Davenport Constants of Finite Abelian Groups

Rui Xiong

May 24, 2018

Abstract

Davenport constant of a finite abelian group G (writing additive) isthe least number n such that for any sequence a1, . . . , an there exists anonempty subsequence which sums zero. Davenport constant has notbeen solved completely, see [1] for more information. In this article, wewill introduce computations of some easy abelian groups.

Contents

1 Background 2

2 Definition 3

3 Davenport constants of cyclic groups 3

4 Davenport constants of p-groups 4

5 Davenport constant of Z/nZ× Z/nmZ 6

1 Background

Davenport constant is a classic topic in additive number theory. It arises fromalgebraic number theory. It is well known that any integer can be factorizedinto products of primes. One sees that it may fail if we extend the ring to alittle bigger one, for example Z[

√−5], since 6 = 2× 3 = (1−

√−5)(1 +

√−5).

It is not difficult to see that a factorization into product of irreducible elementsstill exists for each element (since they are all noetherian), but it may not beunique.

Dedekind proved a nice theorem for algebraic number rings, that the idealsof an algebraic number ring can be uniquely factorized into prime ideals. Andone can define class group for algebraic number ring R to be the quotient ofthe group of fractional ideals by the group of principal fractional ideals. Theclass group is proven to be a finite group for algebraic number rings.

Roughly speaking, class group of some ring measures the distance that aring to have unique factorization property.

Then, let’s go back to the original problem. The factorization into productof irreducible elements can be reformed into the language of ideals. It impliesthat the ideal generated by an irreducible elements can be factorized intoprime ideal(s). Actually, Davenport constant of class group gives an up-boundof the length of an factorization into ideals of any irreducible elements.

See, for example, [4] for more information.

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2 Definition

(1)——Definition of Davenport Constants. For an abelian group G(writing additive), define its Davenport constant

D(G) = min

{n ∈ N :

∀a1, . . . , an ∈ G,∃∅ 6= S ⊆ {1, . . . , n}, such that

∑s∈S as = 0

}That means the Davenport constant of an abelian group is the least numbersuch that any sequence of length n has a sum-zero nonempty subsequence.

As the §1 goes, if an irreducible element x has a decomposition

(x) = p1 . . . pn pi’s are prime ideals

with n > D(class group), then some of pi are product-zero in class group, thatmeans, the product of them is a principal ideal, contradict to the assumptionthat x is irreducible.

(2)——Fact. Each finite abelian group G is isomorphic to direct products ofcyclic group. That is,

G ∼=r⊕

i=1

Z/niZ with n1|n2| . . . |nr (∗)

for some r, n1, . . . , nr ∈ Z. Since by CRT, Z/mnZ = Z/mZ ⊕ Z/nZ if m,nrelative prime. Thus one can write also

G ∼=⊕p∈P

Gp Gp is a direct product of Z/piZ

Where Gp is called the p-torsion part of G.

Thus, we can assume that G is in form of (∗).

3 Davenport constants of cyclic groups

Maybe the cyclic group is the simplest case. Before our proof, let us seewhat happen in case of cyclic group. Let us fix G = Z/nZ. Firstly, we seethat D(G) ≥ n − 1, since the sequence of n − 1 copies of 1 has no nonemptysubsequence sum-zero.

(3)——Davenport constant for cyclic groups. For G = Z/nZ, D(G) =n.

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Proof. As we see before, it suffices to show that any sequences of lengthn have a nonempty subsequence sum-zero. Note that they can be presentedby a1, . . . , an where ai are integers. It is equivalent to show some nonemptysubsequence whose sum is 0 after moduling n. Let

s0 = 0 s1 = a1 s2 = a1 + s2 . . . sn = a1 + . . .+ an

by pigeon hole principle, for some i < j, si ≡ sj mod n. That means, ai+1 +. . .+ aj ≡ 0 mod n. The proof is complete. �

Then, it is necessary to consider what is the relationship between D(G),D(H)and D(G ⊕H). Firstly, it means there is sequence of length D(G) − 1 (resp.D(H)− 1) has no sum-zero subsequences, then we can product the sequence,we have

[D(G)− 1][D(H)− 1] ≤ D(G⊕H)− 1

And when m,n relatively prime and G = Z/nZ, H = Z/mZ, since Z/mnZ =Z/mZ⊕Z/nZ, left hand side is (n− 1)(m− 1) and right hand side is mn− 1.It implies that the equality does not always holds.

A remarkable guess is that for G = Z/nZ and H = Z/mZ with n|m, theequality holds. We will see it in §5.

(4)——Definition. Even more, following [5], if G is in the form of (∗) inFact2, write

M(G) = 1 +

r∑i=1

(ni − 1)

we see that D(G) ≤ M(G) by the discussion above, and D(G) = M(G)for G cyclic. Next, we will prove the equality holds for many situation inprocessing sections. The results of next two sections are all mainly due toOlson—[7] and [8]. However, M(G) = D(G) does not hold always. Evenmore, M(G)− D(G) can be arbitrary large, see [5].

4 Davenport constants of p-groups

A p-gourp is an abelian group of order pn for some prime p and integer n ≥ 1.By Fact2, a p-group G is a direct product of some Z/piZ for some i ≥ 1.

The proof due to Olson [7] used the concept of group ring. It works asgenerating functions.

(5)——Definition of Group Ring. For an additive group G, we can define

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the group ring

Z[G] =

formal sum∑g∈G

niXg : ni ∈ Z

with the nature addition structure. And let nXg ·mXh = nmXg+h, then oper-ator · (as multiplication) extends to whole Z[G] by distributive law. Moreover,we accept the convention that nX0 = n.

(6)——Example. For additive group G, let∑

g∈GXg ∈ Z[G].Then for fixed

h ∈ G, the number of solutions (x1, . . . , xn) in Gn of the equation

x1 + . . .+ xn = h

is the coefficient of Xh in(∑

g∈GXg)n

.

(7)——Davenport constants for p-groups. For p-group

G = Z/pn1Z⊕ . . .⊕ Z/pnrZ n1 ≤ . . . ≤ nr

we have D(G) = 1 +∑r

i=1(pni − 1). Then in language of definition4, thatmeans D(G) = M(G) for p-groups.

Proof. Let d = 1 +∑r

i=1(pni − 1). One have see that D(G) ≥ d. Put

ei = (0, . . . , 0, 1, 0, . . . , 0) ∈ G the 1 is at i-th position

It forms a Z-basis for G. It suffices to show for any sequence of length d havea subsequence sum-zero. Say, x• : x1, . . . , xd ∈ G.

Consider

χx• =

d∏i=1

(1−Xx1)

we claim that the coefficients of χx• are all 0 module p, note that

the coefficient of X0 = the number of odd subsequence sum-zero− the number of even subsequence sum-zero

where ‘even subsequence’ counts the empty subsequence. Thus if the claimholds, there exists some nonempty subsequence sum-zero.

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Since (1 −Xx+y) = (1 −Xx) + Xx(1 −Xy), one can deduce that χx• issum of products of χe• with some element in Z[G], where e• consists of ei’s.So it suffices to deal with the situation x• consisting of ei’s.

Assume that the multiplicity of ei is mi. Since∑r

i=1mi = d = 1 +∑ri=1(pni − 1), thus at least one of i satisfy mi > pni − 1, that is mi ≥ pni .

Now

(1−Xei)pni ≡ 1− (Xei)p

ni= 1−Xpniei = 1−X0 = 0 mod p

since pniei = 0. Thus χe• ≡ 0 mod p. The proof is complete. �

5 Davenport constant of Z/nZ× Z/nmZNow, assume that n ≥ 1 and m ≥ 1. The proof of the following result alsodue to Olson [8] is of more flavor of additive combinatorics. It needs basicknowledges of quotient groups.

(8)——Lemma. For prime p and G = Z/pZ⊕Z/pZ, any sequences of length3p− 2 has a nonempty subsequence sum-zero and of length at most p.

That is, for x1, . . . , x3p−2 ∈ G, there exists nonempty S ⊆ {1, . . . , 3p− 2}such that

∑s∈S as = 0 and |S| ≤ p.

Proof. We can include G into Z/pZ⊕ Z/pZ⊕ Z/pZ by

x = (x′, x′′) 7→ (x′, x′′, 1) =: (x, 1)

then, by Davenport constant for p-gourp theorem7, (xi, 1) have non-emptysubsequence sum-zero. Without loss of generality, let (x1, 1), . . . , (xs, 1) to besuch sequence, thus s ≡ 0 mod p, but s ≤ 3p − 2 implies s = p or 2p. Ifs = p, then we have done. If s = 2p, then, by Davenport constant for p-gourptheorem7 again, one can extract a proper subsequence no longer then p ofx1, . . . , xs sum-zero (since D(Z/pZ⊕Z/pZ) = 2p− 1, delate any element andtake complement if necessary). �

(9)——Davenport constants for Z/nZ×Z/nmZ. For group G = Z/nZ×Z/nmZ, one have D(G) = 1 + (n− 1) + (nm− 1) = M(G).

Proof. Let d = 1 + (n− 1) + (nm− 1) = M(G). Also, we have D(G) ≥ d.It suffices to show the inverse.

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We prove by induction on n. For n = 1, it follows by Davenport constantfor cyclic groups3. Now, assume n ≥ 2. Pick some p|n, and put H = pZ/nZ⊗pZ/nmZ, one see that G/H = Z/pZ⊕ Z/pZ.

For any sequence x1, . . . , xd, since d ≥ 3p − 2. One can extract a sub-sequence of length at most p, say x1, . . . , xs such that x1 + . . . + xs ∈ H.Then consider xp+1, . . . , xd, and continuous our process we will have u− 1 =n+mn

p −2 disjoint subsequence sum-in-H of length at most p. And the remainsare at least 2p − 1, one can extract subsequence with sum-in-H again. Lety1, . . . , yu ∈ H be the sum of them. But u = n+mn

p −1 = (n/p) +m(n/p)−1,one can extract a subsequence of yi by induction hypothesis. The proof iscomplete. �

In the proof of Davenport constant for cyclic groups 3, we proved thateach set of n integers contains some subset the sum of which is a multiple ofn. Here we have proven that each set of 2n − 1 Gaussian integers containssome subset the sum of which is a multiple of n.

Erdos, Ginzburg and Ziv[3] proved that each set of 2n−1 integers containssome subset of n elements the sum of which is a multiple of n. Where theterm ‘set’ can be replaced into ‘sequence’. Alon and Dubiner[2] gave alterna-tive proof using Chevalley-Warning Theorem. For more introduction on thismethod, see [6] P233. Now, we can give a alternative proof.

(10)——Erdos, Ginzburg and Ziv’s Theorem. Each set of 2n − 1integers contains some subset of n elements the sum of which is a multiple ofn.

Proof. For such sequence x1, . . . , x2n−1, consider them in Z/nZ ⊕ Z/nZas (x1, 1), . . . , (x2n−1, 1), then any subsequence sum-zero has length n (sinceone need exact n of 1’s at second indices). �

Actually, Erdos, Ginzburg and Ziv’s original proof resembled the proof ofDavenport constants for Z/nZ× Z/nmZ 9. It is not amazing that it followsby it.

References

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ztzl_kxnt/kxnt_sgxz/201512/t20151218_225337.html.

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[2] N. Alon and M. Dubiner. Zero-sum sets of prescribed size. In Combina-torics, Paul Erdos is eighty, Vol. 1, Bolyai Soc. Math. Stud., pages 33–50.Janos Bolyai Math. Soc., Budapest, 1993.

[3] P. Erdos, A. Ginzburg, and A. Ziv. Theorem in the additive numbertheory. Bull. Res. Counc. Israel Sect. F Math. Phys., 10F(1):41–43, 1961.

[4] Alfred Geroldinger and Imre Z. Ruzsa. Combinatorial number theory andadditive group theory. Advanced Courses in Mathematics. CRM Barcelon-a. Birkhauser Verlag, Basel, 2009. Courses and seminars from the Doc-Course in Combinatorics and Geometry held in Barcelona, 2008.

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