On the non-existence of Abelian Moore Cayley graphs with excess one

Abstract The order of an Abelian Cayley graph of degree 2n and diameter 2 cannot exceed 2n + 2n+ 1, which is the famous Abelian Cayley-Moore bound. Leung and Zhou [J. Combin. Theory Ser. A 171 (2020) Art# 105157] recently shown that such a graph attaining the aforementioned bound exists if and only if n = 1, 2. This note is concerned with the Abelian Cayley graphs of degree 2n and diameter 3, whose order is 2n + 2n + 2, one larger than the Abelian Cayley-Moore bound of degree 2n and diameter 2. Their generating set denoted by S satisfies |S̃| = 2n + 2n+ 1 where S̃ = S ∪ {e}, e being the identity element of the underlying group. For n = 1, 2, it is easy to find examples. For n > 2, several non-existence results for infinitely many values of n are provided by using two methods, which are related to symmetric polynomials theories and algebraic number theory.


Introduction
For positive integers d and k, denote by n d,k the largest order of a graph of maximum degree d and diameter k. The wellknown and widely studied degree-diameter problem is to determine the largest number n d,k . The famous upper bound on the number n d,k is the Moore bound, which gives n d,k ≤ 1 + d + d(d − 1) + · · · + d(d − 1) k−1 for every d, k ≥ 1. Except for k = 1 or d ≤ 2, graphs achieving the Moore bound exist only for k = 2 and d = 3, 7, and possibly 57; see [1,6,8]. Until now, it is not known yet whether a Moore graph of degree d = 57 and diameter k = 2 on 3250 vertices exists. For a summary on the history and development on this topic, we refer the reader to the survey paper of Miller andŠiráň [14].
Let G be a finite group with identity element e and let S ⊂ G be a unit-free, inverse-closed generating set for G (i.e. e / ∈ S, S = S −1 ). The Cayley graph Γ(G, S) for the underlying group G and the generating set S is a graph with vertex set V (Γ) = G and edge set E(Γ) = {{g, h}|g, h ∈ G, g −1 h ∈ S}. As the generating set S is closed, g −1 h ∈ S implies that h −1 g ∈ S and therefore our Cayley graphs are undirected. In particular, when G is Abelian, we call Γ(G, S) an Abelian Cayley graph. For a Cayley graph Γ(G, S), denoting S ∪ {e} byS, it is easy to show that its diameter is k if and only if k is the smallest integer such that all elements of G appear inS k S k = k i=1 s i : s i ∈S for i = 1, 2, ..., k and its degree is |S|. For an Abelian Cayley graph with |S| = 2n and diameter k, denote by A C (2n, k) its largest order. The upper bound on A C (2n, k) is given by The right-hand-side of the above inequality is also called the Abelian Cayley-Moore bound, obtained first by Dougherty and Faber in [7], and we denote its value by M C (2n, k). Obviously, M C (2n, 2) = 2n 2 + 2n + 1. An Abelian Cayley graph whose order meets M C (2n, k) is called an Abelian Cayley-Moore graph.
The problem whether Abelian Cayley-Moore graphs exist or not has been studied extensively. Leung and Zhou [10] proved that Abelian Cayley-Moore graphs of diameter 2 exist if and only if n = 1, 2. As the Abelian Cayley-Moore bound is sometimes hard to be met, several researchers working in this field try to find a larger lower bound of A C (2n, k); see [4, 7, 11-13, 17, 18].
For a Cayley graph Γ(G, S) of degree 2n and diameter k + 1, suppose that G has size M C (2n, k) + λ, where λ is a small positive integer. IfS satisfies |S k | = M C (2n, k), borrowing the terminology of Bannai and Ito [2], we call the parameter λ the excess. We give two Abelian Cayley graphs of diameter 3 and excess 1 as follows, which are also the examples of the graphs we research for n = 1, 2. Example 1.1. Let C m denote the cyclic group of order m generated by g.
• For n = 2, let G = C 14 . Define S = {g ±1 , g ±4 }. The graph Γ(G, S) is depicted in Figure 1. The corresponding packing of Z 2 by Lee spheres of radius 2 is given in Figure 2.  In this paper, we consider the Abelian Cayley graphs Γ(G, S) of degree 2n, diameter 3 and excess 1 (i.e. |G| = M C (2n, 2)+ 1 = 2n 2 + 2n + 2). Such Abelian Cayley graphs also provide almost lattice packings of Z 2 by Lee spheres of radius 2 with a small amount of blank.
The rest of this paper is organized as follows. In Section 2, we review some basic notations on group ring to obtain the necessary and sufficient condition on this problem and translate the problem into several group ring equations. In Section 3, we provide some results about the non-existence of Abelian Cayley graphs for infinitely many values of n by utilizing two approaches. The first approach involves the symmetric polynomials which is similar to the one used in [9] by Kim. The second one is close to the one used in [20] which requires that |G| must have small prime divisors.

A necessary and sufficient condition
For convenience, we assume that G is a multiplicative group with identity element e. The group ring Z[G] denotes a set of formal sums g∈G a g g where a g ∈ Z. For any set A whose elements belong to G (A may be a multiset), we identify A with the group ring element g∈G a g g, where a g is the multiplicity of g appearing in A. The addition of elements in Z[G] is defined as follows:

The multiplication is defined by
where λ ∈ Z. For A = g∈G a g g and t ∈ Z, we define For any A = g∈G a g g and χ ∈ G where G is the character group of G, we define χ(A) = g∈G a g χ(g).
The following inversion f ormula gives us a strategy to calculate a h for all h ∈ G and shows that A is completely determined by its character value χ(A), where χ ranges over G.
, then for all h ∈ G.
Group rings and the associated characters are of great significance to research the difference sets and related topics. See [3,16] and the references within. In the rest of this paper, by abuse of notation, we will use the same symbol to denote a subset in G and the associated element in Z[G]. Moreover, we use |A| to denote the number of distinct elements in A, rather than the counting of elements with multiplicity unless stating it specially.
For the Abelian Cayley graph Γ(G, S) which we research, we can prove that there is only one element of order 2 in G. Proof. Since 2 | |G| and 4 |G|, there is only one element f of order 2 in G. By the order of S and S is inverse-closed, we see that f / ∈ S. Moreover, if f ∈S 2 , then there exist a, b ∈ S such that f = ab = a −1 b −1 , which means |S 2 | < 2n 2 + 2n + 1. However, by the assumption that Γ(G, S) is of diameter 3 and excess 1, |S 2 | = 2n 2 +2n+1 which is a contradiction. Therefore, f / ∈S 2 .
In the rest part of this section, we present a necessary and sufficient condition for the existence of an Abelian Cayley graph Γ(G, S) which we research in the language of group ring equations.
Hence, (a) and (b) are obviously satisfied.
In the language of group ring, we have By computation, we have Therefore, (c) holds.
For the other direction of the proof, we only need to define S = T \ {e} and the verification is straightforward.

Main results
In this section, we intend to apply two approaches to prove non-existence results of T satisfying (a), (b) and (c) in Theorem 2.1. The first one is an imitation of the method provided by Kim in [9]. The second one based on the necessary and sufficient condition mentioned in Section 2 is close to the one used in [20] by Zhang and Zhou which requires |G| having small prime divisors.

A symmetric polynomial approach
First we apply the Kim's approach to this problem. Compared with the original one in [9], we do not need the assumption that |G| is a prime, but n 2 + n + 1 is a prime.
where S t := n i=1 x t i . Since this is also the sum of the 2k-th powers of every element in 2G/H \ {0}, Let a and b be the least positive integers satisfying p | 4 a + 4n + 2 and p | 4 b − 1. Define We prove the following two claims by induction on k.
Claim 1: If 1 ≤ k < p−1 2 is not in X, then S 2k = 0. By induction on k, suppose that S 2k = 0 for each k ≤ k 0 − 1 that is not in X. Assume that k 0 / ∈ X. As X is closed under addition, for each t, either t or k 0 − t is not in X.
Since any integer k for which p | 4 k + 4n + 2 must be of the form a + by whence k ∈ X. This implies p 4 k0 + 4n + 2. By (3) and the induction hypothesis, Hence, S 2k0 = 0. Let e k be the elementary symmetric polynomials with respect to x 2 1 , x 2 2 , ..., x 2 n .
Claim 2: If 1 ≤ k ≤ n < p−1 2 is not in X, then e k = 0. We again prove by induction on k, suppose that e k = 0 for all k ≤ k 0 − 1 not in X and k 0 / ∈ X. Since X is closed under addition, for each 0 < t < k 0 , at least one of t and k 0 − t is not in X. By Claim 1 and the inductive hypothesis, e t = 0 or S 2(k0−t) = 0. Together with Newton identities on x 2 1 , x 2 2 , ..., x 2 n , we have then e k0 = 0.
Note that e n = x 2 1 · · · x 2 n , it is clear that none of x 1 , ..., x n is 0. Then e n = 0. By Claim 2, n ∈ X. This finishes the proof.
By MAGMA [5] program, in 1 < n ≤ 10 5 , n has 10750 choices to make n 2 + n + 1 be a prime. However, only n = 2 and n = 3 are not excluded by Theorem 3.1 in these values. The detailed result is showed in Table 1. Actually, Zhang and Zhou [20, Subsection 3.1] provide a generalization of Kim's approach which requires that 2n 2 + 2n + 1 has a prime divisor larger that 2n + 1.
In [20,Subsection 3.2], the approach to deal with |G| with a small prime divisor was put forward. Next, we use this approach to deal with |G| = 2n 2 + 2n + 2 with a small prime divisor.

The method of Zhang and Zhou
By Lemma 2.2, we know that G has a unique element of order 2 denoted by f . Therefore, for any subgroup H of G, if |H| is even, then H must contain f . Let (·) : G → G/H be the canonical homomorphism. Thenf is the identity element in G/H. For A = g∈G a g g ∈ Z[G], we define A = g∈G a gḡ . Assume a subgroup H ≤ G of even order m, the following two equations must hold for T satisfying Conditions (b) and (c).
Here, for the convenience of description, we use 1 to denote the identity elementē in G/H. Hence (2n−2)ē is simply written as (2n − 2). First, we consider a very special case for which Conditions (b') and (c') hold.
Lemma 3.1. Let K be an Abelian group of order v with identity element e K and S = a · e K + bK ∈ Z[K]. If v and m are positive integers such that a + vb = 2n + 1 and mv = 2n 2 + 2n + 2, and S satisfies then 8n − 7 is a square in Z.
By checking the coefficient of e K in (4), we get a 2 + a − 2n + 2 = 0, which shows that 8n − 7 must be a square in Z.
As 3 is the smallest prime dividing |G|, let us look at the existence of T in G/H which is isomorphic to the cyclic group C 3 of order 3.

Proposition 3.1. Suppose that G/H ∼ = C 3 , 8n − 7 is a non-square. Then there is no T ∈ Z[C 3 ] of size 2n + 1 satisfying Conditions (b') and (c').
Proof. We proceed by way of contradiction, assume that there exists T ∈ Z[C 3 ] satisfying Conditions (b') and (c'). Note that T (2) = T (−1) = T . Thus T = a + bG/H for some a, b ∈ Z ≥0 , and by Condition (c') we have By Lemma 3.1, it contradicts our assumption that 8n−7 is a non-square. Hence, there is no T ∈ Z[C 3 ] satisfying Conditions (b') and (c').
To get stronger non-existence results, we need to apply some konwledge on the algebraic number theory.
Next, we look at v = 7, which is the second smallest possible odd prime dividing |G|. In order to deal with this case, we need to exploit the theory about the decomposition of a prime p into prime ideals in Z[ζ ω ], which can be found in [15]. Proof. By way of contradiction, assume that there exists a subset T ∈ Z[C 7 ] satisfying Conditions (b') and (c'), then for i = 0, 1, 2. Thinking of them as polynomials with variables T (2 i ) , we calculate the resultants of f 0 , f 1 and f 2 to obtain a polynomial h having only one variable T . All above can be done by MAGMA [5]. Furthermore, we factorize h (mod G/H) It is no doubt that h must be congruent to 0 is such that which implies that 8n − 7 is a square in Z[ζ 7 ]. Because 8n − 7 is a non-square in Z, we can assume 8n − 7 has a square divisor 8n − 7 = tk 2 where t is a square-free integer larger than 1 and k ∈ Z. Then t ≡ 1 (mod 8). Obviously t > 7. By Lemma 3.2, the smallest cyclotomic field containing Q( √ 8n − 7) is Q(ζ t ) rather than Q(ζ 7 ), so there is no χ(T ) such that (6) holds. Hence, T 2 + T − 2n + 2 cannot be congruent to 0 modulo G/H, then l ≡ 0 (mod G/H).
Since l as a polynomial in T is of degree 6, we try to take a prime number p and consider the equation χ(l) = 0 modulo p. Let p be primitive modulo v = 7, i.e. p = 3, 5 (mod v). By Lemma 3.3, (p) is a prime ideal in Z[ζ v ]. Substitute X for χ(T ) (mod p) in χ(l) ≡ 0 (mod p) and let its coefficients be calculated modulo p. Then we obtain a polynomial l(X) in F p [X].
By χ(T (p) ) ≡ χ(T ) p (mod p), we can calculate τ i := χ(T (2 i ) ) from τ 1 as follows: Consider the necessary conditions that τ 1 must satisfy. First, by (5), for i = 0, 1, 2. Second, we calculate the coefficients of aḡ by using the inversion formula (2). Let β be an element of order It is obvious that aḡ (mod p) must correspond to an element in F p . In addition, since the size of T is 2n + 1, all aḡ's also satisfy that ḡ∈G/H aḡ ≡ 2n + 1 (mod p).
Based on all the previous necessary conditions, now we provide a method to exclude the existence of T satisfying Conditions (b') and (c'). First we choose a prime p. Then, depending on the value of n modulo p, we divide the calculations into p different cases. In each case, l is a concrete polynomial. To check the necessary conditions, we calculate all the roots of l in F 3 p = F v−1 2 p and then for each root τ 1 , (i) plug it into (7) to get τ i ; (ii) check whether (8) holds for each i; (iii) derive aḡ from (9); (iv) check whether aḡ satisfies aḡ (mod p) ∈ F p and (10).
By our MAGMA program, taking p = 101, for each possible value of n modulo p, we can always find at least one of the necessary conditions not satisfied. Hence there is no T such that Conditions (b') and (c') hold.
In the proof of Theorem 3.2, we need to raise T to T (2) in Thus, we cannot apply this approach to G/H whose order is even. For |G/H| = k i=1 p i where p i are prime numbers, if we can show the non-existence of T in G/H which is of order p i for some i ∈ {1, . . . , k}, then we also have the non-existence result for G/H. Hence we only concentrate on the case in which |G/H| is an odd prime. As n 2 + n + 1 is an odd number for all n ∈ Z, v | 2(n 2 + n + 1) implies v | n 2 + n + 1 for every odd prime v.
The approach used in the proof of Theorem 3.2 can be further applied for larger G/H. The next 5 possible values of a prime dividing n 2 + n + 1 is 13, 19, 31, 37 and 43. Our MAGMA program shows that for G/H ∼ = C 13 , G/H ∼ = C 19 and G/H ∼ = C 31 , we can choose p = 227, p = 241 and p = 881 respectively and follows the steps in Theorem 3.2 to prove that there is no T satisfying Conditions (b') and (c'). For the next odd prime 37 dividing the order of G, our computer is not powerful enough to provide us the univariate polynomial with the variable T by calculating the resultants of 18 pairs of polynomials. Hence, we cannot provide any non-existence result for v = 37 or any larger prime numbers. As the process is more or less the same, we omit it here and present the results directly as follows: Corollary 3.1. Let G be an Abelian group of order 2n 2 + 2n + 2. Suppose that 8n − 7 is not a square in Z. Assume that one collection of the following conditions holds (1) 3, 7, 19 or 31 divides n 2 + n + 1; (2) 13 | n 2 + n + 1, 8n − 11 / ∈ {13k 2 : k ∈ Z}.
There is no T ⊆ G viewed as an element in Z[G] satisfying Conditions (a), (b) and (c).
In Table 1, we list the cardinalities of n excluded by Corollary 3.1 and Theorem 3.1, respectively and the last row indicates the numbers of n to which Corollary 3.1 or Theorem 3.1 can be applied.