1. Introduction
Let 
be a finite additively written group (not necessarily commutative). Let 
be a subset of 
Define 
{
are distinct 
}. For technical reasons we define 
We call 
an additive basis of 
if 
The critical number 
of 
is the smallest integer 
such that every subset 
of 
with 
forms an additive basis of 
was first introduced and studied by Erdős and Heilbronn in 1964 [1] for 
where 
is a prime. This parameter has been studied for a long time and its exact value is known for a large number of groups (see [2-10]).
Following Erdős [1], we say that
is complete if
and incomplete otherwise.
In this paper, we would like to study the following question: What is the structure of a relatively large incomplete set? Technically speaking, we would like to have a characterization for incomplete sets of relatively large size. Such a characterization has been obtained recently for finite abelian groups (see [11-13]). In this paper, we shall prove the following result.
Theorem 1.1. Let
be a finite nilpotent group with order 
where 
is the smallest prime dividing 
Also assume that 
is composite and 
Let
be a subset of 
such that 
If 
is incomplete, then there exist a subgroup 
of order 
and 
such that 
2. Notations and Tools
If 
be a subset of the group
, we shall denote by 
the cardinality of
, by 
the subgroup generated by
. If 
are subsets of
, let 
denote the set of all sums
, where 
Recall the following well known result obtained by Cauchy and Davenport.
Lemma 2.1. Let 
be a prime number. Let 
and 
be non-empty subsets of 
Then
We also use the following well known result.
Lemma 2.2 [14]. Let 
be a finite group. Let 
and 
be subsets of 
such that 
Then
Lemma 2.3 [3]. Let 
be a cyclic group of order
, where 
are primes. Then
Lemma 2.4 [8]. Let 
be a non-abelian group of order 
where 
are distinct primes. Then 
Lemma 2.5 [10]. Let 
be a finite nilpotent group of odd order and let 
be the smallest prime dividing 
If 
is a composite number then 
Lemma 2.6. Let 
be a finite nilpotent group of odd order and let
be the smallest prime dividing 
If 
then 
Proof. Obviously, this follows from Lemmas 2.3-2.5.
Lemma 2.7 [15]. Let 
be a subset of a finite group 
of order
. If 
then 
Lemma 2.8 [16]. Let 
be a noncyclic group. Let 
be a subset 
Then 
Let 
and 
As usual, we write 
We have the following result obtained by Olson.
Lemma 2.9 [5]. Let
be a nonempty subset of 
and 
Let 
Then
We shall also use the following result of Olson.
Lemma 2.10. Let 
be a finite group and let
be a generating subset of 
such that 
Let
be a subset of 
such that 
Then there is
such that
This result follows by applying Lemma 3.1 of [15] to 
Let 
be a subset of
with cardinality
Let 
be an ordering of 
For 
set 
and 
The ordering 
is called a resolving sequence of 
if, for each 
The critical index of the resolving sequence is the largest
such that
generates a proper subgroup of
. Clearly, every nonempty subsets
has a resolving sequence.
We need the following basic property of resolving sequence which is implicit in [5].
Lemma 2.11. Let
be a generating subset of a finite group
such that
and 
Let the ordering
be a resolving sequence of
with critical index 
Then, there is a subset 
such that
and
Proof. This is essentially formula (4) of [5]. By Lemma 2.9 we have
By Lemma 2.10 we have 
for each
On the other hand, by Lemma 2.8 we have
By the definition of
, we have
By taking
we have the claimed inequality.
Lemma 2.12. Let
be a finite group with order 
where 
is the smallest prime dividing 
and 
Let 
be a subset of 
such that 
and 
Then there exists a set 
such that 
and
Proof. Since 
and 
is the smallest prime dividing 
we have 
By Lemma 2.7, 
Clearly, we may partition 
such that 
and 
We consider two cases.
Case 1. 
Set
. By Lemma 2.10, there is
such that
It follows 
by Lemma 2.9.
Since 
we have, by Lemma 2.2,
.
Case 2.
.
By Lemma 2.2,
. Put
. By Lemma 2.10, there is
, such that
.
Therefore,
By Lemma 2.2,
implies
.
In both cases, one of the sets 
verifies the conclusion of the lemma. This completes the proof.
Lemma 2.13. Let
, where 
is the smallest prime dividing
If
and 
then 
Proof. Set
and
.
First, let us show that
. Assume the contrary that 
We have
Since
, we have
a contradiction to 
Second, let us show that
.
Assume the contrary. Since
,
, we have 1) 
On the other hand, since
, we have
.
Then, 
A contradiction to (1). Therefore, we have
This completes the proof.
Lemma 2.14. Let
be a finite group with order
. Let
be a proper subgroup of
and
a subset of 
If 
and 
is a primethen
.
Moreover, if 
then there is
such that 
Proof. By 
we shall mean
, where 
is the canonical morphism. Put
.
From our assumption we have 
By Lemma 2.1, we have
It follows that 
Assume now
. If there is 
such that 
say 
then 
By Lemma 2.1, we have
a contradiction to 
Then there is 
such that
3. Proof of Theorem 1.1
Proof. By Lemma 2.12 there exists a set 
such that
, 
and
(2)
We have
Therefore 
generates 
By Lemma 2.11, there is
such that
verifying
(3)
Let 
be the subgroup generated by
and let
be the smallest prime dividing
.
Put 
Set
By (2) and (3), we have 
By Lemma 2.13, we have
By Lemma 2.6, we get 
Since 
we see easily that 
is a prime. Since
is incomplete, we have
By Lemma 2.14, 
We have
which implies
and 
Hence,
By Lemma 2.14, there exist a subgroup
of order
and 
such that
4. Acknowledgements
The authors would like to thank the referee for his/her very useful suggestions. This work has been supported by the National Science Foundation of China with grant No. 11226279 and 11001035.