Beiträge zur Algebra und Geometrie Contributions to Algebra and Geometry Volume 37 (1996), No. 1, 103-118.

Blocks for Symmetric Groups and their Covering Groups and Quadratic Forms

Karin Erdmann and Gerhard O. Michler Mathematical Institute, Oxford University 24-29 St. Giles, Oxford OX1 3LB, England Institut für Experimentelle Mathematik, University of Essen Ellernstr. 29, D-45326 Essen, Germany

Introduction.

Let p be a prime number, K a field of characteristic p and let S_{n} be the group of permutations of degree n. We are interested in the p -blocks of the group algebra KS_{n}. It is conjectured that for p ≥ 5 and all n the group algebra KS_{n} has p -blocks of all possible defects; that is of any defect d where p^{d} divides n! . This is equivalent to the conjecture that for all n the group algebra KS_{n} has a p -block of defect zero, see [5], [6]. It appears to be folklore that the existence of blocks of defect zero should be true for large n, and this is supported by work of [4], [1], [2] by means of modular forms. We consider an elementary approach which is motivated by representation theory.

It is well-known that the irreducible characters of S_{n} are parametrized by partitions of n; and moreover the “Nakayama conjecture” holds: Irreducible characters χ^{λ} and χ^{μ} belong to the same p -block if and only if λ and μ have the same p -core, see [3; 6.1.42]. In particular χ^{λ} belongs to a block of defect zero if and only if λ is a p -core. One of the (combinatorial) tools to study p -modular representations of symmetric groups is to represent partitions on an abacus with p runners, see Ch. 1 or [3; 2.7]. In fact one of the possible proofs of the Nakayama conjecture uses the abacus representation. More recently it was proved in [9] that the number of Morita equivalence classes of p -blocks of symmetric groups with a fixed defect group is finite; for this work the abacus representation was crucial. (This was one of our motivations.)

Here we study p -blocks of defect zero, by exploiting the abacus method. Our work is aimed at giving an elementary approach to the existence conjecture, and to show how this is related to problems of representing integers as sums of squares. Suppose a partition is represented on an abacus with p runners. Then it belongs to a p -block of defect zero if and only if it is not possible to move a bead upwards on its runner. Hence the abaci

104 K.Erdmann, G.O.Michler: Blocks for Symmetric Groups …

0138-4821/93 $ 2.50 \begin{matrix}c \\ ◯ \end{matrix}1996 Heldermann Verlag, Berlin representing a p -core are completely characterized by the set of p -tuples of integers ≥ 0 giving the number of beads per column. (For a list without repetitions one may take p - tuples where the first number is zero, which corresponds to the abacus with first column being empty, that is, the beads represent the first column-hook lengths.)

The problem is that for such a partition it is not obvious which integer N it represents. Our first result is to determine closed formulae for the number N, each formula is a sum of a quadratic form and a linear form. The variables are the number of beads per column. (As a Corollary we also get the form representing N of [2].)

Then we discuss change of variables and we use an orthogonal system of variables to prove that the conjecture holds for p = 5 and p = 7.

We also consider spin blocks, by a similar approach. The same question arises, namely for which n does a covering group Ŝ_{n} of S_{n} have a spin block over characteristic p of defect zero. It was proved in [7] that the spin blocks of a covering group Ŝ_{n} can be parametrized by “p -bar cores” which are certain partitions of n. Moreover, p -bar partitions λ may also be represented on an abacus in a way such that Λ is a p -bar core if and only if it is not possible to slide a bead upwards on a runner and in addition either column i or column p_{i} is empty. However, the rule for representing a partition is different here; for details see [7].

For a p -bar partition λ with a given abacus representation we determine a closed form for the integer n with λ ⊢ n. This is also the sum of a quadratic form and a linear form. We are able to get a complete answer for spin cores when p = 5 and p = 7. For p = 5 there are exceptions, and we have a complete characterization of integers n such that Ŝ_{n} has a 5-spin block of defect zero, see 2.1. If p = 7 then for every n, the covering group Ŝ_{n} has 7-spin blocks of defect zero (and hence spin blocks of all possible defects).

Our results for spin blocks are new. Moreover it seems that methods using modular forms cannot be used here in the same way as for symmetric groups.

In Chapter 1 we describe the abacus representations of partitions and determine the quadratic and linear forms. In the second chapter we deal with spin cores. In Chapter 3 we describe the diagonalization and in Chapter 4 we study p -cores of symmetric groups for p = 5 and p = 7. The last section contains the number theoretic results.

1. Preliminaries

1.1 Let λ be a partition of n, so λ = (λ_{1}, λ_{2}, …, λ_{k}), where the λ_{i} are non-negative decreasing integers whose sum is n. Take an integer s ≥ k and set λ_{m} = 0 for k+1 ≤ m ≤ s then the s-element set of β-numbers for λ is by definition the set {β_{1}, β_{2}, …, β_{s}} where

β_{i} = (λ_{i} − i) + s.

(1 ≤ i ≤ s), see [3; 2.7]. Thus β_{1} > β_{2} … > β_{s} ≥ 0 and the the partition λ may be reconstructed from the set of β-numbers, for any such s.

Fix a positive integer p (which at the moment need not be a prime). We wish to represent λ on an abacus with p runners. The runners, which go vertical (and which we also call columns) are labelled 0, 1, …, p − 1. The rows of the abacus are numbered 1, 2, … A bead placed at position (j, i) where 0 ≤ i ≤ p − 1, 1 ≤ j represents the number (j − 1)p + i.

An abacus display for λ with s beads, where s ≥ k, is an abacus which represents the s-element set of β-numbers for λ. Observe that given an abacus display, one can reconstruct λ. Namely, λ_{i} is equal to the total number of gaps before the i -th bead. In particular, the last part of λ is non-zero if and only if there is no bead at position zero.

A partition is a p -core if and only it is not possible to slide a bead up on its runner, in some abacus display; if so then this holds in all possible abacus displays. Hence an abacus display of a p -core may be identified with a p -tuple

Γ = (y_{0}, y_{1}, …, y_{p−1})

where y_{i} is the number of beads on the i -th runner.

For example, let p = 5 and λ = (4, 1^{2}). The abacus display on an abacus with three beads is identified with Γ = (0, 2, 1, 0, 0). If we take four beads then we get Γ = (1, 0, 2, 1, 0).

In general, we identify the core with any such Γ. In particular taking all such abacus displays where the runner 0 is empty gives a bijection between all p -cores and the set of all p -tuples y⃗ = (0, y_{1}, …, y_{p−1}) with y_{i} ≥ 0.

1.2 Now assume that p is a prime number. We consider p -blocks of the group algebra S_{n}. As usual, we say that a partition λ belongs to a p -block if the irreducible character χ^{λ} parametrized by λ belongs to this block. It is well-known that the partitions λ, μ belong to the same p -block if and only if they have the same p -core; this is the Nakayama conjecture. In particular, a partition represents a block of defect zero if and only if it is a p -core. Hence there is a bijection between blocks of defect zero and p -cores.

1.3 We fix a prime p. Suppose n ∈ N, then we say that n is an abacus number if there is an abacus which is a p -core representing n.

1.4 We describe now some abacus operations. Suppose Γ = (y_{0}, y_{1}, …, y_{p−1}) is a core abacus representing n. Then n is also represented by the core abacus

(y_{p−1} + 1, y_{0}, …, y_{p−2}).

We have the following general result which we will use later.

1.5 Lemma Suppose the integer n has an abacus representation Γ = (0, y_{1}, …, y_{p−1}). Then n + y_{p−1} + 1 also is an abacus number, represented by the abacus

(y_{p−1} + 1, 0, y_{1}, …, y_{p−2}).

In particular if such Γ exists with y_{p−1} = 0 then n + 1 is an abacus number.

Proof. The abacus given represents the same integer as the abacus Γ′ = (y_{p−1} + 1, 0, y_{1}, …, y_{p−2}). Moving all beads of the first column to the right gives a core abacus representing n + y_{p−1} + 1.

1.6 Proposition Let N be the integer which is represented by the abacus Γ = (y_{0}, y_{1}, y_{2}, …, y_{p−1}). Then

(∗)  N = \frac{p}{2} ∑_{i=0}^{p−1} y_{i}^{2} + ∑_{i=0}^{p−1} iy_{i} − \frac{s(s + p − 1)}{2} .

Alternatively, N = q(y⃗) + ℓ(y⃗) where

(i) q(y⃗) = \frac{p−1}{2}(∑_{i=0}^{p−1} y_{i}^{2}) − \frac{1}{2} ∑_{i≠j} y_{i}y_{j} and

(ii) ℓ(y⃗) = ∑_{i=0}^{p−1}(i − \frac{p−1}{2})y_{i}.

Proof. Recall that the l-th bead in Γ which is in the k -th row and the i -th column represents the β-number

β_{l} = (λ_{l} − l) + s

so that (λ_{l} − l) + s = (k − 1)p + i, for 1 ≤ l ≤ s (see 1.1). Consider a fixed column Γ_{i} of Γ. We have

∑_{β_{l}∈Γ_{i}}(λ_{l} − l) = ∑_{k=1}^{y_{i}}((k − 1)p + i − s) = (y_{i} − 1)y_{i}p/2 + iy_{i} − sy_{i}.

It follows that

∑_{l=1}^{s} λ_{l} = ∑_{l=1}^{s}(λ_{l} − l) + ∑_{l=1}^{s} l = (∑_{i=0}^{p−1} ∑_{β_{l}∈Γ_{i}}(λ_{l} − l)) + s(s + 1)/2

and by the above this is equal to

(∑_{i=0}^{p−1}((y_{i} − 1)y_{i}p/2 + iy_{i} − sy_{i})) + s(s + 1)/2

= ((p/2) ∑_{i=0}^{p−1} y_{i}^{2}) + (∑_{i=0}^{p−1} y_{i}i) − (∑ y_{i})((p/2) + s) + s(s + 1)/2

and since ∑ y_{i} = s the last two terms give −s(s + p − 1)/2. To obtain the alternative formula, one substitutes s = ∑_{i=0}^{p−1} y_{i}.

1.6.1 Remark In [2], a different quadratic form for the abacus numbers was obtained. This formula follows easily from 1.6.

Take an abacus Γ as in 1.6 where s is chosen such that y_{i} > 0 for all i and also that s = s̄p for some s̄ ∈ N. Now define n_{i} := y_{i} − s̄; then clearly ∑_{i=0}^{p−1} n_{i} = 0. Substituting these n_{i} into 1.6 (∗) gives that

N = (p/2) ∑_{i=0}^{p−1} n_{i}^{2} + ∑_{i=0}^{p−1} in_{i}.

1.7 We complete squares in the formula (∗) of 1.6. We get

\frac{p}{2}(y_{i}^{2} + \frac{2iy_{i}}{p}) = \frac{(py_{i} + i)^{2}}{2p} − \frac{i^{2}}{2p} .

Using the standard formula for sums of squares gives

(1.7 ∗)  N = \frac{1}{2p} ∑_{i=0}^{p−1} z_{i}^{2} − \frac{(p − 1)(2p − 1)}{12} − \frac{s(s + p − 1)}{2}

where z_{i} = py_{i} + i.

1.8 Now we will determine a quadratic form for p -bar cores of spin blocks, for covering groups of symmetric groups. It was proved in [7], that spin blocks also may be parametrized by p -cores, but of a slightly different kind. Here one represents the parts of a partition on the abacus with p runners (instead of β-numbers). The p -cores are then exactly those partitions which are represented as follows:

(i) The position 0 of the abacus is empty;
(ii) It is not possible to slide a bead up its runner;
(iii) For 1 ≤ j ≤ p − 1, at least one of column j or column p − j is empty.

We say that n is a bar abacus number if there is such an abacus which is a p -core representing n. We identify such a core with a p -tuple

Γ = (0, l_{1}, …, l_{p−1})

with l_{j}l_{p−j} = 0, 1 ≤ j ≤ p − 1. Noting that here a bead of the abacus represents some part of λ we get that λ is a partition of N where

N = ∑_{i=1}^{p−1}((l_{i} − 1)l_{i}p/2 + l_{i}i) = (p/2) ∑_{i=1}^{p−1} l_{i}^{2} + ∑_{i=1}^{p−1} l_{i}(i − (p/2)).

1.8.1 We complete squares. First we have

(p/2)l_{i}^{2} + l_{i}(i − p/2) = \frac{1}{8p}(2pl_{i} + (2i − p))^{2} − \frac{1}{8p}(2i − p)^{2}.

This gives the formula

N = 1/8p ∑_{i=1}^{p−1}(2pl_{i} + (2i − p))^{2} − (p − 1)(p − 2)/24.

But then since l_{i}l_{p−i} = 0 half of the terms are always zero. Therefore we prefer the following alternative: Set

X_{i} = \{ \begin{matrix}2pl_{i} + (2i − p) \\ 2pl_{p−i} + (p − 2i) \end{matrix} \begin{matrix}l_{i} ≠ 0 \\ l_{i} = 0 \end{matrix}

and then

N = \frac{1}{8p} ∑_{i=1}^{(p−1)/2} X_{i}^{2} − \frac{(p − 1)(p − 2)}{48} .

2. Abacus numbers for spin blocks when p = 5 and p = 7.

Assume first that p = 5 where char K = 5. Then the group algebra KŜ_{n} may or may not have spin blocks of defect zero. Our result gives a complete answer when this is the case.

2.1 Theorem Let G = Ŝ_{n}. Then the following are equivalent.

(a) G has a 5-spin block of defect zero.

(b) There are integers X, Y such that 10(4n + 1) = X^{2} + Y^{2}.

(c) Each prime factor r of 10(4n + 1) with r ≡ 3 (mod 4) occurs to an even power as a factor of 10(4n + 1).

Proof. It is well-known that (b) and (c) are equivalent, see for example [8; 9.2]. We will establish the equivalence of (a) and (b).

Recall that the 5-bar cores are in 1–1 correspondence with 5-tuples of integers Γ = (0, l_{1}, l_{2}, l_{3}, l_{4}) such that l_{i} ≥ 0 and l_{1}l_{4} = 0 = l_{2}l_{3}. Given such Γ, it corresponds to a spin block of defect zero of Ŝ_{n} where

n = 1/40(X_{1}^{2} + X_{2}^{2}) − 1/4, X_{i} = \{ \begin{matrix}10l_{i} + (2i − 5), \\ 10l_{5−i} + (5 − 2i), \end{matrix} \begin{matrix}l_{i} ≠ 0 \\ l_{i} = 0 \end{matrix}

Hence (a) is true if and only if there is a solution of

10(4n + 1) = X^{2} + Y^{2}

where X ≡ ±3 (mod 10) and Y ≡ ±1 (mod 10). The proof is therefore complete if we show that

2.2 Suppose n is an integer such that the equation 10(4n + 1) = X^{2} + Y^{2} has a solution in integers. Then there is also a solution where X ≡ ±3 (mod 10) and Y ≡ ±1 (mod 10).

Proof. By 5.2 (with m = 5(4n + 1)) there are integers X, Y which are ≢ 0 (mod 5) satisfying

10(4n + 1) = X^{2} + Y^{2}.

Then X, Y must be odd. Since the numbers are ≢ 5 (mod 10) they can only be ≡ ±1 or ±3 (mod 5). Suppose X, Y are both ≡ ±1 (mod 10), say, then X^{2} + Y^{2} ≡ 2 (mod 10), a contradiction. Hence, without loss of generality X ≡ ±3 (mod 10) and Y ≡ ±1 (mod 10), as required.

Now we consider the case when p = 7. Here we have the following result.

2.3 Theorem Let G = Ŝ_{n} for n ≥ 1. Then G has always at least one 7-spin block of defect zero.

Proof. Let n ≥ 1. We must show that there is a 7-bar core λ with λ ⊢ n. That is, we require some Γ = (0, l_{1}, …, l_{6}) with l_{i} ≥ 0 and l_{i}l_{7−i} = 0 and

n = N(Γ) = \frac{1}{56} ∑_{i=1}^{3} X_{i}^{2} − \frac{30}{48}, X_{i} = \{ \begin{matrix}14l_{i} + (2i − 7) \\ 14l_{7−i} + (7 − 2i) \end{matrix} l_{i} ≠ 0 ; otherwise

Such Γ exists if and only if the equation

(∗)  56n + 35 = ∑_{i=1}^{3} X_{i}^{2}

has integer solutions X_{i} such that X_{1} ≡ ±5 (mod 14), X_{2} ≡ ±3 (mod 14) and X_{3} ≡ ±1 (mod 14).

Since 56n+35 is not of the form 4^{a}(8b+7) we know that (∗) has integer solutions. By 5.3 there is also a solution where the X_{i}^{2} are ≡ 1, 2, 4 (mod 7). So we may set X_{1} = 7x + 5, X_{2} = 7y + 3 and X_{3} = 7z + 1. Since 56n + 35 ≡ 3 (mod 8), the X_{i} are odd and hence x, y, z are even. This completes the proof.

3. Diagonalization

Let k be a positive integer, and let z_{1}, z_{2}, …, z_{k} be indeterminates. Consider the quadratic form

q(z⃗) = m(∑_{i=1}^{k} z_{i}^{2}) − (1/2)(∑_{i≠j} z_{i}z_{j})

where m is a non-zero integer. Let A be the corresponding symmetric matrix. Then A has eigenvalues

λ_{1} = m − (k − 1)/2, λ_{2} = λ_{3} = … λ_{k} = m + (1/2).

(An eigenvector for eigenvalue λ_{1} is given by (1, 1, …, 1)^{t}, and eigenvectors for the other eigenvalues are for example (1, 0, …, −1, 0, …).) We assume that the λ_{i} are non-zero; and we are interested in the case when k is a 2-power, say k = 2^{d}.

3.1 Under these hypotheses, it is easy to diagonalize the quadratic form over Q. Namely there is a symmetric integral matrix P whose columns are orthogonal eigenvectors of A, and where all entries of P are ±1. (For example, one may take the character table of an elementary abelian 2-group.) As a convention, we take (1, 1, …, 1) as the first row of P. It follows that P^{2} = P^{t}P = 2^{d}I. We change variables; let w⃗ = z⃗P; so that z⃗ = 2^{−d}(w⃗P); then it is clear that

q(w⃗) = 2^{−d}(∑_{i=1}^{k} λ_{i}w_{i}^{2}).

3.2 We consider now a linear form ℓ(z⃗) = a_{1}z_{1} + a_{2}z_{2} + … a_{k}z_{k} with a_{i} ∈ Z. We wish to express ℓ(z⃗) in terms of the variables w_{i} in 3.1. Suppose p⃗_{i} is the i -th row vector of P for 1 ≤ i ≤ k and let a⃗ = (a_{1}, …, a_{k}). Then it follows easily that

ℓ(w⃗) = 2^{−d}(∑_{i=1}^{k} μ_{i}w_{i})

where μ_{i} = ⟨a⃗, p⃗_{i}⟩. Here ⟨−, −⟩ is the usual inner product on R^{k}.

Consider the number C := ∑_{i=2}^{k} μ_{i}^{2} (we will need it later.) We claim that C = [2^{d}(∑_{i=1}^{k} a_{i}^{2})] − (∑_{i=1}^{k} a_{i})^{2}. This holds because

∑_{i=1}^{k} μ_{i}^{2} = ⟨(μ_{i}), (μ_{i})⟩ = ⟨a⃗P, a⃗P ⟩ = ⟨a⃗P^{2}, a⃗⟩ = 2^{d}⟨a⃗, a⃗⟩,

and μ_{1} = ∑_{i=1}^{k} a_{i}; recall that p⃗_{1} = (1, 1, …, 1).

3.3 Consider now the function N : Z^{k} → Z given by N(z⃗) = q(z⃗) + ℓ(z⃗). We change variables as described above and obtain

N(w⃗) = q(w⃗) + ℓ(w⃗) = 2^{−d}(∑_{i=1}^{k} λ_{i}w_{i}^{2} + ∑_{i=1}^{k} μ_{i}w_{i}).

Recall that the eigenvalues λ_{2}, ..., λ_{k} are all equal, set λ = λ_{2}. Now define

(∗)  N_{1} = 2^{−d}(λ_{1}w_{1}^{2} + μ_{1}w_{1})

and let ρ := N − N_{1}. We complete squares and obtain for ρ the formula

(∗∗)  ρ = 2^{−d−2}λ^{−1}(∑_{i=2}^{k}(2λw_{i} + μ_{i})^{2} − C).

where C = ∑_{i=2}^{k} μ_{i}^{2}, see 3.2.1.

3.4 From now we assume that k = 4 and that m = (p − 1)/2 where p is a prime ≥ 5. Then λ = p/2 and λ_{1} = (p − 4)/2. The formulae in 3.3 become now

N_{1} = [(p − 4)/8]w_{1}^{2} + (μ_{1}/4)w_{1}

and

ρ = (1/8p)(∑_{i=2}^{4}(pw_{i} + μ_{i})^{2} − C).

3.5 Consider now w⃗ = z⃗P; then z⃗ = (1/4)w⃗P. We say that the values w_{i} are admissible if the corresponding z_{i} are non-negative integers. We investigate now in detail conditions for the w_{i} to be admissible. For k = 4 we may take the variables w_{i} as follows.

\begin{matrix}w_{1} = z_{1} + z_{2} + z_{3} + z_{4}, & w_{2} = z_{1} + z_{2} − z_{3} − z_{4}, \\ w_{3} = z_{1} − z_{2} − z_{3} + z_{4}, & w_{4} = z_{1} − z_{2} + z_{3} − z_{4}. \end{matrix}

Then we have that

\begin{matrix}z_{1} = (1/4)(w_{1} + w_{2} + w_{3} + w_{4}), & z_{2} = (1/4)(w_{1} + w_{2} − w_{3} − w_{4}), \\ z_{3} = (1/4)(w_{1} − w_{2} − w_{3} + w_{4}), & z_{4} = (1/4)(w_{1} − w_{2} + w_{3} − w_{4}). \end{matrix}

We obtain now the following criterion.

3.5.1 Lemma The numbers w_{i} are admissible if and only if they satisfy the following conditions:

(i) w_{1} ≡ w_{i} (mod 2) and ∑ w_{i} ≡ 0 (mod 4); (ii) w_{1} + w_{i} ≥ |w_{j} + w_{k}| for {i, j, k} = {1, 2, 3}.

4. Abacus numbers for symmetric groups when p = 5 and p = 7.

4.1 Assume first that p = 5. We consider cores for the symmetric group. We take as an abacus for a general core Γ = (0, y_{1}, ..., y_{4}). This represents the integer N(Γ) = q(y⃗)+ℓ(y⃗); where

q(y⃗) = 2(∑_{i=1}^{4} y_{i}^{2}) − (1/2)(∑_{i≠j} y_{i}y_{j})

and

ℓ(y⃗) = −y_{1} + y_{3} + 2y_{4}.

This was proved in 1.6. For the change of variables described in Chapter 3 (with y_{i} = z_{i}) we must set

a_{1} = −1, a_{2} = 0, a_{3} = 1, a_{4} = 2

(with the notation of 3.2). Therefore using 3.5 and 3.2 we obtain that

\begin{matrix}μ_{1} = 2, \\ μ_{3} = a_{1} − a_{2} − a_{3} + a_{4} = 0, \end{matrix} \begin{matrix}μ_{2} = a_{1} + a_{2} − a_{3} − a_{4} = −4, \\ μ_{4} = a_{1} − a_{2} + a_{3} − a_{4} = −2. \end{matrix}

Hence C = 20. Consequently we have

N_{1} = (1/8)w_{1}^{2} + (1/2)w_{1}

and

ρ = N − N_{1} = 1/40((5w_{2} − 4)^{2} + (5w_{3})^{2} + (5w_{4} − 2)^{2} − 20).

Note that the variable w_{1} is the total number of beads of the abacus.

4.2 Theorem Assume p = 5. Then every integer n ≥ 1 is an abacus number. That is, the group algebra KS_{n} has at least one 5-block of defect zero, for any n ≥ 1.

Actually we prove something slightly stronger. Namely there is a block of defect zero whose core has number of parts ≡ −1 (mod 4) if n ≥ 105.

Proof. Given any integer n ≥ 1; then there are unique integers k and n_{0} with k, n_{0} ≥ 0 such that

(1) n = (2k^{2} + k) + n_{0} with 0 ≤ n_{0} < 4k + 3. This is clear, noting that 2(k + 1)^{2} + (k + 1) − (2k^{2} + k) = 4k + 3. We will now show that n can be represented on a core abacus with 4k − 1 beads, if k > 6.

Let w_{1} = 4k − 1; then N_{1} = 2k^{2} + k − (3/8). Define now m = m(n) to be the integer (2) m = 40(N − N_{1}) + 20 = 40n_{0} + 35 Our aim is to show that there are integers w_{2}, w_{3}, w_{4} such that m = (5w_{2} − 4)^{2} + (5w_{3})^{2} + (5w_{4} − 2)^{2} and moreover such that the w_{i} are admissible, in the sense of 3.5.1.

Since m ≡ 3 (mod 8) and m ≡ 0 (mod 5), we know by 6.1 that there is a solution in integers of

(∗)  m = X^{2} + Y^{2} + Z^{2}

with X ≡ 1 (mod 10), Y ≡ 5 (mod 10) and Z ≡ 3 (mod 10). Now set

X = 5w_{2} − 4 Y = 5w_{3} Z = 5w_{4} − 2 where w_{2} = 2r_{2} + 1

where w_{3} = 2r_{3} + 1
where w_{4} = 2r_{4} + 1.

Then the w_{i} are odd, hence ≡ w_{1} (mod 2). By 3.5.1 we also require that ∑ w_{i} ≡ 0 (mod 4), that is, ∑_{i=2}^{4} r_{i} ≡ 1 (mod 2). Suppose that instead we have ∑ r_{i} ≡ 0 (mod 2). Observe that Y = 10r_{3} + 5 = −(10(−r_{3} − 1) + 5). Replace r_{i} by r′_{i} where

r′_{3} = −r_{3} − 1, r′_{j} = r_{j} otherwise.

Then ∑ r′_{i} ≡ 1 (mod 2), as required.

It remains to investigate the condition w_{1} +w_{i} ≥ |w_{j} +w_{k}|. Without loss of generality, assume that w_{1} + w_{2} < |w_{3} + w_{4}|. We assume in (1) that 0 ≤ n_{0} < 4k + 3, hence

35 ≤ m = 40n_{0} + 35 < 40(4k + 3) + 35 = 160k + 155.

We use now Lemma 4.3. This gives that |r_{i}| < \frac{6}{10}k since k > 6. Then we get

w_{1} + w_{2} = (4k − 1) + (2r_{2} + 1) < |2(r_{3} + r_{4}) + 2|

which implies that

4k < 2 + \frac{18}{5}k

and k ≤ 5, contrary to the hypothesis.

This proves that any integer n with n ≥ 2 · 7^{2} + 7 = 105 is an abacus number. For integers n ≤ 104 one may directly find a core abacus representing n (see also 4.3.1).

4.3 Lemma Suppose X = 10r_{2} + 1, Y = 10r_{3} + 5 and Z = 10r_{4} + 3 with integers r_{i}, and assume that k is an integer such that

X^{2} + Y^{2} + Z^{2} < 160k + 155.

Then |r_{i}| < \frac{6}{10}k unless k ≤ 6.

Proof. The worst case to consider is the possibility that (−r_{3}) ≥ \frac{6}{10}k. If so then

(6k − 5)^{2} ≤ (10r_{3} + 5)^{2} < 160k + 155.

Hence (6k)^{2} − 220k < 130 and therefore

(6k − \frac{55}{3})^{2} < 130 + (\frac{55}{3})^{2} < \frac{(65)^{2}}{9}

and k < \frac{40}{6} and consequently k ≤ 6.

4.3.1 Remark With the notation of 4.2, suppose n = 2k^{2} + k + n_{0} and 0 ≤ n_{0} ≤ 2k + 1. Then we have that 0 ≤ 40n_{0} + 35 ≤ 80k + 105. Now one can prove that, with X, Y, Z as in 4.3, if X^{2} + Y^{2} + Z^{2} ≤ 80k + 105 then |r_{i}| ≤ \frac{6}{10}k unless k ≤ 4. Hence for k = 5, 6 one deduces that n is an abacus number if

2k^{2} + k ≤ n ≤ 2k^{2} + 3k + 1.

This leaves for each of these values for k only 2k + 2 integers to be considered; and also the integers n ≤ 55.

4.4 Now assume that p = 7. We will consider cores for the symmetric group with abacus representations of the form Γ = (0, y_{1}, y_{2}, y_{3}, 0, y_{5}, 0). This is a partition of n where n = N(Γ) = q(y⃗) + ℓ(y⃗) where

q(y⃗) = 3 ∑_{i} y_{i}^{2} − \frac{1}{2} ∑_{i≠j} y_{i}y_{j}

and

ℓ(y⃗) = −2y_{1} − y_{2} + 2y_{5}.

We change variables as in Ch. 3. Set

z_{i} = \{ \begin{matrix}y_{i} \\ y_{5} \end{matrix} \begin{matrix}i = 1, 2, 3 \\ i = 4 \end{matrix}

and w⃗ = z⃗P, and with the notation of 3.2 we have (a_{1}, a_{2}, a_{3}, a_{4}) = (−2, −1, 0, 2). It follows then with P and μ_{i} as in 3.5 that

μ_{1} = −1, μ_{2} = −5, μ_{3} = 1, μ_{4} = −3.

Hence we get the constant C = 35. (This was chosen because C ≡ 0 (mod 7).) Consequently N_{1} = 3/8w_{1}^{2} + (−1/4)w_{1} and

N − N_{1} = \frac{1}{56}((7w_{2} − 5)^{2} + (7w_{3} + 1)^{2} + (7w_{4} − 3)^{2} − 35).

4.5 Theorem Assume p = 7, then every integer n ≥ 1 is an abacus number. That is, the group algebra KS_{n} has at least one 7-block of defect zero, for any n ≥ 1.

Proof. We use the abacus representation as in 4.4. By 1.5 it suffices to consider odd integers n.

Given an odd integer n ≥ 1, there are integers k and n_{0} ≥ 0 such that

(1) n = 6k^{2} − 4k + n_{0} with 0 < n_{0} < 12k + 2. This is clear noting that [6(k + 1)^{2} − 4(k + 1)] − (6k^{2} − 4k)] = 12k + 2. We will now show that n can be represented on a core abacus with 4k − 1 beads if k > 12.

Let w_{1} = 4k − 1 then we have N_{1} = 6k^{2} − 4k + (5/8) and hence N − N_{1} = n_{0} − (5/8). Define now m = m(N) to be the integer

(2) m = 56(N − N_{1}) + 35 = 56n_{0}. Our aim is to show that there are integers w_{2}, w_{3}, w_{4} such that

m = (7w_{2} − 5)^{2} + (7w_{3} + 1)^{2} + (7w_{4} − 3)^{2}

and moreover such that the w_{i} are admissible, in the sense of 3.5.1.

Since n_{0} is odd, m is not of the form 4^{a}(8b + 7). So there is a solution in integers of the equation

(∗)  m = X^{2} + Y^{2} + Z^{2}.

By 5.3, there is also a solution with X, Y, Z ≢ 0 (mod 7). Since m ≡ 0 (mod 7), the equation (∗) has modulo 7 essentially one non-trivial solution, namely X ≡ ±5 (mod 7), Y ≡ ±1 (mod 7) and Z ≡ ±3 (mod 7).

Define integers w_{i} by setting X = 7w_{2} −5, Y = 7w_{3} +1 and Z = 7w_{4} −3. Since m ≡ 0 (mod 8), we must have that X, Y, Z are even and hence the w_{i} are odd. Set w_{i} = 2r_{i} + 1, we will show that these w_{i} are admissible.

Clearly w_{i} ≡ w_{1} (mod 2). Recall that w_{1} = 4k − 1, hence ∑ w_{i} ≡ 2(r_{2} + r_{3} + r_{3} + 1) (mod 4). Let u = 7r_{2} + 1, v = 7r_{3} + 4 and w = 7r_{4} + 2. Then we have that

u^{2} + v^{2} + w^{2} ≡ 0 (mod 2)

(since 4(u^{2} +v^{2} +w^{2}) = m ≡ 0 (mod 8)). Now calculating mod 2 we have u^{2} ≡ (r_{2}^{2} +1)^{2} ≡ r_{2} + 1, v^{2} ≡ r_{3}^{2} ≡ r_{3} and w ≡ r_{4}^{2} ≡ r_{4} and hence r_{2} + 1 + r_{3} + r_{4} ≡ 0 (mod 2), as required.

It remains to investigate the condition w_{1} + w_{i} ≥ |w_{j} + w_{k}|. Assume that, say, w_{1} + w_{2} < |w_{3} + w_{4}|. We assume in (1) that 0 ≤ n_{0} < 12k + 2, hence

0 ≤ m = 56n_{0} < 56(12k + 2) = 21 · 32k + 7 · 16.

From Lemma 4.6 below we get |r_{i}| < \frac{8}{14}k for k > 12. This implies now that 4k < \frac{24}{7}k + 2 and k ≤ 3, contrary to the hypothesis.

This proves that every integer n with n ≤ 13^{2} · 6 − 13 · 4 is an abacus number. For the remaining integers one may directly find a core abacus representing n (see also 4.6.1).

4.6 Lemma Suppose X = 14r_{2} + 2, Y = 14r_{3} + 8 and Z = 14r_{4} + 4, with r_{i} ∈ Z, such that

X^{2} + Y^{2} + Z^{2} < 21 · 32k + 7 · 16.

Then |r_{i}| < \frac{8}{14}k unless k ≤ 12.

Proof. The worst possibility arises when (−r_{3}) ≥ \frac{8}{14}k. Then

(8k − 8)^{2} < 21 · 32k + 7 · 16.

We complete squares and deduce

(8k − 50)^{2} < 16 · 11 + 4 · (25)^{2} = 4 · 669.

Hence 8k − 50 < 2 · 26 and k < \frac{102}{8} and hence k ≤ 12.

4.6.1 (1) With the notation of 4.5, suppose n = 6k^{2} − 4k + n_{0} where 0 < n_{0} ≤ 6k + 1. Then 0 < 56n_{0} ≤ 21 · 16k + 56. One may prove that if X, Y, Z are as in 4.6 and

X^{2} + Y^{2} + Z^{2} < 21 · 16k + 56

then |r_{i}| ≤ \frac{8}{14}k unless k ≤ 7. So for 8 ≤ k ≤ 12 one obtains by this method also (using 1.6) that n is an abacus number when

6k^{2} − 4k ≤ n ≤ 6k^{2} + 2k + 2.

(2) A further improvement is obtained by taking the number of beads w_{1} to be of the form w_{1} = 4k + 1. Then N_{1} = 6k^{2} + 2k + \frac{1}{8}. We express now an integer n in the form n = 6k^{2} + 2k + n_{0} with 0 ≤ n_{0} < 12k + 8. Then we have N − N_{1} = n_{0} − \frac{1}{8}. Therefore if we set

m = 56(N − N_{1}) + 35 = 56n_{0} + 28

then we obtain similarly as in 4.5 that for at least two successive values of n_{0} the equation

m = X^{2} + Y^{2} + Z^{2}

has an admissible solution for k not too small. If we restrict ourselves to 0 ≤ n_{0} < 6k + 4 then 56n_{0} + 28 < 21 · 16k + 72. One may prove that if X, Y, Z are as in 4.6 and X^{2} + Y^{2} + Z^{2} < 21 · 16k + 9 · 28 then |r_{i}| ≤ \frac{8}{14}k unless k ≤ 7. So for 8 ≤ k ≤ 12 one obtains that n is an abacus number if

6k^{2} + 2k ≤ n ≤ 6k^{2} + 8k + 4.

Combining (1) and (2) gives all integers n with n > 280.

5. Representing integers as a sum of three squares

First we will prove the results required for p = 5.

5.1 Lemma Suppose m ≡ 3 (mod 8) and ≡ 0 (mod 5). Then the equation

m = X^{2} + Y^{2} + Z^{2}

has a solution in integers such that X ≡ 1 (mod 10), Y ≡ 5 (mod 10) and Z ≡ 3 (mod 10).

Proof. The conditions on m ensure that there are integers u, v, w ∈ Z such that m = u^{2} + v^{2} + w^{2}.

(1) We claim that there is such a representation in which not all of u, v, w are ≡ 0 (mod 5). This is clear if m ≢ 0 (mod 25). Now suppose m = 25m_{1}, then m_{1} ≡ 3 (mod 8) and hence there are integers u, v and w with m_{1} = u^{2} + v^{2} + w^{2} where u ≢ 0 (mod 5), by the induction hypothesis. Now consider the representations

\begin{aligned}25m_{1} & = (3u + 4v)^{2} + (4u − 3v)^{2} + 25w^{2} \\ 25m_{1} & = (3v + 4u)^{2} + (4v − 3u)^{2} + 25w^{2}. \end{aligned}

We claim that one of 3u + 4v, 3v + 4u must be ≢ 0 (mod 5). Otherwise we would have 3u + 4v ≡ 0 (mod 5) and u ≡ 2v (mod 5) and in addition 3v + 4u ≡ 0 (mod 5) and v ≡ 2u (mod 5); and consequently u ≡ 4u (mod 5) and u ≡ 0 (mod 5), contrary to the hypothesis. Hence m has the required representation, and (1) is proved.

Take now a representation where not all of u, v, w are ≡ 0 (mod 5). The squares modulo 5 are {0, 1, 4}; therefore to obtain u^{2} + v^{2} + w^{2} ≡ 0 (mod 5) we must have that {u^{2}, v^{2}, w^{2}} = {0, 1, 4}. Hence without loss of generality,

(2) v^{2} ≡ 0, w^{2} ≡ 4 and u^{2} ≡ 1 (mod 5).

(3) Let u, v, w be as before, then u, v and w are odd: Squares are ≡ 0, 4 or 1 (mod 8); and we are given that u^{2} + v^{2} + w^{2} ≡ 3 (mod 8). Suppose that u, say, is even. Then u^{2} + v^{2} + w^{2} is either ≡ 4 + v^{2} + w^{2} or ≡ v^{2} + w^{2} (mod 8) and v^{2} + w^{2} ≡ 7 or 3 (mod 8), which is not possible.

Now we have that v is odd and ≡ 0 (mod 5), so v ≡ 0 (mod 10). Moreover w ≡ ±2 (mod 5) and w is odd, so w ≡ ±3 (mod 10). Similarly u ≡ ±1 (mod 5) and u is odd, so u ≡ ±1 (mod 10). This completes the proof.

5.2 Lemma Suppose m is an odd integer such that the equation

2m = X^{2} + Y^{2}

has a solution in integers. Then there is a solution in which X, Y ≢ 0 (mod 5).

Proof. This is clear if m ≢ 0 (mod 25). Now suppose m = 25m_{1}, and assume X, Y are some solution. Suppose X = 5X_{1}, then also Y = 5Y_{1} and 50m_{1} = 25(X_{1}^{2} + Y_{1})^{2}, and hence 2m_{1} = X_{1}^{2} + Y_{1}^{2}.

By the induction hypothesis, there are u, v such that 2m_{1} = u^{2} + v^{2} with u, v ≢ 0 (mod 5). Then

\begin{matrix}(3u + 4v)^{2} + (4u − 3v)^{2} = 2m, and \\ (3v + 4u)^{2} + (4v − 3u)^{2} = 2m. \end{matrix}

It follows that at least one of 3u + 4v, 3v + 4u must be ≢ 0 (mod 5), as in 5.1. Say X = 3u + 4v ≢ 0 (mod 5), then Y = (4u − 3v) ≢ 0 (mod 5) as well since m ≡ 0 (mod 5), and this gives a required solution.

Now we study the equations occuring when p = 7.

5.3 Lemma Suppose m ∈ Z and m ≡ 0 (mod 7). Assume that the equation

m = u^{2} + v^{2} + w^{2}

has a solution in Z. Then there is also a solution where u^{2}, v^{2} and w^{2} are ≡ 1, 2 and 4 (mod 7).

Proof. By inspection, there is only one non-trivial solution for m ≡ u^{2} + v^{2} + w^{2} (mod 7). Therefore it suffices to show that there is a solution where at least one of u, v, w is ≢ 0 (mod 7). We proceed by induction on m. The statement is clear if m ≢ 0 (mod 49). Now suppose m = 49m_{1}. We assume that m is a sum of three squares, therefore m is not of the form 4^{a}(8b + 7); and then m_{1} is not of this form too. Therefore by induction there are integers u, v, w with m_{1}^{2} = u^{2} + v^{2} + w^{2} and u ≢ 0 (mod 7). Now consider the two expressions

\begin{matrix}49m_{1} = & (6u + 3v − 2w)^{2} + (3u − 2v + 6w)^{2} + (2u − 6v − 3w)^{2} \\ 49m_{1} = & (6u + 3w − 2v)^{2} + (3u − 2w + 6v)^{2} + (2u − 6w − 3v)^{2}. \end{matrix}

Suppose both 6u + 3v − 2w and 6u + 3w − 2v are ≡ 0 (mod 7). Then it follows that also the difference 5v − 5w ≡ 0 (mod 7); that is, v ≡ w (mod 7). Consequently 6u + v ≡ 0 (mod 7) and 6u − v ≡ 0 (mod 7) and it follows that u ≡ 0 (mod 7), a contradiction.

Acknowledgement. This work was supported by the SERC (via a Visiting Fellowship of the second author and also via the Warwick Algebra Symposium 1991), and by the Graduierten-Kolleg Essen. The authors are grateful to the hospitality of the Mathematics Departments at Oxford, Warwick and of the Institut für Experimentelle Mathematik at Essen. The authors are grateful to H. G. Rück for valuable help.

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Received April 10, 1995

MSC 1991: 20C30, 11E25; 20C20