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Davenport Constants of Finite Abelian Groups

Rui Xiong

May 24, 2018

Abstract

Davenport constant of a finite abelian group G (writing additive) is the least number n such that for any sequence a1, . . . , an there exists a nonempty subsequence which sums zero. Davenport constant has not been solved completely, see [1] for more information. In this article, we will 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 from algebraic number theory. It is well known that any integer can be factorized into products of primes. One sees that it may fail if we extend the ring to a little 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 elements still exists for each element (since they are all noetherian), but it may not be unique.

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

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

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

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

2

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 number such 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, that means, the product of them is a principal ideal, contradict to the assumption that x is irreducible.

(2)——Fact. Each finite abelian group G is isomorphic to direct products of cyclic 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,n relative 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 see what happen in case of cyclic group. Let us fix G = Z/nZ. Firstly, we see that D(G) ≥ n − 1, since the sequence of n − 1 copies of 1 has no nonempty subsequence sum-zero.

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

3

Proof. As we see before, it suffices to show that any sequences of length n have a nonempty subsequence sum-zero. Note that they can be presented by a1, . . . , an where ai are integers. It is equivalent to show some nonempty subsequence 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, the equality holds. We will see it in §5.

(4)——Definition. Even more, following [5], if G is in the form of (∗) in Fact2, 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 in processing sections. The results of next two sections are all mainly due to Olson—[7] and [8]. However, M(G) = D(G) does not hold always. Even more, 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 as generating functions.

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

4

the group ring

Z[G] =

formal sum∑ g∈G

niX g : 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∈GX g ∈ 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∈GX g )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(p ni − 1). Then in language of definition4, that

means D(G) = M(G) for p-groups.

Proof. Let d = 1 + ∑r

i=1(p ni − 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 have a 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 claim holds, there exists some nonempty subsequence sum-zero.

5

Since (1 −Xx+y) = (1 −Xx) + Xx(1 −Xy), one can deduce that χx• is sum 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 +∑r i=1(p

ni − 1), thus at least one of i satisfy mi > pni − 1, that is mi ≥ pni . Now

(1−Xei)p ni ≡ 1− (Xei)p

ni = 1−Xp

niei = 1−X0 = 0 mod p

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

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

(8)——Lemma. For prime p and G = Z/pZ⊕Z/pZ, any sequences of length 3p− 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-empty subsequence sum-zero. Without loss of generality, let (x1, 1), . . . , (xs, 1) to be such sequence, thus s ≡ 0 mod p, but s ≤ 3p − 2 implies s = p or 2p. If s = p, then we have done. If s = 2p, then, by Davenport constant for p-gourp theorem7 again, one can extract a proper subsequence no longer then p of x1, . . . , xs sum-zero (since D(Z/pZ⊕Z/pZ) = 2p− 1, delate any element and take 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.

6

We prove by induction on n. For n = 1, it follows by Davenport constant for 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