Remarks on singular Cayley graphs and vanishing elements of simple groups

Let $\Gamma$ be a finite graph and let $A(\Gamma)$ be its adjacency matrix. Then $\Gamma$ is {\it singular} if $A(\Gamma)$ is singular. The singularity of graphs is of certain interest in graph theory and algebraic combinatorics. Here we investigate this problem for Cayley graphs ${\rm Cay}(G,H)$ when $G$ is a finite group and when the connecting set $H$ is a union of conjugacy classes of $G.$ In this situation the singularity problem reduces to finding an irreducible character $\chi$ of $G$ for which $\sum_{h\in H}\,\chi(h)=0.$ At this stage we focus on the case when $H$ is a single conjugacy class $h^G$ of $G.$ Here the above equality is equivalent to $\chi(h)=0$. Much is known in this situation, with essential information coming from the block theory of representations of finite groups. An element $h\in G$ is called vanishing if $\chi(h)=0$ for some irreducible character $\chi$ of $G.$ We study vanishing elements mainly in finite simple groups and in alternating groups in particular. We suggest some approaches for constructing singular Cayley graphs.


Introduction
Let Γ be a finite graph and let A(Γ) be its adjacency matrix. Then Γ is singular if A(Γ) is a singular matrix. Alternatively, Γ is singular if and only if its spectrum contains the eigenvalue 0. All graphs in this paper are undirected, without loops and without multiple edges; for all definitions please see Section 2.
Singular graphs play a significant role in graph theory, and there are many applications in physics and chemistry, see Section 2. While the literature on graph spectra is vast it is not likely that a general theory of graph singularity per se will emerge. Some progress however can be made for graphs which admit a group of automorphisms that is transitive on the vertices of the graph. In some cases the singularity problem then can be solved using techniques from ordinary character theory. The main purpose of this paper is to investigate these applications of character theory in graph theory.
In the following G denotes a finite group and H denotes a connecting set in G. This is a subset of G such that (i) H does not contain the identity element 1 of G, (ii) H = H −1 := {h −1 | h ∈ H} and (iii) H generates G, that is, H does not lie in any proper subgroup of G. From these data the Cayley graph Γ = Cay(G, H) with vertex set V = G and connecting sets H can be defined, see Section 2. Here Γ is a regular graph of degree |H| and the group G acts transitively on the vertices of Γ. Note though that a graph may be the Cayley graph of more than one group and connecting set. In this paper we specify the singularity problem to Cayley graphs Cay(G, H) when the connecting set H is G-invariant, that is, H is a union of conjugacy classes of G. In this case the following theorem reduces the singularity problem to a problem of character theory. Comments: 1. Burnside's theorem on character zeros [3, §32, Exercise 3] shows that every character χ of degree > 1 takes the value χ(h) = 0 for some h ∈ G. Hence for every non-abelian group there exists a singular Cayley graph.
2. Obviously, if χ(h) = 0 for all h ∈ H then Cay(G, H) is singular. In addition, χ(h) = 0 if and only if χ(h −1 ) = 0. Therefore, for constructing singular Cayley graphs it suffices to specify an irreducible character χ of G and a set X generating G such that χ takes the value 0 on X. Then, setting H = ∪ g∈G g(X ∪ X −1 )g −1 , we conclude that H is a connecting set and so Cay(G, H) is singular.
If the character table of a group G is available explicitly (which is the case for sporadic simple groups, say) then one can determine in principle all singular Cayley graphs Cay(G, M ∪ M −1 ) for G-invariant M .
In general we have to look at elements g in G that take the value 0 for certain irreducible characters. Following [4] we say that g is non-vanishing if χ(g) = 0 for every irreducible character χ of G, otherwise we call g vanishing. Vanishing group elements are of particular interest in the block theory of finite groups. We postpone our comments on this matter until Section 4. Here we limit ourselves to the following well known special case. Let |G| denote the order of G. If p is a prime then |G| p is the p-part of |G|, that is, |G|/|G| p is coprime to p. The element g in G is p-singular if p divides the order of g. Lemma 1.2. [3, Theorem 86.3] Let G be a finite group whose order is divisible by the prime p and let χ an irreducible character of G. Suppose that χ(1), the degree of χ, is a multiple of |G| p . Then χ(g) = 0 for every p-singular element g ∈ G.
In what follow p is a prime. The irreducible characters of degree divisible by |G| p are refered as those of defect 0 (or of p-defect 0 if p is not clear from the context). From this lemma and Theorem 1.1 we obtain the following general result on singular Cayley graphs. At the first sight such characters do not appear to be a common phenomenon. However, this is not so as the following result shows: Proposition 1.4. [5, Corollary 2] Let G be a non-abelian finite simple group and let p > 3 be a prime dividing |G|. Then G has an irreducible character of defect 0. This remains true for p = 2, 3 unless G is a sporadic group (with known exceptions) or an alternating group.
It follows by Lemma 1.2 that in a non-abelian simple group G any element of order divisible by the prime p > 3 vanishes at some irreducible character of G. Hence Theorem 1.1 applies and combining Proposition 1.4 with Theorem 1.1 we get Theorem 1.5. Let p > 3 be a prime. Let G be a non-abelian simple group and M ⊂ G \ {1} a G-invariant subset consisting of p-singular elements. Then the Cayley graph Cay(G, M ∪ M −1 ) is singular. This remains true for p = 2, 3 unless G is an alternating group or a sporadic simple group.
The exceptions in this theorem are genuine. They can be detected easily for alternating groups A n with n = 7, 11, 13 by inspection of the character tables. However, for arbitrary n the problem of describing all non-vanishing elements in A n is still open. In any case, Theorem 1.5 yields many examples of singular Cayley graphs.
We first state some elementary results which yield a variety of singular Cayley graphs when G = A n . For g ∈ G let (c 1 , . . . , c k ), with c 1 ≥ c 2 ≥ · · · ≥ c k , be the cycle lengths of g, in the sense that g has k cycles where the longest cycle is of length c 1 , the second longest of length c 2 , and so on. Theorem 1.6. Let G = A n with n > 4 be the alternating group.  Note that R 1 consists of all elements of G = A n fixing exactly one point of the natural Gset, whereas R 2 consists of elements fixing at most one point and having no 2-and 4-cycles in their cycle decomposition.
An element g ∈ G is called real if g −1 is conjugate to g. In symmetric groups all elements are real. This is not the case for alternating groups. From Theorem 1.6 we deduce the following The proof of Theorem 1.6 is based on the Murnahgan-Nakayama formula for computing the values of irreducible characters of symmetric groups (see Section 4). In general, the problem of describing, for a given irreducible character χ of a given group G, the set {g ∈ G : χ(g) = 0} seems to be intractable, even when G is a symmetric or alternating group. However, the block theory of group characters supplies powerful tools for approaching this problem. In particular, we use block theory to prove Theorems 1.8 and 1.9 below.
Recall that for every prime p dividing the order of a finite group G the set of all irreducible characters of G is partitioned into blocks (or p-blocks to be accurate) and each block B determines a p-subgroup of G, defined up to conjugacy in G. This group is the defect group of B. If χ ∈ B then χ(g) = 0 whenever the p-part of g is not contained in a defect group of B. To use this fact, it is important to know the blocks with smallest defect groups. If p > 3 then this smallest defect group is {1} for A n , see [5], and this yields Proposition 1.4 above. (Note that blocks with trivial defect group are called blocks of defect 0.) For G = A n and p = 2 the smallest defect groups can be easily determined (see Section 5), whereas for p = 3 this is still an open problem. This is discussed in [1, Theorem 2.1], where, for |G| 3 = 3 a and n = 7, the order of a smallest defect group is bounded from above by 3 (a−1)/2 . We improve this bound to 3 (a−1)/3 , see Proposition 6.5. Now we turn to the simplest (in a sense) version of the singularity problem for Cayley graphs: we assume that H = C ∪ C −1 where C = {1} is a single conjugacy class in G. Theorem 1.5 resolves this version of the problem for Cayley graphs of the shape Cay(G, H) with G simple, except when G = A n and when the elements of C have order 2 α 3 β for some α and β. Theorem 1.7 reduces the problem to the case where C = C −1 , that is, where H is a single conjugacy class. Below we state some partial results. One of them is the following: Theorem 1.8. Let G = S n or A n with n ≥ 5 and n = 7, 11, and let ω(G) be the set of element orders of G. Let ω 2,3 (G) be the set of all numbers in ω(G) that are not divisible by any prime p > 3. Then G contains a vanishing element of order m for every 1 = m ∈ ω 2,3 (G).
More can be said about the possible choices of h as an element of order m. By Theorem 1.5 there are no restrictions unless m is of the shape 2 α 3 β for some α and β. In that latter case h can be chosen as any element of order m fixing a least number of members of the natural set {1, ..., n}, see Theorem 6.8. Theorem 1.9. Let G = A n with n > 4 and let g ∈ G. Suppose that 2|g| and 3|g| are not in ω(G). Then g is vanishing unless n = 7.
This statement is new only for |g| = 2 α 3 β , otherwise it follows from Lemma 1.2 and Proposition 1.4. It is not true that all elements satisfying the condition in Theorem 1.9 vanish at the same character of G. But if αβ = 0 then this is the case, see Corollary 3.2 and Proposition 6.4.
Notation: Our notation for finite simple groups agrees with the Atlas [2]. In particular, A n means the alternating group on n letters, and S n is the symmetric group. The underlying set is often denoted by Ω n , and it can be idetified with {1, . . . , n}. For a set M ⊂ S n the support of M is supp(M ) := {x ∈ Ω n : gx = x for some g ∈ M }. In the other words, supp(M ) is the complement in Ω n of the set of the elements fixed by M .
If G is a group then we write |G| for the order of G; if p is a prime then |G| p is the p-part of |G|, equivalently, the order of a Sylow p-subgroup of G. For non-zero integers m, n we denote the g.c.d. of m, n by (m, n). If g ∈ G then |g| is the order of g. The identity element of G is denoted by 1. For h ∈ G we write h G for the conjugacy class of h in G. We write Irr G for the set of all irreducible characters of G. If χ is a character and M ⊂ G then χ(M ) = 0 means that χ(g) = 0 for all g ∈ M .

Singularity of Graphs and Cayley Graphs
Let Γ = (V, E) be a graph with vertex set V and edge set E. Let n := |V |. Two distinct vertices u and v are adjacent to each other, denoted u ∼ v, if and only if {u, v} ∈ E.
Let F be a field of characteristic 0. Then we denote by FV the vector space over F with basis V. This is a permutation module for the automorphism group of Γ. The natural inner product on FV is given by The adjacency map α : FV → FV is the linear map given by Therefore α is symmetric with respect to this inner product. The matrix of α with respect to the basis V is the adjacency matrix A = A(Γ) of Γ. Since A is symmetric A is diagonalizable when F = R and so all eigenvalues are real. The spectrum of Γ are the eigenvalues λ 1 , λ 2 , ..., λ n of A. The singularity of graphs plays a significant role in several parts of mathematics and applications. It would be impossible to review the vast literature on graph eigenvalues in this paper. In representation theory and finite incidence geometry the containment of one permutation character in another often is easiest to establish by showing that a certain graph is non-singular or that its nullity is bounded in a particular way. A famous example is the theorem of Livingstone and Wagner about the representations of a permutation group G on the k-and (k + 1)-subsets of the set on which G acts.
We mention also the significance of graph singularity in systems analysis, physics and chemistry, see for instance the survey article [6]. Essentially, when modelling a discrete mechanical system (Hamiltonians) it is often necessary to work out a linear approximation of an operator where the constituents of the system and the relationships between them are represented by a finite graph. Many characteristics and observables of the systemits energy for instance -then typically involve the spectrum of this graph. This is one of the principles that underpins spectroscopy and Hückel Theory in chemistry [14]. In such applications the singularity of a molecular graph of a feasible compound typically indicates that the compound is highly reactive, unstable, or nonexistent, see [6].
In this paper we concentrate on the singularity of graphs whose automorphism group is transitive on vertices. This includes in particular Cayley graphs for which we now give the basic definitions.
Let G denote a finite group with identity element 1. Then the subset H of G is a connecting set provided the following holds: (i) 1 does not belong to H, Suppose now that H is a connecting set. Then define the graph Γ = (V, E) with vertex set V = G by calling two vertices u and v ∈ G adjacent, In particular, Γ is regular of degree |H|. Similarly, HHv is the set of all vertices of distance ≤ 2 from v, and so on. The radius r(Γ) of Γ, as a graph invariant, is useful for studying generating sets in a group. Evidently, r = r(Γ) is the least number r > 0 such that H r := {h 1 h 2 · · · h r | h i ∈ H} is equal to G. This invariant is a subject of intensive study by group theorists.
Let g ∈ G. Then the right multiplication x → xg for x ∈ G is an automorphism of Γ as is easy to see. Therefore the right-regular representation of G on itself provides an injective homomorphism G → Aut(Cay(G, H)) for any connecting set H. Cayley graphs are characterized by this property:

. (Sabidussi) The graph Γ is isomorphic to a Cayley graph if and only if
Aut(Γ) contains a subgroup that acts regularly on the vertices of Γ.
Next consider the left multiplication x → g −1 x for g and x ∈ G. By contrast, this does not yield an automorphism of Γ in general. It is easy to show that x → g −1 x is an automorphism of Cay(G, H) if and only if gH = Hg. This is relevant for this paper as we are dealing with connecting sets that are unions of conjugacy classes.
If Γ = Gay(G, H) then the space FG is the underlying space of the group algebra of G over F. Then α = h∈H ρ(h) is an element of the group algebra, which is in the center of it whenever H is a union of conjugacy classes.
Let ρ i and α i denote the restriction of ρ and α to E i , respectively. Thus H) is singular, say λ 1 = 0, then every irreducible representation ρ 1,i appearing in ρ 1 satisfies h∈H χ 1,i (h) = 0, where χ 1,i is the character of ρ 1,i . Conversely, if χ j,i is an irreducible character with h∈H χ j,i (h) = 0 then ρ j,i appears in ρ and so there is some E j on which λ j = 0. ✷ Finally we consider a connected graph Γ = (V, E) which admits a vertex transitive group G of automorphisms. In this case we construct an associated Cayley graph Γ * := Cay(G, H) as follow. Fix a vertex v ∈ V and let C be its stabilizer in G, with c := |C|. In view of Sabidussi's theorem we may assume that c > 1. Next let H : Also, H generates G, this follows from the transitivity of G on vertices and the connectedness of Γ. Therefore we have a Cayley graph Γ Comment: We see that the singularity problem for vertex transitive graphs can be reduced -in principle at least -to the nullity problem for Cayley graphs. The theorem can also be used to construct singular graphs: any graph with a vertex transitive but not vertex regular group of automorphisms yields a singular Cayley graph with the same group of automorphisms.

Elementary observations on zeros of alternating groups characters
The comments in Section 1 suggest to pay particular attention to the alternating groups.
In fact, the reasonings in this paper are mostly concerned with these groups. In this section we collect a number of well known facts about characters of alternating groups and prove some results on the zeros of some of their irreducible characters.
We first recall certain notions of the representation theory of S n . It is well known that the irreducible characters of S n are in bijection with the Young diagrams, and also with the partitions of n. So we write φ Y for the irreducible representation or the irreducible character of S n corresponding to the Young diagram Y . For the Young diagram Y we write |Y | for the number of boxes in it. A subdiagram of Y is a Young diagram of S m for m < n which is contained in Y as a subset with the same top left hand corner. A box in Y is called extremal if there is no box either below or to the right of it. The set of all extremal boxes form the rim of Y.
The notion of a hook in a Young diagram Y is common knowledge, see [9, page 55]. The number of boxes in a hook is called the length of it. A hook of length m is called an m-hook.
The leg of a hook is the set of all boxes below the first row and ends in its foot. The number of the boxes in the leg is the leg length. The arm of the hook is its horizontal part, it ends in the hand of the hook, the right furthest box in the arm. Both foot and hand of the hook belong to the rim of Y. Below we need the Murnahgan-Nakayama formula [9, 2.4.7]. It expresses the character value of an irreducible character χ Y in combinatorial terms. Let g ∈ S n and g = ab where a is an m-cycle and where b ∈ S n−m ⊂ S n is the permutation induced by g on the points fixed by a. The Murnahgan-Nakayama rule is the induction formula where the sum runs over all m-rims ν of Y and where i is the leg size of ν. (If no m-rim exists then we have χ(g) = 0 by convention.) As an illustration, we state the following Lemma 3.1. Let G = A n or S n with n ≥ 7 and let M ⊂ G be the subset of all elements whose cycle decomposition has a cycle of length greater than 2 √ n + 2. Then χ(M ) = 0 for some irreducible character χ of G.
Proof: Let m be the minimal number i such that i 2 > n, so m > √ n. Let g ∈ M and let c(g) be maximal length of a cycle in the cycle decomposition of g. Then c(g) > 2 √ n + 2. If In all cases the hook lengths of Y does not exceed 2m − 1. By the Murnahgan-Nakayama formula (3) we have χ Y (g) = 0 whenever c(g) ≥ 2m. As c(g) > 2 √ n + 2 ≥ 2(m − 1) + 2, the result follows. ✷ We view χ as a character of S n . Let g = g 1 b where g 1 is a cycle of size c 1 and the cycle lengths of b are (c 2 , . . . , c k ). By the Murnahgan-Nakayama rule (3) where ν runs over the c 1 -rims of Y and i is the leg length of ν. (If no c 1 -rim exists then χ(g) = 0.) Set r = c 1 . It is clear from the diagram shape that an r-rim is either a part of the first row (and then [2]. In each case there is exactly one way to delete ν, so the sum has at most one term. Suppose first that n − r < 7. There is no way to remove an r-rim to obtain Y 1 of size 6. So χ(g) = 0 if n − r = 6. Let n − r ≤ 5; then c 2 + · · · + c k = n − r ≤ 5, and hence Proof: The permutation character is of the shape π = 1 + χ with χ irreducible. By assumption therefore χ(g) = π(g) − 1 = 0.  (1) The cycle decomposition of g has a cycle of even length.
(2) The cycle decomposition of g has two cycles of equal odd length. Note that the fixed points are counted as cycles of length 1, so this includes any permutation that has two or more fixed points.
(3) All the cycles of g have distinct odd lengths c 1 , . . . , c k and k i=1 (c i − 1)/2 is even. In other words, the number of c i 's that are congruent 3 to modulo 4 is even.
Proof: Clearly g is conjugate to g −1 in S n . Note that the S n -conjugacy class of g is an A n -conjugacy class if and only if there is an odd permutation that centralizes g, these are the conditions 1) and 2). In the remaining case, a cycle of odd length 2ℓ + 1 is inverted by an element of sign (−1) ℓ , and this gives the condition 3). Here the S n -conjugacy classes of g split in A n , but g is conjugate to g −1 in A n . ✷ Proof of Theorem 1.7. If n ≥ 7 then any non-real element satisfies the assumption of Lemma 3.3, whence the result. If n ≤ 6 then either n = 6, |g| = 5 or n = 4, |g| = 3. In these cases the result follows from Lemma 3.4. Finally, the claim that H is connected follows immediately if n > 4 as A n is simple. If n = 4 then H consists of all elements of order 3, so the claim follows by inspection of normal subgroups of A 4 . ✷ Proposition 3.6. Let G be a simple group and suppose that g ∈ G is non-real. Then g is vanishing.

Blocks in symmetric groups and vanishing elements
In this section we expose some part of representation theory of symmetric groups that is needed for the remainder of the paper. Recall the notation from the end of Section 1. Let G be a finite group. For every prime p dividing |G| the irreducible characters of G partition into p-blocks. To every p-block there corresponds a conjugacy class of p-subgroups of G and each of them is called a defect group of the block. If p d is the order of a defect group then d is called the defect of the block. In particular, blocks of defect 0 are those whose defect groups consist of one element. In addition, G has a p-block of defect 0 if and only if there is an irreducible character of G which has p-defect 0. See for instance Navarro [13] or Curtis and Reiner [3], Chapter VII, for general theory of blocks. We use the following well known fact: Corollary 15.49] Let G be a finite group and let g ∈ G be a p-singular element. Let g = g p h = hg p , where g p , h ∈ g , g p is a p-element and |h| is coprime to p. Let χ be an irreducible character of G. Suppose g p is not contained in any defect group for the p-block containing χ. Then χ(g) = 0, in particular g is a vanishing element of G.
To use this lemma, one needs to know the defect groups of the p-blocks of S n (for p = 2 and 3). These are described in [9, Theorems 6.2.39 and 6.2.45] for any prime p. However, first we discuss a special case of blocks and characters of defect 0.
Remark: It is not true that if G does not have an irreducible character of 3-defect 0 then there exists a 3-singular element g ∈ G such that χ(g) = 0 for every irreducible character χ of G, see the character table of Suz. Recall that every finite group G has a unique maximal normal nilpotent subgroup F (G), called the Fitting subgroup of G. Lemma 4.2 can be extended to non-simple groups as follows.  Let p be a prime, let n ≥ p be a natural number and n = a 0 + a 1 p + · · · + a k p k its p-adic expansion. Let h be a p-element of S n whose cycle structure is 1 a 0 p a 1 (p 2 ) a 2 ...(p k ) a k . Let χ be an irreducible character of S n such that p divides χ(1). Then χ(h) = 0. This remains true for A n provided h ∈ A n . (See Remark following [12, Th 4.2]).

4.2.
Blocks of Symmetric groups. Let p be a prime. Every diagram that does not contain a p-hook is called a p-core. For instance, 2-cores are the diagrams of triangle shape [k, k − 1, . . . , 1]; in particular, S n has no 2-core of size n unless n = 1 + ... + k = k(k + 1)/2 for some integer k > 0. Every diagram Y contains a unique p-core subdiagramỸ which is maximal subject to condition |Y | ≡ |Ỹ | (mod p). The key result of block theory of symmetric groups states that two irreducible characters are in the same block if their Young diagram yield the sameỸ [9, 6.1.21]. There is a simple algorithm to obtainỸ as follows.
If Y has no p-hook then Y =Ỹ . Otherwise remove arbitrary p-rim to obtain a subdiagram Y 1 . If Y 1 has a p-hook, remove some p-rim from Y 1 to obtain a subdiagram Y 2 and so on. The process stops if and only if one gets a subdiagramỸ which is a p-core. By [9, Theorems 2.7.16], this final subdiagramỸ is unique (independently from the p-hooks choice), and called the p-core of Y . Thus, |Ỹ | = |Y | − pb for some uniquely determined integer b ≥ 0, and this b is called the p-weight of Y (see [9, p. 80]). Note that the p-weight of a diagram is 0 if and only if the diagram is a p-core.
The following well known fact follows easily from the dimension formula for irreducible characters in terms of hooks:  Recall that the defect groups of a block are unique up to conjugacy. Here the group S pb is a natural subgroup of S n in the sense that this permutes pb elements of Ω n , and fixes the remaining elements. (Note that if b = 0, then S pb is meant to be the identity group, and if n = pb then S pb = S n .) Moreover, the character of S n−pb corresponding toỸ is of defect 0.  It is easy to construct irreducible characters with given p-core C (provided n − |C| is a multiple of p): For our purpose we are interested in the defect groups rather than in blocks themselves. Moreover, we can fix a defect group in every conjugacy class of defect groups in such a way that these defect groups form a chain with respect of inclusion. In fact, if the defect groups D, D ′ are Sylow p-subgroups in S pb , S pb ′ , resp., and b < b ′ , then we can assume D ⊂ D ′ . (For this, one can order the elements of Ω and choose S pb to be the subgroup fixing elementwise the last n − pb elements of Ω.) Therefore, with this ordering of defect groups it is meaningful to speak of the minimal defect group of S n , that is, the one with least possible b. Recall that the defect groups of a block are conjugate, and if D is one of them then the defect of a block is the number d such that |D| = p d . So a minimal defect group is a defect group of a block of minimal defect. Note that the maximal defect group is always a Sylow p-subgroup of S n . By Lemma 4.2, if p > 3 then the minimal defect group is trivial (that is, the group of one element). (Formally, the lemma is stated for A n but it remains true for S n .) If p > 2 then the defect groups of A n are exactly the same as those of S n ; if p = 2 then the defect groups of A n are of shape D ∩ A n for a defect group D of S n . Moreover, if χ is an irreducible character of S n reducible as that of A n then irreducible constituents belong to blocks whose defect groups are conjugate in S n , and hence have the same support. (This follows from [13, 9.26, 9.2].) Lemma 4.13. Let B be a 2-block (resp., 3-block) of S n of non-zero defect. Then B contains an irreducible character that remains irreducible under A n .
Proof: It is well known that the characters labeled by non-symmetric diagrams are irreducible under A n . So we show that B contains a character whose Young diagram is not symmetric.
Let Y = [l 1 , . . . , l k ] be the 2-core (resp. 3-core) diagram determined by B. As B is not of defect 0, n = |Y |, so n − |Y | = 2b (resp. 3b) for some integer b > 0. If l 1 ≥ k then the diagram is not symmetric and the character labeled by Y 1 belongs to the block B (see Lemma 4.12). If l 1 < k (so p = 3) then take for Y 1 the diagram obtained from Y by adding 3b boxes to the 1st column, and conclude similarly. ✷ Lemma 4.14. Let G = S n or A n , g ∈ G, and let D 2 and D 3 be defect groups of a 2-block, or a 3-block, respectively. Suppose that |g| = 2 α 3 β and |supp(g)| > |supp(D 2 )|+ |supp(D 3 )|.
Then χ(g) = 0 for some irreducible character χ of G.
Proof: Let G = S n and let g = g 2 g 3 = g 3 g 2 where g 2 is a 2-element and g 3 is a 3element of G. It is easy to observe that |supp(g)| ≤ |supp(g 2 )| + |supp(g 3 )|. So either |supp(g 2 )| > |supp(D 2 )| or |supp(g 3 )| > |supp(D 3 )|. In the former case g 2 is not conjugate to an element of D 2 , so χ(g) = 0 for every irreducible character χ in a 2-block with defect group D 2 , by Lemma 4.1. Similarly, consider the latter case. Here the result follows for G = S n . If G = A n then the result follows from that for S n and the fact stated prior Lemma refns7. ✷ Comments: In view of Lemma 4.14, it is desirable to determine the minimal defect group for every n and p = 2 or 3. If p = 2 then the number bp = 2b must be of shape k(k + 1)/2, so the minimal defect group of S n is a Sylow 2-subgroup of S 2b , where k is the maximal integer such that n − 2b = k(k + 1)/2 for some b > 0. If p = 3 we only prove the existence of a defect group D with |supp(D)| ≤ 2 √ n + 4 (Lemma 6.1), which is sufficient for our purpose.

5.
Minimal defect group of S n for p = 2 The following lemma is obvious in view of the above comments. Let d = |supp(D)| and let m be the maximal integer such that m(m + 1)/2 ≤ n. Set a = n − m(m + 1)/2, so a ≤ m. If a is even then d = a ≤ m. If a = 0 then d = 0, and the lemma is trivial. So we assume a > 0.
Suppose that a is odd. Then T m is not a 2-core of any diagram of S n . So consider T m−1 .
Then n − |T m−1 | = a + m. If m is odd then a + m is even and then d = a + m ≤ 2m. Let m even. Then a ≤ m − 1. Then T m−1 is not a 2-core of any diagram of S n . Consider T m−2 .
Therefore, d ≤ 3m − 2 in any case. We claim that 3m − 2 < 3 √ 2n − 20. Indeed, this is equivalent to (3m − 2) 2 < 18n − 60, or 9m 2 − 6m + 64 < 18n. As m(m + 1)/2 < n, we have 9m 2 + 9m < 18n, it suffices to show that 9m 2 − 6m + 64 < 9m 2 + 9m, or 64 < 15m. This is true if m ≥ 5. This holds if n ≥ 15. As S 15 has a block of defect 0, we are left with n = 14. If n = 14 then m = 4, |T m | = 10 and d = 4 < 3 √ 8, as claimed. ✷ Recall that for g ∈ S n and a prime p we denote by g p the element such that g = g p h, where h ∈ g and |h| is not a multiple of p. One observes that |supp(g p )| is the sum of cycle lengths divisible by p in the cycle decomposition of g.
Then there is a 2-block B of G such that χ(R) = 0 for every irreducible character χ of B.
Proof: Let T be the triangle diagram of maximal size |T | ≤ n such that n − |T | is even.
Suppose first that |T | = n. Then, by Lemma 4.8, S n has a character χ of 2-defect 0. Recall that χ is a unique character in a block it is contained in. So for G = S n the statement follows from Lemma 1.2. If G = A n then χ| An is the sum of two irreducible characters of degree χ(1)/2; moreover, each of them is of 2-defect 0 (indeed, an irreducible character χ of A n is of 2-defect 0 if and only if χ(1) is a multiple of |S n | 2 ; as |S n | 2 = 2|A n | 2 , it follows that χ(1)/2 is a multiple of |A n | 2 ). This implies the result for A n .
Let |T | < n and let Y be any diagram of size n containing T . Let χ be the irreducible character of S n labeled by Y . Then χ belongs to a block B, say, whose defect group D satisfies |supp(D)| = n − |T |. By Lemma 5.2, |supp(g 2 )| ≥ 3 √ 2n − 20 > n − |T |, so g 2 is not conjugate to an element of D. By Lemma 4.1, χ(g) = 0 for every irreducible character χ of B, whence the result for S n . It is known that the defect groups of blocks of A n to which the irreducible constituents of χ| An belongs are D ∩ A n and conjugate in S n (see comments prior Lemma 4.13). So g 2 is not conjugate to an element of D ∩ A n , and the result follows as above for S n .
By Lemma 4.13, Y can be chosen non-symmetric, so χ is irreducible under restriction to A n . ✷ We say that the element g ∈ G ⊆ S n has maximal support if |supp(g)| ≥ |supp(h)| whenever h ∈ G and |h| = |g|. One easily observes that if g is of maximal support in G = A n and |g| is even then |supp(g)| ≥ n − 3. Proof: Let g ∈ G be a 2-element of maximal support. Then |supp(g)| ≥ n − 3. If n > 13 then n − 3 ≥ 3 √ 2n − 20, and the result follows from Lemma 5.3. For n ≤ 13 and n = 5, 6, 8, 10, 12 the result follows as A n has a 2-block of defect 0. So we are left with n = 9, 13, which can be inspected by the character table of G in [2]. Let G = S n . Then we have to deal also with the cases with n = 5, 8, 12. If n = 12 the all elements of maximal support are in A 12 . The cases with n = 5, 8 follows by inspection. For the use in Section 7 we compute the minimal numbers in the set {|supp(D)| : D is a defect group of a 2-block of S n , 13 < n < 34}. Note that these are equal to n − t, where t is the maximal number of shape m(m + 1)/2 such that n − t is even.
Examples. Lemma 6.1. Let G = S n , n > 13. Then G has a 3-block whose defect group support size is at most 2 √ n + 4.

Proof:
We have to show that there is a 3-block with defect group D, say, such that |supp(D)| ≤ 2 √ n + 4. We can assume that S n has no block of defect 0 (otherwise D = 1 and the statement is trivial).

Remark:
The bound 2 √ n + 4 in Lemma 6.1 is not sharp.
For g ∈ G = S n denote by g 3 the element such that g = g 3 h = hg 3 , where g 3 is a 3-element and |h| is coprime to 3. Proof: By Lemma 6.1, there is a 3-block of S n whose defect group support is at most 2 √ n + 4. So g 3 cannot lie in any defect group of this block. By Lemma 4.1, χ(g) = 0 for any irreducible character in this block and any g ∈ M . (Recall that the defect groups of a block are conjugate, so their supports are of the same size.) As the defect groups of 3-blocks of A n are the same as those of S n , the result follows. ✷ For further use we record in Table 2 the list of maximal sizes of a core for 3-blocks of S n for n < 51 provided S n has no 3-block of defect 0. (In the latter case the size of some core equals n.) The numbers n to be inspected are in the first row of the table, see Corollary 4.3 and the comment following it. In the table C is a core of maximal size |C| ≤ n with n ≡ |C| (mod 3), so n − |C| = |supp(D)|, where D is a minimal defect group of S n for p = 3. The table shows that for every n < 51 there is an irreducible character χ such that χ(g) = 0 whenever g is a 3-element of order at least 9. Note that if n = 51 then |C| = 42 and n − |C| = 9. Proof: This immediately follows by inspection of Table 2. Proof: Let g ∈ M . Observe that |supp(g)| > n/3. If n 3 ≥ 2 √ n + 2 then the result follows from Lemma 3.1. Suppose that n 3 < 2 √ n + 2. Then n < 47.
If 27 ≤ n ≤ 46 then e = 27 > 16 > 2 √ n + 2. For 9 ≤ n ≤ 26 we have e = 9 and |supp(g)| > 9. Acording with Table 2, there is a 3-block of defect group support at most 6, so g is not conjugate to any such defect group. Therefore, χ(g) = 0 for any irreducible character in this block, whence the result. If n = 5, 6, 8 then the groups A n , S n have a 3-block of defect 0, so we are left with n = 7 as claimed. ✷ Proposition 6.5. Let G = S n or A n with n > 4, n = 7, and let |G| = 3 a m, where m is not a multiple of 3. Then there is a 3-block of defect at most (a − 1)/3.
Proof: Suppose first that n 3 ≥ |supp(D)|, where D is a defect group of some 3-block B of G. Then 3|D| 3 ≤ |G| 3 for n = 7. Indeed, G contains a 3-subgroup X, say, isomorphic to the wreath product of D with A 3 , and |X| = 3|D| 3 . Let |D| = 3 r , so r is the defect of the block B. Then |X| = 3 3r+1 and 3r + 1 ≤ a. This implies r ≤ (a − 1)/3. Suppose that n ≥ 58. Then n 3 > 2 √ n + 4. By Lemma 6.1 there is a 3-block of G whose defect group satisfies |supp(D)| ≤ 2 √ n + 4. So the result follows. Let n < 58. By Table 2, n 3 ≥ |supp(D)| unless n = 7. ✷ 6.1. Elements of maximal support whose order is divisible by 3. Let G = S n or A n and g ∈ G. Recall that g is of maximal support in G if |supp(g)| ≥ |supp(h)| for any element h of the same order as g. In other words, the number of fixed points of such an h does not exceed the number of fixed points of g.
Note that for g in G to be of maximal support depends on whether G = S n or A n . Say, if g ∈ A 6 is an involution (a double transposition) then g is of maximal support in A 6 but not in S 6 . The following is easily verified: Lemma 6.6. Let G = S n or A n and g ∈ G. Suppose that g is of maximal support.
Let G = A n . If χ| An is irreducible then we are done. Otherwise, χ| An is the sum of two irreducible characters of equal degree. If D = 1 then χ is of defect 0, which is equivalent to saying that χ(1) is a multiple of |S n | 3 = |A n | 3 (see [3,Theorem 86.3]). Suppose that D = 1, that is, the block χ belongs to is not of defect 0. Then, by Lemma 4.13, this block has an irreducible character χ ′ labeled by a non-symmetric Young diagram which is therefore irreducible under restriction to A n . ✷ Obviously, Theorem 1.8 follows from the following result: Theorem 6.8. Let G = S n or A n with n > 4 and let g ∈ G. Suppose that g is of maximal support in G. Then χ(g) = 0 for some irreducible character of G, unless n = 7 or 11.
The cases with n < 13 can be examined from their character tables [2]. If 14 > n > 4 and n = 7, 11 then A n has a block of 3-defect 0. The non-vanishing elements g ∈ G = A 11 are in class 2B and 3A in notation of [2], and those in 3A are not of maximal support. Let G = A 7 . Then an element of maximal support is in class 3B or 6A, the latter is nonvanishing, see [2]. If n = 13 then the character table is available in the GAP library. One observes that the only non-vanishing elements in A 13 are in classes 2A, 2B, 3A, and hence not of maximal support. This completes the proof. ✷ Denote by ω(G) the set of element orders in the group G.
If n = 17, 16, 14 then G has a 3-block of defect 0, and if n = 15, 12 then G has a 2-block of defect 0. If n ≤ 13 then the result follows by inspection of the character table of G. ✷