Tiny zero-sum sequences over some special groups

Theorem 1

Let p be a prime. Then, t ( G ) = η ( G ) for the following groups.

1. G ≅ C 2 α r [1, Theorem 3.5],

2. G ≅ C m 3 with m = 3 α or m = 5 β [2, Theorem 1.7],

3. G ≅ C p α ⊕ C p α [1, Corollary 3.5], and

4. G ≅ C 2 ⊕ C 2 p [3, Theorem 1.4].

Theorem 2

Let p be a prime number with p ≥ 5 . Then, t ( G ) = η ( G ) = 3 p + 4 for G ≅ C 3 ⊕ C 3 p .

Lemma 3

[14, Corollary 4.4] Let m , n ∈ ℕ with m | n . Then, η ( C m ⊕ C n ) = 2 m + n − 2 and D ( C m ⊕ C n ) = m + n − 1 .

Lemma 4

[10, Theorem 5.4.5] Let S ∈ ℱ ( C p ) be a sequence of length p − 1 . If S is zero-sum free, then S = g p − 1 for some g ∈ C p \ { 0 } .

Lemma 5

Let S ∈ ℱ ( C p ) be a sequence of length 2 p − 1 . If S contains no two disjoint non-empty zero-sum subsequences, then S = g 2 p − 1 for some g ∈ C p \ { 0 } .

Proof

Let T be an arbitrary subsequence of S such that | T | = p − 1 . Then, | S T − 1 | = p = D ( C p ) . Therefore, S T − 1 contains a non-empty zero-sum subsequence. It follows from the hypothesis of the lemma that T is zero-sum free. Hence, T = g p − 1 for some g ∈ C p \ { 0 } by Lemma 4. Now the result follows from the arbitrariness of the choice of T.□

Lemma 6

Let G ≅ C 3 2 and suppose that S is a minimal zero-sum sequence over G.

1. If | S | = 4 , then S = g 1 2 g 2 g 3 for pairwise distinct elements g 1 , g 2 , g 3 ∈ G \ { 0 } . In particular, h ( S ) = 2 .

2. If | S | = 5 , then S = g 1 2 g 2 2 g 3 for pairwise distinct elements g 1 , g 2 , g 3 ∈ G \ { 0 } .

Proof

Note that for every short zero-sum sequence W over G, we have h ( W ) = 1 or W = h 3 for some h ∈ G . Let S be a minimal zero-sum sequence of length ≥ 4 over G. Then, h ( S ) ≤ 2 .

Assume to the contrary that | supp ( S ) | ≥ 4 . Suppose g 1 , g 2 , g 3 , g 4 ∈ supp ( S ) are pairwise distinct elements. It follows by η ( G ) = 7 that g 1 2 g 2 2 g 3 2 g 4 2 have a short zero-sum subsequence T with h ( T ) = 1 , whence T is a short zero-sum subsequence of S, a contradiction. Thus, | supp ( S ) | ≤ 3 .

1. Suppose | S | = 4 . It follows by h ( S ) ≤ 2 and | supp ( S ) | ≤ 3 that S must be of the form S = g 1 2 g 2 g 3 or S = g 1 2 g 2 2 , where g 1 , g 2 , g 3 ∈ G \ { 0 } are pairwise distinct elements. If S = g 1 2 g 2 2 , then g 1 g 2 is a zero-sum subsequence of S, a contradiction. Thus, the assertion follows.

2. Suppose | S | = 5 . The assertion follows by h ( S ) ≤ 2 and | supp ( S ) | ≤ 3 . □

Lemma 7

Let G ≅ C 3 2 be a finite abelian group and let S be a sequence of length 6 over G. If S does not contain a short zero-sum subsequence, then S = g 1 2 g 2 2 g 3 2 for pairwise distinct elements g 1 , g 2 , g 3 ∈ G \ { 0 } .

Proof

Since S has no short zero-sum subsequence, we obtain h ( S ) ≤ 2 and 0 ∉ supp ( S ) . Assume to the contrary that there exists g ∈ supp ( S ) such that v g ( S ) = 1 . It follows by η ( G ) = 7 that S ⋅ g has a short zero-sum subsequence T, whence v g ( T ) = v g ( S ⋅ g ) = 2 . Let T = g 2 h , where h ∈ supp ( S ) . Then, h = g and hence v g ( T ) = 3 , a contradiction. Therefore, v g ( S ) = 2 for every g ∈ supp ( S ) and the assertion follows.□

Claim A

If | S p | = 0 and there exists a subsequence W 0 of S 3 p such that | W 0 | = p − 1 and W 0 3 divides S 3 p , then S has a tiny zero-sum subsequence.

Proof of Claim A

Since | S p | = 0 , we have | S 3 | + | S 3 p | = 3 p + 4 . If | S 3 | ≥ 7 , then the assertion follows by t ( H 1 ) = η ( H 1 ) = η ( C 3 2 ) = 7 . Therefore, we may assume | S 3 | ≤ 6 and | S 3 p | ≥ 3 p − 2 . Suppose S 3 p ⋅ ( W 0 3 ) − 1 = g 1 ⋅ … ⋅ g t , where t ∈ ℕ and g 1 , … , g t ∈ G . Let W i = W 0 ⋅ g i for all i ∈ [ 1 , t ] . Then, for each i ∈ [ 1 , t ] , we have | W i | = p , and hence there exists a non-empty subsequence T i of W i such that φ 2 ( T i ) has sum zero. Therefore, the sequence U = σ ( T 1 ) ⋅ … ⋅ σ ( T t ) ⋅ S 3 is a sequence over H 1 of length t + | S 3 | = 7 , whence U has a short zero-sum subsequence. Therefore, there exist a subset I ⊂ [ 1 , t ] and a subsequence S ′ of S 3 such that S ′ ⋅ Π i ∈ I T i is zero-sum and 1 ≤ | I | + | S ′ | ≤ 3 . It follows that S ′ ⋅ Π i ∈ I T i is a non-empty zero-sum subsequence of S and k S ′ ⋅ Π i ∈ I T i = | S ′ | 3 + ∑ i ∈ I | T i | 3 p ≤ | S ′ | + | I | 3 ≤ 1 . □

Claim B

If | S 3 | = 3 , | S 3 p | ≥ 3 p , and there exists a subsequence W 0 of S 3 p such that | W 0 | = p − 2 and W 0 3 divides S 3 p , then S has a tiny zero-sum subsequence.

Proof of Claim B

Suppose S 3 p ⋅ ( W 0 3 ) − 1 = g 1 ⋅ … ⋅ g t , where t ∈ [ 6 , 7 ] and g 1 , … , g t ∈ G . Let W i = W 0 g i for all i ∈ [ 1 , t ] . We distinguish four cases.

Suppose there exist distinct i , j ∈ [ 1 , t ] such that neither φ 2 ( W i ) nor φ 2 ( W j ) is zero-sum free, say i = 1 and j = 2 . Then, there exist non-empty subsequences T 1 of W 1 and T 2 of W 2 such that φ 2 ( T 1 ) and φ 2 ( T 2 ) are zero-sum. Moreover, as η ( H 2 ) = η ( C p ) = p there are non-empty subsequences T 3 of W 0 g 3 g 4 and T 4 of W 0 g 5 g 6 such that φ 2 ( T 3 ) and φ 2 ( T 4 ) are zero-sum. Now consider the sequence U = σ ( T 1 ) ⋅ … ⋅ σ ( T 4 ) ⋅ S 3 , which is a sequence of length 7 over H 1 , whence U has a short zero-sum subsequence. Therefore, there exist a subset I ⊂ [ 1 , 4 ] and a subsequence S ′ of S 3 such that S ′ ⋅ ∏ i ∈ I T i is zero-sum and 1 ≤ | I | + | S ′ | ≤ 3 . It follows that S ′ ⋅ ∏ i ∈ I T i is a non-empty zero-sum subsequence of S and k S ′ ⋅ ∏ i ∈ I T i = | S ′ | 3 + ∑ i ∈ I | T i | 3 p ≤ | S ′ | + | I | 3 ≤ 1 .

Suppose | S 3 p | = 3 p + 1 and there is precisely one i ∈ [ 1 , 7 ] , say i = 1 , such that φ 2 ( W i ) is not zero-sum free. Let V 1 be a non-empty subsequence of W 1 with φ 2 ( V 1 ) is zero-sum and let V 2 = W 0 g 2 g 3 , V 3 = W 0 g 4 g 5 , and V 4 = W 0 g 6 g 7 . Since φ 2 ( W i ) is zero-sum free for every i ∈ [ 2 , 7 ] , it follows by Lemma 4 that | supp ( φ 2 ( W 0 ⋅ g 2 ⋅ … ⋅ g 7 ) ) | = 1 , whence φ 2 ( V k ) is zero-sum for every k ∈ [ 1 , 4 ] . Now consider the sequences U = σ ( V 1 ) ⋅ … ⋅ σ ( V 4 ) ⋅ S 3 , which is a sequence of length 7 over H 1 , whence U has a short zero-sum subsequence. Therefore, there exist a subset I ⊂ [ 1 , 4 ] and a subsequence S ′ of S 3 such that S ′ ⋅ ∏ i ∈ I V i is zero-sum and 1 ≤ | I | + | S ′ | ≤ 3 . It follows that S ′ ⋅ ∏ i ∈ I V i is a non-empty zero-sum subsequence of S and k S ′ ⋅ ∏ i ∈ I V i = | S ′ | 3 + ∑ i ∈ I | V i | 3 p ≤ | S ′ | + | I | 3 ≤ 1 .

Suppose | S 3 p | = 3 p + 1 and φ 2 ( W i ) is zero-sum free for all i ∈ [ 1 , 7 ] . Then, by Lemma 4 we have | supp ( φ 2 ( W 0 ⋅ g 1 ⋅ … ⋅ g 7 ) ) | = 1 . We claim that | supp ( g 1 ⋅ … ⋅ g 7 ) | = 1 . Assume to the contrary that there are two distinct k 1 , k 2 ∈ [ 1 , 7 ] , say k 1 = 1 and k 2 = 2 , such that g 1 ≠ g 2 . Let V 1 = W 0 g 1 g 3 , V 2 = W 0 g 4 g 5 , V 3 = W 0 g 6 g 7 , and V 4 = W 0 g 2 g 3 . Then, φ 2 ( V k ) is zero-sum for every k ∈ [ 1 , 4 ] . Now consider the sequences U = σ ( V 1 ) ⋅ σ ( V 2 ) ⋅ σ ( V 3 ) ⋅ S 3 and U ′ = σ ( V 4 ) ⋅ σ ( V 2 ) ⋅ σ ( V 3 ) ⋅ S 3 . If U has a short zero-sum subsequence, then there exist a subset I ⊂ [ 1 , 3 ] and a subsequence S ′ of S 3 such that S ′ ⋅ ∏ i ∈ I V i is zero-sum and 1 ≤ | I | + | S ′ | ≤ 3 . It follows that S ′ ⋅ ∏ i ∈ I V i is a non-empty zero-sum subsequence of S and k S ′ ⋅ ∏ i ∈ I T i = | S ′ | 3 + ∑ i ∈ I | V i | 3 p ≤ | S ′ | + | I | 3 ≤ 1 , then we are done. If U ′ has a short zero-sum subsequence, then we also get a tiny zero-sum subsequence of S. Hence, we can assume that both U and U ′ have no short zero-sum subsequence, whence by Lemma 7 and the structures of U and U ′ we obtain σ ( V 1 ) = σ ( V 4 ) and hence g 1 = g 2 , a contradiction. Thus, | supp ( g 1 ⋅ … ⋅ g 7 ) | = 1 and so we have ( W 0 g 1 ) 3 | S 3 p . Hence, by Claim A we have done.

Suppose | S 3 p | = 3 p and there are at most one i ∈ [ 1 , 6 ] such that φ 2 ( W i ) is not zero-sum free. We may assume that φ 2 ( W j ) is zero-sum free for all j ∈ [ 2 , 6 ] . Then, by Lemma 4 we have | supp ( φ 2 ( W 0 ⋅ g 2 ⋅ … ⋅ g 6 ) ) | = 1 . If φ 1 ( g 1 ⋅ … ⋅ g 6 ) has a short zero-sum subsequence, then φ 1 ( S 3 p ) has pairwise disjoint p − 1 short zero-sum subsequences. Together with | S p | = 1 and η ( H 2 ) = p , we can find a tiny zero-sum subsequence of S. Otherwise, by Lemma 7 we have φ 1 ( g 1 ⋅ … ⋅ g 6 ) = h 1 2 h 2 2 h 3 2 , where h 1 , h 2 , h 3 ∈ H 1 are pairwise distinct. After renumbering if necessary, we may assume φ 1 ( g 1 g 2 ) = h 1 2 , φ 1 ( g 3 g 4 ) = h 2 2 , and φ 1 ( g 5 g 6 ) = h 3 2 . Let V 1 = W 0 g 1 g 2 , V 2 = W 0 g 3 g 4 , V 3 = W 0 g 5 g 6 , V 4 = W 0 g 3 g 5 , and V 5 = W 0 g 4 g 6 . Then, φ 2 ( V 2 ) , φ 2 ( V 3 ) , φ 2 ( V 4 ) , and φ 2 ( V 5 ) are all zero-sum sequences. Furthermore, there exists a non-empty subsequence T 1 of V 1 such that φ 2 ( T 1 ) is zero-sum. Consider the sequences U = σ ( T 1 ) ⋅ σ ( V 2 ) ⋅ σ ( V 3 ) ⋅ S 3 and U ′ = σ ( T 1 ) ⋅ σ ( V 4 ) ⋅ σ ( V 5 ) ⋅ S 3 , which are both sequences of length 6 over H 1 . If U or U ′ has a short zero-sum subsequence, then the corresponding subsequence of S is a tiny zero-sum sequence. Otherwise, by Lemma 7 we know that both U and U ′ have the same structures. Together with σ ( V 4 ) = σ ( V 5 ) , we obtain σ ( V 2 ) = σ ( V 4 ) , and then g 4 = g 5 . However, φ 1 ( g 4 ) = h 2 , φ 2 ( g 5 ) = h 3 , and h 2 ≠ h 3 , which is a contradiction.□

Claim C

Suppose m + | S p | = η ( H 2 ) − 1 = p − 1 . Let h ∈ supp ( φ 1 ( S 3 p ) ) with v h ( φ 1 ( S 3 p ) ) > 3 and let g 1 , g 2 ∈ supp ( S 3 p ) . If φ 1 ( g 1 ) = φ 1 ( g 2 ) = h , then g 1 = g 2 .

Proof of Claim C

Assume to the contrary that g 1 ≠ g 2 . Note that S 3 p = U 1 ⋅ … ⋅ U m ⋅ U 0 , where U 0 = S 3 p ( U 1 ⋅ … ⋅ U m ) − 1 with h ( U 0 ) ≤ 2 . Since the decomposition only depends on φ 1 ( S 3 p ) , without loss of generality we may assume g 1 | U 1 and g 2 ∤ U 1 .

If g 2 | U 0 , then let U 1 ′ = U 1 g 2 g 1 − 1 . Thus, both sequences σ ( U 1 ) ⋅ … ⋅ σ ( U m ) ⋅ S p and σ ( U 1 ′ ) ⋅ … ⋅ σ ( U m ) ⋅ S p are zero-sum free and of length p − 1 . It follows by Lemma 4 and p − 1 ≥ 4 that σ ( U 1 ) = σ ( U 1 ′ ) and hence g 1 = g 2 , a contradiction.

If g 2 ∤ U 0 , say g 2 | U 2 , then let U 1 ′ = U 1 g 2 g 1 − 1 and U 2 ′ = U 2 g 1 g 2 − 1 . Thus, both sequences σ ( U 1 ) ⋅ σ ( U 2 ) ⋅ … ⋅ σ ( U m ) ⋅ S p and σ ( U 1 ′ ) ⋅ σ ( U 2 ′ ) ⋅ … ⋅ σ ( U m ) ⋅ S p are zero-sum free and of length p − 1 . It follows by Lemma 4 and p − 1 ≥ 4 that σ ( U 1 ) = σ ( U 1 ′ ) and hence g 1 = g 2 , a contradiction.□

Case 1

| S 3 | = 4 , | S p | = 1 , and | S 3 p | = 3 p − 1 .

By inequalities (7) and (8), we obtain that n = 2 , i.e., there are exactly two pairwise disjoint subsequences V 1 , V 2 of S 3 p such that σ ( φ 2 ( V i ) ) = 0 ∈ H 2 and | V i | ≤ p for i = 1 , 2 . If | V 1 | + | V 2 | ≤ 2 p − 1 , then | S 3 p ( V 1 V 2 ) − 1 | ≥ p and so there exists one more subsequence disjoint with V 1 and V 2 , a contradiction. Therefore, | V 1 | = | V 2 | = p , | S 3 p ( V 1 V 2 ) − 1 | = p − 1 , and φ 2 ( S 3 p ( V 1 V 2 ) − 1 ) are zero-sum free over H 2 . By Lemma 4, we may assume that φ 2 ( S 3 p ( V 1 V 2 ) − 1 ) = b p − 1 , φ 2 ( V 1 ) = b 1 p , and φ 2 ( V 2 ) = b 2 p , where b , b 1 , b 2 ∈ H 2 \ { 0 } . Note that both b p − 1 b 1 p and b p − 1 b 2 p contain no two disjoint zero-sum subsequences over H 2 . It follows from Lemma 5 that b = b 1 = b 2 . Hence, we may assume that φ 2 ( S 3 p ) = b 3 p − 1 .

Next, we assert that S 3 p = g 3 p − 1 for some g ∈ G , which implies that m + | S p | ≥ 3 p − 1 3 + 1 ≥ p , a contradiction to inequality (3). Assume to the contrary that there exist g 1 , g 2 ∈ supp ( S 3 p ) such that g 1 ≠ g 2 . Let T 1 be a subsequence of S 3 p ( g 1 g 2 ) − 1 with length p − 1 and let T 2 be a subsequence of S 3 p ( T 1 g 1 g 2 ) − 1 with length p. Then, W 1 = σ ( T 1 g 1 ) ⋅ σ ( T 2 ) ⋅ S 3 and W 2 = σ ( T 1 g 2 ) ⋅ σ ( T 2 ) ⋅ S 3 are both sequences over H 1 of length 6. It follows by our assumptions S has no tiny zero-sum subsequence that W 1 and W 2 have no short zero-sum subsequence. Hence, by Lemma 7 σ ( T 1 g 1 ) = σ ( T 1 g 2 ) , i.e., g 1 = g 2 , a contradiction.