INVERSE ZERO-SUM PROBLEMS 1. Introduction Let G be an additive finite abelian group with

37 Pages·2007·0.32 MB·English

INVERSE ZERO-SUM PROBLEMS
WEIDONG GAO, ALFRED GEROLDINGER AND WOLFGANG A. SCHMID
1. Introduction
Let G be an additive finite abelian group with exponent exp(G) = n.
We define some central invariants in zero-sum theory: Let
• D(G) denote the smallest integer l ∈ N such that every sequence S
over G of length |S| ≥ l has a zero-sum subsequence.
• η(G) denote the smallest integer l ∈ N such that every sequence
S over G of length |S| ≥ l has a zero-sum subsequence T of length
|T| ∈ [1,n].
• s(G) denote the smallest integer l ∈ N such that every sequence
S over G of length |S| ≥ l has a zero-sum subsequence T of length
|T| = n.
All these three invariants have been studied since the 1960s initiated by
the works of P. Erd˝os et al. (see [17, 36], and for more detailed historical
information see [23, 24]). For groups of rank at most two the precise values
of all three invariants are well known (see Theorem 2.4). In groups of higher
rank precise values are known only in very special cases (see [10, 24, 3] and
the introduction of Section 3).
The investigation of inverse problems has a long tradition in combinato-
rial number theory (see [34]), and more recently it has been promoted by
applications in the theory of non-unique factorizations (see [30]). In the
present paper we study the inverse problems associated to the above in-
variants. More precisely, we investigate the structure of sequences of length
D(G)−1 (η(G)−1 or s(G)−1 respectively) that do not have a zero-sum
subsequence (of the required length). For cyclic groups these questions are
completely settled. Indeed, the answer for the invariants D(G) and η(G) is
straightforward (see Theorem 2.1), and the inverse problem for the invari-
ant s(G), with G cyclic, gave rise to a great variety of investigations (see
[5, 8, 18, 9, 19, 4, 38, 26, 31, 29]). In this paper we study the problems for
groups G of the form G = Cr, with n,r ≥ 2, where the emphasis is laid on
n
groups of rank two.
2000 Mathematics Subject Classification. 11P70, 11B50.
1
2 WEIDONG GAO, ALFRED GEROLDINGER AND WOLFGANG A. SCHMID
Consider the following two properties.
Property C. Every sequence S over G of length |S| = η(G) − 1 that
has no zero-sum subsequence of length in [1,n] has the form S = Tn−1 for
some sequence T over G.
Property D. Every sequence S over G of length |S| = s(G) − 1 that
has no zero-sum subsequence of length n has the form S = Tn−1 for some
sequence T over G.
Property C was first considered by P. van Emde Boas and Property D by
W. Gao (see [16, 21]). At the beginning of Section 3 we discuss the state of
knowledge on these properties. In [24, Conjecture 7.2] it is conjectured that
everygroupG = Cr,wherer ∈ Nandn ∈ N ,hasPropertyD.InTheorem
n ≥2
3.2 we show that Property C and Property D are both multiplicative, and
thusthisconjectureisessentiallyreducedtothecaseofelementaryp-groups.
Let G = C ⊕C with n ≥ 2. It is conjectured that every minimal zero-
n n
sum sequence S over G of length |S| = D(G) contains some element with
multiplicity n−1 (for several equivalent conditions see [30, Theorem 5.8.7]).
If this conjecture is true, then G has Property C (see [23, Theorem 6.2] and
[24, Theorem 6.7.2.(b)]). In Theorem 4.1 we show that, if ε > 0 and n is a
sufficiently large prime, then such a sequence S contains one element with
multiplicity greater than n1/4−ε. The proof rests on a variety of addition
theorems, among them the Theorem of Dias da Silva–Hamidoune (which
settles the Erd˝os–Heilbronn conjecture).
In Section 5 we study the analogue of Property D for sets (that is, for
squarefree sequences) in the group G = C ⊕ C , where p is a prime. W.
p p
Gao and R. Thangadurai proved that the maximal size of a set S ⊂ C ⊕C ,
p p
for primes p ≥ 67, without a subset of size p that sums to zero, is 2p − 2
(see [28]). In Theorem 5.1 we completely determine the structure of all
extremal sets, and moreover, we slightly improve the bound to p ≥ 47. The
proof of the inverse result is analogue to the proof of the direct result, but
additionally employs some (recent) ideas from the study of Brakemeier’s
function due to A. Bialostocki, M. Lotspeich [5] and F. Hennecart [31] (see
Lemma 5.9).
A crucial idea in all our work on groups of rank two is to find suitable
epimorphisms to cyclic groups, to use results known for cyclic groups and
then to shift the information back to the groups of rank two. We make use
of classical results, such as the Cauchy–Davenport Theorem, and list the
other needed results on cyclic groups at the end of Section 2 (see Theorems
INVERSE ZERO-SUM PROBLEMS 3
2.1 to 2.3). Of course we need the solutions of the original direct problems,
which are summarized in Theorem 2.4.
Throughout this article, let G be an additive finite abelian group.
2. Notations and some main tools
Our notations and terminology is consistent with [24] and [30]. We briefly
gather some key notions and fix the notations concerning sequences over
finite abelian groups. Let N denote the set of positive integers, P ⊂ N the
set of all prime numbers and let N = N∪{0}. For real numbers a,b ∈ R
0
we set [a,b] = {x ∈ Z | a ≤ x ≤ b}, and we denote by bac the largest
integer that is less than or equal to a, and by dae the smallest integer that
is greater than or equal to a.
Throughout, all abelian groups will be written additively. For n ∈ N, let
C denote a cyclic group with n elements. We have
n
∼
G = C ⊕...⊕C ,
n1 nr
where r = r(G) ∈ N is the rank of G, n ,...,n ∈ N are integers with
0 1 r
1 < n | ... | n and n = exp(G) is the exponent of G. Let s ∈ N. An
1 r r
s-tuple (e ,...,e ) of elements of G is said to be independent if e 6= 0 for
1 s i
all i ∈ [1,s] and, for every s-tuple (m ,...,m ) ∈ Zs,
1 s
m e +...+m e = 0 implies m e = ... = m e = 0.
1 1 s s 1 1 s s
An s-tuple (e ,...,e ) of elements of G is called a basis if it is independent
1 s
and G = he i⊕...⊕he i.
1 s
Let F(G) be the free abelian monoid, multiplicatively written, with basis
G. The elements of F(G) are called sequences over G. We write sequences
S ∈ F(G) in the form
Y
S = gvg(S), with v (S) ∈ N for all g ∈ G.
g 0
g∈G
We call v (S) the multiplicity of g in S, and we say that S contains g,
g
if v (S) > 0. S is called squarefree if v (S) ≤ 1 for all g ∈ G. The unit
g g
element 1 ∈ F(G) is called the empty sequence. A sequence S is called a
1
subsequence of S if S |S in F(G) (equivalently, v (S ) ≤ v (S) for all
1 g 1 g
g ∈ G), anditiscalleda proper subsequence ofS ifitisasubsequencewith
1 6= S 6= S. If a sequence S ∈ F(G) is written in the form S = g ·...·g ,
1 1 l
we tacitly assume that l ∈ N and g ,...,g ∈ G.
0 1 l
4 WEIDONG GAO, ALFRED GEROLDINGER AND WOLFGANG A. SCHMID
For a sequence
Y
S = g ·...·g = gvg(S) ∈ F(G),
1 l
g∈G
we call
X
|S| = l = v (S) ∈ N the length of S,
g 0
g∈G
h(S) =max{v (S) | g ∈ G} ∈ [0,|S|]
g
the maximum of the multiplicities of S,
supp(S) = {g ∈ G | v (S) > 0} ⊂ G the support of S,
g
l
X X
σ(S) = g = v (S)g ∈ G the sum of S,
i g
i=1 g∈G
nX (cid:12) o
Σ (S) = g (cid:12) I ⊂ [1,l] with |I| = k
k i (cid:12)
i∈I
the set of k-term subsums of S, for all k ∈ N,
[ [
Σ (S) = Σ (S), Σ (S) = Σ (S),
≤k j ≥k j
j∈[1,k] j≥k
and
Σ(S) = Σ (S) the set of (all) subsums of S.
≥1
The sequence S is called
• zero-sumfree if 0 ∈/ Σ(S),
• a zero-sum sequence if σ(S) = 0,
• a minimal zero-sum sequence if it is a non-empty zero-sum sequence
and every proper subsequence is zero-sumfree,
• a short zero-sum sequence if it is a zero-sum sequence of length |S| ∈
[1,exp(G)].
Throughout the paper, we tacitly use the following argument: If g ∈ G
with ord(g) = exp(G) = n, then S = g ·...·g has zero-sum subsequence
1 l
of length n if and only if the shifted sequence S0 = (g − g) · ... · (g − g)
1 l
has a zero-sum subsequence of length n.
Every group homomorphism ϕ: G → H extends to a homomorphism
ϕ: F(G) → F(H) where ϕ(S) = ϕ(g ) · ... · ϕ(g ). Obviously, ϕ(S) is
1 l
a zero-sum sequence if and only if σ(S) ∈ Ker(ϕ). If m ∈ N, m|n and
1
ϕ: G → G is the multiplication by m (defined by ϕ(g) = mg for every
g ∈ G), then Ker(ϕ) ∼= Cr and ϕ(G) ∼= C ⊕···⊕C .
m n1/m nr/m
Nowwegathertheresultsoncyclicgroupsthatwillbeneededthroughout
this paper.
INVERSE ZERO-SUM PROBLEMS 5
Theorem 2.1. Let G be cyclic of order n ≥ 3 and S ∈ F(G) a zero-
sumfree sequence of length |S| ≥ (n + 1)/2. Then there exists some g ∈
supp(S) such that v (S) ≥ 2|S|−n+1. In particular, D(G) = n and the
g
following statements hold:
(a) If |S| = n−1, then S = gn−1.
(b) If |S| = n−2, then either S = gn−3(2g) or S = gn−2.
(c) If |S| = n−3, then S has one of the following forms:
gn−5(2g)2, gn−4(2g), gn−4(3g), gn−3.
Proof. This is due to J.D. Bovey, P. Erd˝os and I. Niven (see [7] for the
original paper and also [30, Theorem 5.4.5.2]). (cid:3)
Theorem 2.2. Let G be prime cyclic of order p ∈ P and S ∈ F(G).
1. Let v (S) = 0 and |S| = p. Then Σ (S) = G, and in particular,
0 ≤h(S)
S has a zero-sum subsequence of length at least p−h(S).
2. Let k ∈ [2,p − 1], |S| ≥ 2p − k and h(S) ≤ p − k. Then S has a
zero-sum subsequence of length p.
3. Let p ≥ 13, v (S) = 0, k ∈ [(p + 5)/2,p − 4], |S| ≥ 2p − k − 2 and
0
h(S) ≤ k. Then S has a zero-sum subsequence T of length |T| ∈
[p+2−k,p−2].
Proof. 1. and 2. See [28, Lemma 2.6 and Theorem 2.7].
3. Clearly, S has a subsequence S0 of length |S0| = p and with h(S0) ≤
k −2. By 1., S0 has a zero-sum subsequence T of length |T| ≥ p−h(S0) ≥
p+2−k. If |T| ≤ p−2, then we are done. If |T| ≥ p−1, then h(T) ≤ p−4
andTheorem2.1implythatT isnotaminimalzero-sumsubsequence. Thus
it has a proper zero-sum subsequence T0 of length |T0| ∈ [(p−1)/2,p−2] ⊂
[p+2−k,p−2]. (cid:3)
Theorem 2.3. Let G be prime cyclic of order p ∈ P, S ∈ F(G) a
squarefree sequence and k ∈ [1,|S|].
1. |Σ (S)| ≥ min{p,k(|S|−k)+1}.
k
2. If k = b|S|/2c, then |Σ (S)| ≥ min{p,(|S|2 +3)/4}
k
√
3. If |S| = b 4p−7c+1 and k = b|S|/2c, then Σ (S) = G.
k
Proof. This is due to J.A. Dias da Silva and Y. ould Hamidoune (see [11] for
the original paper, and also [2] and [34, Theorems 3.4 and 3.8]). Obviously,
2. and 3. are special cases of 1., which will be needed repeatedly in this
form. (cid:3)
6 WEIDONG GAO, ALFRED GEROLDINGER AND WOLFGANG A. SCHMID
Theorem 2.4. Let G = C ⊕C with 1 ≤ n |n . Then
n1 n2 1 2
s(G) = 2n +2n −3, η(G) = 2n +n −2 and D(G) = n +n −1.
1 2 1 2 1 2
Proof. The result on D(G) goes back to D. Kruyswijk and J.E. Olson, and
the result on s(G) is based on C. Reiher’s result that s(C ⊕C ) = 4p−3
p p
(see [35] and [30, Theorem 5.8.3]). (cid:3)
3. Properties C and D in groups of the form Cr
n
Let G = Cr with n ≥ 2,r ∈ N and and suppose that Property D
n
holds. Then, by definition, there exists some c(G) ∈ N such that s(G) =
c(G)(n−1)+1. Moreover, a simple argument shows that Property C holds
(see [24, Section 7]) and that η(G) = (c(G)−1)(n−1)+1 (see [13, Lemma
2.3]). For r = 1 we have c(G) = 2 and for r = 2 we have c(G) = 4 (see
Theorem 2.4). In case of higher ranks bounds for c(G) were given by N.
Alon and M. Dubiner (see [1]) and then by Y. Edel, C. Elsholtz and S.
Kubertin (see [32, 15, 13, 12]).
We make use of the simple fact that η(Cr) = 2r and s(Cr) = 2r + 1
2 2
(see [30, Corollary 5.7.6]). It follows from the very definition that Cr satis-
2
fies Property D, and a straightforward argument shows that Cr satisfies
3
Property D (see [13, Lemma 2.3.3] and the subsequent discussion). How-
ever, in general only very little is known about Property D. If r = 2 and
n ∈ {2,3,5,7}, thenGhasPropertyDby[37], andforfirstresultsingroups
of higher rank we refer to [25]. Property C was first studied for groups of
rank two in connection with investigations of the Davenport constant of
groups having rank three (see [16, 20], and also [24, Theorem 7.9]).
Theorem 3.2 shows that both Property C and Property D are multi-
plicative, provided that the c(·) invariants of all involved groups coincide
(in contrast to this assumption see the investigation of C3 in [25]). This re-
6
sult generalizes former work (see [21] and [27, Theorem 1]), and reduces the
conjecture, that all groups of the form Cr satisfy the discussed properties,
n
to the case of elementary p-groups (under the above assumption on the c(·)
invariants).
We start with a lemma that is used in the proof of the following theorem.
It is closely related to [27, Theorem 2].
Lemma 3.1. Let G = Cr with n ≥ 2, r ∈ N and let S ∈ F(G) be a
n
sequence of length |S| = η(G)+n−2 that has no zero-sum subsequence
of length n. If G has Property C and h(S) ≥ b(n−1)/2c, then S = Tn−1
for some T ∈ F(G).
INVERSE ZERO-SUM PROBLEMS 7
Proof. Suppose that G has Property C and let g ∈ G such that v (S) ≥
g
b(n − 1)/2c. We assume g = 0 and set S = 0vT with v = v (S) and
0
T ∈ F(G). Since S has no zero-sum subsequence of length n, it follows that
T has no zero-sum subsequence of length in [n−v,n].
Assume to the contrary that v ≤ n−2. Since |T| ≥ η(G)+(n−2)−v ≥
η(G), T has a short zero-sum subsequence, and let T be a short zero-sum
1
subsequence of maximal length. Then we get
|T | ≤ n−1−v and |T−1T| ≥ η(G)−1.
1 1
Since h(T−1T) ≤ h(S) = v ≤ n−2 and G has Property C, it follows that
1
T−1T has a short zero-sum subsequence T . Since T has maximal length,
1 2 1
we obtain that |T T | ≥ n+1 and hence
1 2
n+1 n−3 n+1
≤ |T | ≤ n−1−v < n−1− = ,
1
2 2 2
a contradiction.
Therefore we have v = n−1 and |T| = η(G)−1. Since T has no short
zero-sum subsequence and G has Property C, it follows that S has the
required form. (cid:3)
Theorem 3.2. Let G = Cr with m,n,r ∈ N and let c ∈ N.
mn
1. If both Cr and Cr have Property D and
m n
s(Cr )−1 s(Cr)−1 s(Cr )−1
m = n = mn = c,
m−1 n−1 mn−1
then G has Property D.
2. If both Cr and Cr have Property C and
m n
η(Cr )−1 η(Cr)−1 η(Cr )−1
m = n = mn = c,
m−1 n−1 mn−1
then G has Property C.
3. If Cr has Property D, Cr has Property C, s(Cr) = η(Cr)+n−1,
m n n n
s(Cr )−1 s(Cr)−1 s(Cr )−1 nrb(n−1)/2c+c
m = n = mn = c and m ≥ ,
m−1 n−1 mn−1 cn
then G has Property D.
Proof. The proofs of all three assertions are based on the inductive method.
We need the following terminology. Let k ∈ N, s(Cr) = c(k −1)+1 and
k
S ∈ F(Cr). We say that S is of Type D if |S| = c(k −1), S has no zero-
k
sum subsequence of length k, and S = Tk−1 for some T ∈ F(Cr). Thus the
k
group Cr has Property D if and only if every sequence of length c(k − 1)
k
that has no zero-sum subsequence of length n is of Type D. Similarly, if
η(Cr) = c(k −1)+1 and S ∈ F(Cr), then we say that S is of Type C if
k k
|S| = c(k − 1), S has no short zero-sum subsequence, and S = Tk−1 for
8 WEIDONG GAO, ALFRED GEROLDINGER AND WOLFGANG A. SCHMID
some T ∈ F(Cr). Again, the group Cr has Property C if and only if every
k k
sequence of length c(k − 1) that has no short zero-sum subsequence is of
Type C.
Obviously we may assume that m,n ≥ 2. Let ϕ: G → G denote the
multiplication by m. Then Ker(ϕ) ∼= Cr and ϕ(G) = mG ∼= Cr.
m n
1. Let S ∈ F(G) be of length |S| = c(mn−1) such that S has no zero-
sum subsequence of length mn. We have to show that S is of Type D. Since
v (S) ≤ mn−1 for every g ∈ G, it suffices to show that |supp(S)| = c.
g
Since |S| = n(c(m − 1)) + c(n − 1), it follows that S admits a product
decomposition
S = S ·...·S S0,
1 c(m−1)
where S ,...,S ,S0 ∈ F(G) and, for every i ∈ [1,c(m−1)], ϕ(S ) has
1 c(m−1) i
sum zero and length |S | = n (see [30, Lemma 5.7.10]). Since S has no
i
zero-sum subsequence of length mn, ϕ(S0) has no zero-sum subsequence of
length n. Clearly, we have
|ϕ(S0)| = |S0| = |S|−nc(m−1) = c(n−1),
and thus ϕ(S0) is of Type D whence in particular |supp(ϕ(S0))| = c. We
continue with the following assertion.
A1. |supp(ϕ(S))| = |supp(ϕ(S0))|.
Proof of A1. It suffices to verify that for every i ∈ [1,c(m−1)] there is
some h ∈ supp(ϕ(S0)) such that ϕ(S ) = hn. Assume to the contrary that
i
there is some i ∈ [1,c(m−1)] for which this does not hold. We assert that
ϕ(S S0) is divisible by a product of two zero-sum subsequences of length n.
i
This implies that ϕ(S) is divisible by a product of c(m−1)+1 zero-sum
subsequences of length n, whence S has a zero-sum subsequence of length
mn, a contradiction. We distinguish two cases.
CASE 1: supp(ϕ(S0))∩supp(ϕ(S )) 6= ∅.
i
Let h ∈ supp(ϕ(S0)) ∩ supp(ϕ(S )). Then hn|ϕ(S S0). Since ϕ(S )
i i i
is a zero-sum sequence of length n and distinct from hn, it follows that
|supp(h−1ϕ(S ))| ≥ 2. Therefore h−nϕ(S S0) is not of Type D, and thus it
i i
has a zero-sum subsequence of length n.
CASE 2: supp(ϕ(S0))∩supp(ϕ(S )) = ∅.
i
Let h ∈ supp(ϕ(S )) and U a zero-sum subsequence of hϕ(S0) of length n.
i
Then v (U) ≥ 1 and |supp(h−1U)| ≥ 2. Consequently supp(U−1hϕ(S0)) =
h
(cid:0) (cid:1)
supp(ϕ(S0)) and therefore we obtain that |supp (U−1hϕ(S0))(h−1ϕ(S )) | >
i
|supp(ϕ(S0))|. Therefore U−1ϕ(S S0) is not of Type D, and thus it has a
i
zero-sum subsequence of length n.
INVERSE ZERO-SUM PROBLEMS 9
It remains to show that
|ϕ−1(h)∩supp(S)| = 1 for every h ∈ supp(ϕ(S)),
since then we obtain that
|supp(S)| = |supp(ϕ(S))| = |supp(ϕ(S0))| = c.
(cid:0) (cid:1)
Let h ∈ supp ϕ(S ·...·S ) , and assume to the contrary that there are
1 c(m−1)
two distinct elements g,g0 ∈ supp(S) such that ϕ(g) = ϕ(g0) = h. By A1
we may suppose that g|S ·...·S and g0|S0, say g|S . Since S has no
1 c(m−1) 1
zero-sum subsequence of length mn, the sequence σ(S )·...·σ(S ) ∈
1 c(m−1)
F(Ker(ϕ))hasnozero-sumsubsequenceoflengthmwhenceitisofTypeD.
We consider S0 = g−1g0S . Then ϕ(S0) = ϕ(S ) and hence σ(S ) 6= σ(S0) ∈
1 1 1 1 1 1
Ker(ϕ). Thus the sequence σ(S0)σ(S )·...·σ(S ) ∈ F(Ker(ϕ)) is not
1 2 c(m−1)
of Type D (note that in case m = 2 we have c = |Ker(ϕ)|) whence it has a
zero-sum subsequence of length m, a contradiction.
(cid:0) (cid:1)
It remains to verify that supp(ϕ(S)) = supp ϕ(S ·...·S ) . We set
1 c(m−1)
c0 = |supp(ϕ(S ·...·S ))| and check that c = c0.
1 c(m−1)
By A1 we have |supp(ϕ(S))| = c and the above argument shows that
(cid:0) (cid:1)
|supp(S ·...·S )| = |supp ϕ(S ·...·S ) |. Thus it follows that
1 c(m−1) 1 c(m−1)
X
c(mn−1) = |S| = v (S)
g
g∈supp(S)
X
= v (S)
g
g∈supp(S1·...·Sc(m−1))
(cid:0) (cid:1)
+ |supp(ϕ(S))|−|supp(ϕ(S ·...·S ))| (n−1)
1 c(m−1)
≤ c0(mn−1)+(c−c0)(n−1)
whence
(c−c0)(mn−1) ≤ (c−c0)(n−1) and c = c0.
2. Let S ∈ F(G) be of length |S| = c(mn−1) such that S has no zero-
sum subsequence of length in [1,mn]. We have to show that S is of Type C.
Since v (S) ≤ mn−1 for every g ∈ G, it suffices to show that |supp(S)| = c.
g
Since |S| = n(c(m − 1)) + c(n − 1), it follows that S admits a product
decomposition
S = S ·...·S S0,
1 c(m−1)
where S ,...,S ,S0 ∈ F(G) and, for every i ∈ [1,c(m − 1)], ϕ(S )
1 c(m−1) i
has sum zero and length |S | ∈ [1,n] (see [30, Lemma 5.7.10]). Since S
i
has no zero-sum subsequence of length in [1,mn], ϕ(S0) has no zero-sum
subsequence of length in [1,n] and every sequence ϕ(S ) is a minimal zero-
i
sum sequence. Clearly, we have
|ϕ(S0)| = |S0| ≥ |S|−nc(m−1) = c(n−1),
10 WEIDONG GAO, ALFRED GEROLDINGER AND WOLFGANG A. SCHMID
and thus in fact |S | = n for every i ∈ [1,c(m−1)] and furthermore ϕ(S0)
i
is of Type C whence in particular |supp(ϕ(S0))| = c. We continue with the
following assertion.
A2. |supp(ϕ(S))| = |supp(ϕ(S0))|.
Proof of A2. It suffices to verify that for every i ∈ [1,c(m−1)] there is
some h ∈ supp(ϕ(S0)) such that ϕ(S ) = hn. Assume to the contrary that
i
there is some i ∈ [1,c(m−1)] for which this does not hold. We assert that
ϕ(S S0) is divisible by a product of two zero-sum subsequences of length in
i
[1,n]. This implies that ϕ(S) is divisible by a product of c(m−1)+1 zero-
sum subsequences of length in [1,n], whence S has a zero-sum subsequence
of length in [1,mn], a contradiction. We distinguish two cases.
CASE 1: supp(ϕ(S0))∩supp(ϕ(S )) 6= ∅. We recall that |ϕ(S )| = n. The
i i
remaining argument is analogue to the according one in 1.
CASE 2: supp(ϕ(S0))∩supp(ϕ(S )) = ∅.
i
Let h ∈ supp(ϕ(S )) and U a zero-sum subsequence of hϕ(S0) of length
i
in [1,n]. If |U| < n, then
|U−1ϕ(S S0)| ≥ |S0|+1 = c(n−1)+1 = η(Cr),
i n
and therefore U−1ϕ(S S0) has a zero-sum subsequence of length in [1,n].
i
Suppose that |U| = n, and note that v (U) ≥ 1 and |supp(h−1U)| ≥
h
2. This implies that supp(U−1hϕ(S0)) = supp(ϕ(S0)) and thus we get
(cid:0) (cid:1)
|supp (U−1hϕ(S0))(h−1ϕ(S )) | > |supp(ϕ(S0))|. Therefore U−1ϕ(S S0) is
i i
not of Type C, and thus it has a zero-sum subsequence of length in [1,n].
It remains to show that
|ϕ−1(h)∩supp(S)| = 1 for every h ∈ supp(ϕ(S)),
since then we obtain that
|supp(S)| = |supp(ϕ(S))| = |supp(ϕ(S0))| = c.
(cid:0) (cid:1)
Let h ∈ supp ϕ(S ·...·S ) , and assume to the contrary that there
1 c(m−1)
are two distinct elements g,g0 ∈ supp(S) such that ϕ(g) = ϕ(g0) = h.
By A2 we may suppose that g|S · ... · S and g0|S0, say g|S .
1 c(m−1) 1
Since S has no zero-sum subsequence of length in [1,mn], the sequence
σ(S )·...·σ(S ) ∈ F(Ker(ϕ)) has no zero-sum subsequence of length
1 c(m−1)
in [1,m] whence it is of Type C. We consider S0 = g−1g0S . Then ϕ(S0) =
1 1 1
ϕ(S ) and hence σ(S ) 6= σ(S0) ∈ Ker(ϕ). Thus the sequence σ(S0)σ(S )·
1 1 1 1 2
...·σ(S ) ∈ F(Ker(ϕ)) is not of Type C (note that in case m = 2 we
c(m−1)
have c = |Ker(ϕ)|−1) whence it has a zero-sum subsequence of length in
[1,m], a contradiction.