≥ n − W · . . . · W n − · c [2]1 · d i · d j is a product-onesubsequence of S of length 4 n , a contradiction. Thus we must have v ≤ n +12 . Suppose that n ≥
5. Then u + v ≤ n +52 ≤ n , and since u + v is odd, it follows that u + v ≤ n − x ≥ n , and therefore W · . . . · W n is a product-one subsequence of S of length 4 n , a contradiction.Suppose that n = 3. Then u ≤ v ≤
2. If u + v ≤
5, then 4 x ≥
12, whence S has a product-one subsequence of length 12, a contradiction. Thus we must have that u = 5 and v = 2, and then4 x = (6 n − − u − v = 10, a contradiction. SUBCASE 2.2. r + s + t ≥ u ≤ n . If 2 r + 4 x + 2 s + 2 t ≥ n , then S must have a product-one subsequence of length4 n . We assume that 2 r + 4 x + 2 s + 2 t ≤ n −
2. Then 2 n + 1 ≥ u + v ≥ (6 n − − (4 n −
2) = 2 n + 1,and hence u = n and v = n + 1. Then we may assume by renumbering if necessary that d = 1 G and d = α n . If n ≥
5, then there exist distinct i, j ∈ [3 , v ], say i = 3 and j = 4, such that d d / ∈ { G , α n } .Since | R G · ( d d ) | = n + 1, it follows by mapping under ψ that there exists an ordered subsequence T of R G · d · d having even length such that π ∗ ( T ) ∈ { G , α n } . Since R G has no product-one subsequenceof length 2, we obtain that either T or T · d · d is a product-one subsequence of S having even length atleast 4. Since 2 r + 4 x + 2 s + 2 t = 4 n − r + s + t ≥
1, it follows that S has a product-one subsequenceof length 4 n , a contradiction. If n = 3, then we obtain that { α, α } ⊂ Π ( R G ) as argued in the case n = 3 of CASE 1 . By renumbering if necessary, we may assume that d ∈ { α, α } and d ∈ { α , α } . Itfollows that { α, α } ⊂ Π ( R H ), whence R has a product-one subseqeunce T of length 4 with | T G | = 2.Since 2 r + 2 x + 2 s + 2 t = 10 and r + s + t ≥
1, it follows that S has a product-one subseqeunce of legnth12, a contradiction.Suppose that u ≥ n + 1. Then h ( R G ) ≥
2, and by renumbering if necessary, we may assume that c = c . If t = 0 and v ≤
1, then 2 r + 2 s + 4 x ≥ n − ≥ n −
1, and hence S must have a product-onesubsequence of length 4 n , a contradiction. Thus either t ≥ v ≥
2. By mapping under ψ , [2, Theorem1.3] ensures that there exists an ordered sequence T of c · . . . · c u · d · d (or c · . . . · c u · b [2]1 if v ≤ π ∗ ( T ) ∈ { G , α n } . Since R G has no product-one subsequence of length 2,we obtain that either T or T · c [2]1 is a product-one subsequence of S having even length at least 4. Since2 r + 4 x + 2 s + 2 t ≥ n − r + 4 x + 2 s + 2( t − ≥ n − ≥ n − v ≤
1) and r + s + t ≥
1, itfollows that S has a product-one subsequence of length 4 n , a contradiction.2. Let n = 2. If | S G | = 1, then | S H | = 10, and hence Lemma 2.5 ensures that S has the desiredstructure. We assume that | S G | ≥
2. Let g , g ∈ G be such that g · g | S G . Then | S · ( g · g ) [ − | = 9,and by Theorem 1.1.2, there exists a product-one subsequence T | S · ( g · g ) [ − of length 4. Then | S · T [ − | = 7 and | ( S · T [ − ) G | ≥ | g · g | ≥
2. Then Lemma 4.1.2 ensures that S · T [ − = ( α r ) [3] · ( α r τ ) [3] · x , where r , r ∈ [0 ,
3] such that r is even and x ∈ { α, α , α r τ } with r r (mod 2). If α r ∈ supp( T ),then ( α r ) [4] | S , and we obtain by Lemma 4.1.2 that S · (cid:0) ( α r ) [4] (cid:1) [ − = x · ( α r ′ ) [3] · ( α r τ ) [3] , where r ′ ∈ [0 ,
3] is even. Thus T = α r · ( α r ′ ) [3] is a product-one sequence, which implies that r ′ = r . Therefore S = x · ( α r ) [7] · ( α r τ ) [3] has the desired structure. Suppose that α r / ∈ supp( T ). Let g ∈ supp( T )be any element. Then S · ( α r τ · x · g ) [ − has length 8, and by Theorem 1.1.2, it has a product-onesubsequence T of length 4. Then S · T [ − has no product-one subsequence of length 4, and Lemma4.1.2 ensures that | supp( S · T [ − ) | = 3. Since T = ( α r ) [4] , we have α r · α r τ · x · g | S · T [ − , whence g ∈ { x, α r τ } . Thus T ∈ (cid:8) x [4] , ( α r τ ) [4] , ( x · α r τ ) [2] (cid:9) . If T = x [4] (or T = ( x · α r τ ) [2] respectively),then ( α r ) [2] · ( x · α r τ ) [2] · x [2] (or ( α r ) [2] · ( x · α r τ ) [2] · ( α r τ ) [2] respectively) is a product-one subsequenceof S of length 8, a contradiction. Therefore T = ( α r τ ) [4] , and S has the desired structure. (cid:3) Proof of Theorem 1.3.
1. (b) ⇒ (a) (1) Suppose that α, τ ∈ G are generators, H = h α i , and S =( α r ) [2 n − · ( α r ) [2 n − · α r τ , where r , r , r ∈ [0 , n −
1] with gcd( r − r , n ) = 1. It follows that S H = g [2 n − · h [2 n − with g, h ∈ H and ord( gh − ) = 2 n . Then S H has no product-one subsequence oflength 2 n , and thus the assertion follows. RD ˝OS-GINZBURG-ZIV CONSTANT 15 (2) By (1), it suffices to verify that a sequence S with | S G | ≥ S = ( α t ) [3] · ( α t τ ) [3] · α t , where t , t , t ∈ [0 ,
3] such that t is even and t is odd. Assume to the contrary that S has a product-one subsequence T of length 4. Since( α t ) [3] · α t is not product-one sequence, we must have | T G | = 2, and thus either T = ( α t ) [2] · ( α t τ ) [2] or T = α t · α t · ( α t τ ) [2] . But they are nor product-one sequences. By the similar argument, we obtainthat the remaining sequence has no product-one subsequence of length 4.(a) ⇒ (b) Let n ≥ s ( G ) = 4 n by Theorem 1.1.2. Thus the assertion follows by Lemmas4.1 and 2.4.2. (b) ⇒ (a) (1) The assertion follows from Lemma 2.5.(2) By (1), it suffices to verify that a sequence S with | S G | ≥ S = ( α t ) [3] · ( α t τ ) [7] · α t , where t , t , t ∈ [0 ,
3] such that t iseven and t is odd. Assume to the contrary that S has a product-one subsequence T of length 8. Then wehave either | T G | = 4 or | T G | = 6. If | T G | = 4, then T = ( α t ) [3] · ( α t τ ) [4] · α t , but it is not product-onesequence. Thus | T G | = 6, and hence we have either T = ( α t ) [2] · ( α t τ ) [6] or T = α t · α t · ( α t τ ) [6] . Butthey are not product-one sequences. By the similar argument, we obtain thet the remaining sequenceshave no product-one subsequence of length 8.(a) ⇒ (b) Let n ≥
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