ON THE IRREDUCIBLE REPRESENTATIONS OF THE ALTERNATING GROUP WHICH REMAIN IRREDUCIBLE IN CHARACTERISTIC p

We consider the problem of which ordinary irreducible representations of the alternating group An remain irreducible modulo a prime p. We solve this problem for p = 2, and present a conjecture for odd p, which we prove in one direction.


Introduction
An interesting question for any finite group is to ask which ordinary irreducible representations of that group remain irreducible in characteristic p. For the symmetric group S n this amounts to classifying the irreducible Specht modules, and this problem was solved several years ago, through the combined efforts of James, Mathas, Lyle and the author [JM2,JM3,L1,F1,F2]; when p is odd, the irreducible Specht modules are precisely those labelled by JM-partitions. In this paper we address the case of the alternating group A n . The author considered this problem in [F3], solving it completely in the case p = 2 and presenting a conjectured solution for odd p. In the present paper we prove this conjecture.
As with many problems concerning the representation theory of the alternating group, our technique is to translate the problem to one about the symmetric group, using elementary Clifford theory to transfer results between the two settings. For the problem at hand, this translation was done in [F3], where the main problem for the alternating groups was reduced to the question of which Specht modules labelled by self-conjugate partitions have exactly two composition factors (with multiplicity). So the present paper is concerned entirely with the representation theory of the symmetric group.
Our main result when p 5 is that the self-conjugate partitions labelling Specht modules with composition length 2 are the partitions which we called R-partitions in [F3]; these have a simple description in terms of hook lengths in the Young diagram. The fact that the corresponding Specht modules have composition length 2 was shown in [F3], so the task undertaken in this paper is to prove the converse. The same applies in the case p = 3, where the classification takes a slightly different form.
Our basic strategy involves applying the results of Brundan and Kleshchev concerning Robinson's i-restriction functors e i . We suppose λ is a self-conjugate partition which is not an R-partition or a JM-partition; then we must show that the Specht module S λ has at least three composition factors. By removing all the removable nodes of residue 0 from the Young diagram of λ, one obtains a self-conjugate partition λ ▽0 , and the Brundan-Kleshchev results imply that the composition length of S λ is at least that of S λ ▽0 ; so by induction, we may assume that either λ ▽0 is an R-partition or a JM-partition, or λ ▽0 = λ. Similarly, for any i 0 we can define a self-conjugate partition λ ▽±i by repeatedly removing removable nodes of residues i and p − i from λ, and we can make the same inductive assumption about λ ▽±i . This restricts the possibilities for λ considerably. In fact, we can strengthen this inductive argument using James's Regularisation Theorem, which gives an explicit composition factor D λ rest of the Specht module S λ , for any λ. It is very helpful for our purposes to be able to tell, for a given partition λ, when there is an r such that e r i S λ 0 but e r i D λ rest = 0. An important new result in this paper (Proposition 4.9) is an explicit combinatorial criterion for this; it is surprising to the author that this does not seem to have been discovered before.
This inductive argument deals with most cases. Several of the remaining cases are eliminated with using the theory of Rouquier blocks, whose decomposition numbers are very well understood. By developing the associated abacus combinatorics we exploit these results, together with the classification of irreducible Weyl modules, to show that our main theorem holds for the so-called p-quotient-separated partitions.
These arguments leave just one family of Specht modules to deal with, which we approach using the theory of homomorphisms between Specht modules. Establishing the existence of a non-zero homomorphism S λ → S ν shows that S λ and soc(S ν ) share a composition factor, and we use this in a certain special case to show that S λ has at least three composition factors. The Specht homomorphism that we use is constructed as the composition of two well-known homomorphisms, namely the one-node Carter-Payne homomorphism and the regularisation homomorphism. However, we have considerable work to do in showing that the composition is non-zero in our particular situation. To do this, we give a result describing the least dominant tableau occurring when the regularisation homomorphism is expressed in terms of semistandard homomorphisms; as a by-product, this gives a new proof of the (non-trivial) fact that the regularisation homomorphism is non-zero.
This homomorphism result, together with a small lemma concerning the Mullineux map (which describes the effect of the functor − ⊗ sgn on simple modules), is enough to complete the proof. We conclude the paper with a simple corollary which shows that the only irreducible representations of A n remaining irreducible modulo every prime are the one-dimensional representations.
Acknowledgement. The author was inspired to re-visit this problem at an Oberwolfach miniworkshop on the representation theory of the symmetric groups; he is very grateful to Suzanne Danz and David Hemmer for the invitation to this workshop, and to David Hemmer for subsequent discussions on this problem.

Representation theory of the symmetric group
In this section, we recall some essential background on the representation theory of the symmetric group. Throughout this paper n is a non-negative integer and F is a field of characteristic p; we use the convention that the characteristic of a field is the order of its prime subfield, so p ∈ {2, 3, 5, . . . } ∪ {∞}.

Partitions and Specht modules
A composition of n is a sequence λ = (λ 1 , λ 2 , . . . ) of non-negative integers summing to n. If n is not specified, we write |λ| for the sum of the terms of λ. When writing compositions, we usually omit trailing zeroes and group together consecutive equal parts with a superscript; the unique composition of 0 is denoted ∅. A composition which is weakly decreasing is called a partition. We often identify a composition λ with its Young diagram, which is the set (r, c) ∈ N 2 | c λ r .
Elements of the Young diagram of λ are called nodes of λ; more generally, a node is any element of N 2 .
We adopt the English convention for drawing Young diagrams, in which λ is drawn with left-justified rows of boxes of lengths λ 1 , λ 2 , . . . successively down the page.
If λ is a partition, the conjugate partition λ ′ is defined by or, in terms of Young diagrams, by reflecting along the main diagonal. λ is self-conjugate if λ ′ = λ. λ is p-restricted if λ i − λ i+1 < p for all i 1, and p-regular if λ ′ is p-restricted.
A node (r, c) of a partition λ is removable if it can be removed from λ to leave a smaller partition (i.e. if c = λ r > λ r+1 ), while a node (r, c) not in λ is an addable node of λ if it can be added to λ to give a larger partition.
The p-residue of a node (r, c) is the residue of c − r modulo p (or simply the integer c − r, when p = ∞). If a node has residue i, we call it an i-node.
If λ is a composition of n, let M λ denote the Young permutation module for FS n corresponding to λ, as defined in [J2, §4]. If λ is a partition, let S λ denote the Specht module corresponding to λ. If λ is p-restricted, then S λ has a simple socle D λ , and the modules D λ afford all the irreducible representations of FS n as λ ranges over the set of p-restricted partitions of n. When p = ∞, we have D λ = S λ , so the characters χ λ of the Specht modules give all the ordinary irreducible characters of S n .
Remark. It is slightly more traditional to label the simple FS n -modules by p-regular partitions: if λ is p-regular, then S λ has a simple cosocle D λ , and these modules also afford all the irreducible representations of FS n . It is well known how to convert from one convention to the other; we have chosen the p-restricted convention in this paper because it aligns better with some of the references that we cite.
We shall also briefly need to consider Weyl modules for the Schur algebra S(n, n), for which we refer to the book by Green [G]. We let ∆ λ denote the Weyl module (also called the standard module) labelled by the partition λ of n, and L λ its unique irreducible quotient.

James's Regularisation Theorem
The main aim in this paper is to consider Specht modules with very few composition factors. A very helpful fact in this endeavour when p < ∞ is that we know an explicit composition factor of every Specht module. This result is James's Regularisation Theorem, which we phrase here in terms of p-restricted partitions.
Suppose l 0. Define the lth ramp in N 2 to be the set of nodes (r, c) for which c − 1 + (p − 1)(r − 1) = l. If l < m, we say that ramp m is later than ramp l. If λ is a partition, the p-restrictisation of λ is the p-restricted partition λ rest obtained by moving all the nodes in each ramp as far to the left within that ramp as possible. (λ rest is simply called the p-restriction of λ in [FLM], but we introduce the slightly absurd term restrictisation here to avoid confusion with restriction in the sense of restricting to subgroups, which we shall consider a great deal. The linguistically sensitive reader may rest assured that we shall use this term as little as possible.) Example. Take λ = (8, 6, 2, 1 2 ) and p = 3. Then λ rest = (6, 5, 4, 2, 1), as we can see from the following Young diagrams, in which we label each node with the number of the ramp in which it lies. 0 1 2 3 4 5 6 7 2 3 4 5 6 7 4 5 6 8 , 0 1 2 3 4 5 2 3 4 5 6 4 5 6 7 6 7 8 . Now we can state part of James's Regularisation Theorem, translated to the p-restricted convention.

The p-core and p-weight of a partition
If λ is a partition, the rim of λ is defined to be the set of all nodes (r, c) of λ such that (r + 1, c + 1) is not a node of λ. Given any node (r, c) of λ, the (r, c)-rim hook of λ is the connected portion of the rim running from the node (r, λ r ) down to the node (λ ′ c , c). The (r, c)-hook length of λ is the number of nodes in the (r, c)-rim hook, i.e. λ r − r + λ ′ c − c + 1. If p < ∞, then we say that λ is a p-core if none of the hook lengths of λ is divisible by p, or equivalently if none of the hook lengths equals p.
If λ is an arbitrary partition, the p-core of λ is obtained as follows. Choose a node (r, c) of λ such that the (r, c)-hook length equals p, and delete the (r, c)-rim hook from λ; repeat until a p-core is obtained. This p-core is independent of the choice of rim hook deleted at each stage, and hence so is the number of rim hooks deleted; this number is called the p-weight of λ.

The sign representation
Let sgn denote the one-dimensional sign representation of S n . This gives rise to a functor − ⊗ sgn : FS n -mod → FS n -mod, which takes simple modules to simple modules. The effect of this functor on Specht modules is well-known; let M * denote the dual of a module M.
Since every simple FS n -module is self-dual [J2, Theorem 11.5] and every Specht module is indecomposable when p 3 [J2, Corollary 13.18], Theorems 2.1 and 2.2 imply the following. Lemma 2.3 [F3, Proof of Theorem 4.1]. Suppose p 3, and λ is a self-conjugate partition such that S λ has exactly two composition factors S, T. Then S ⊗ sgn T S. Now we consider simple modules. Since S ⊗ sgn is simple whenever S is, there is an involution m p on the set of p-restricted partitions of n, defined by D λ ⊗ sgn D m p (λ) . This involution is known as the Mullineux map, since Mullineux [M] gave (albeit without proof) the first of several known recursive combinatorial descriptions of the map. We do not give this algorithm here, since for this paper we just need the following simple result concerning m p .
Proof. We use the work of Ariki et al. [AKT] which addresses the relationship between the labellings of simple modules by p-restricted partitions and by Littelmann paths. Following Kreiman et al. [KLMW], they define for each p-restricted partition λ a p-core roof(λ), which has the following two properties: This cannot be congruent to 1 modulo p, since then the (1, 1)-hook length of roof(λ) would be divisible by p, contradicting the fact that roof(λ) is a p-core.
It would be very interesting to see a more direct proof of Lemma 2.4 using Mullineux's algorithm, for example.

JM-partitions
Now we describe the partitions which label irreducible Specht and Weyl modules when p 3. Define the p-power diagram of a partition λ to be the diagram obtained by filling the (r, c)-box in the Young diagram of λ with the p-adic valuation of the (r, c)-hook length, for each node (r, c) of λ. Say that λ is a p-JM-partition (or simply a JM-partition) if the following property holds: every non-zero entry in the p-power diagram is either equal to all the other entries in the same row or equal to all the other entries in the same column.
Example. The partition (19, 11, 2 3 , 1 2 ) is a 3-JM-partition, as we see from its 3-power diagram: A similar result applies for irreducible Weyl modules (without the restriction p 3). Observe that a JM-partition is p-restricted if and only if every non-zero entry in its p-power diagram is equal to every entry in the same row. A similar statement applies to p-regular JM-partitions (which are also known as Carter partitions).
Theorem 2.6 [JM2, Theorem 4.5]. The Weyl module ∆ λ is irreducible if and only if λ is a p-restricted JM-partition.

R-partitions
We now describe the self-conjugate partitions introduced in [F3] which (we claim) label the irreducible representations of the alternating group that remain irreducible in characteristic p.
Given a partition λ, construct the p-power diagram of λ as above. Say that λ is an R-partition if λ is self-conjugate and there is a distinguished node (r, r) of λ such that: • the (r, r)-entry in the p-power diagram is non-zero, and • any non-zero entry in the p-power diagram other than the (r, r)-entry is either equal to all the entries in its row or equal to all the entries in its column.
R-partitions were studied in detail in [F3]; later, we shall cite some results describing abacus displays for R-partitions. For now, we recall the two distinct types of R-partitions described in [F3, §4.2]. Suppose λ is an R-partition, with distinguished node (r, r).
1. λ is an R-partition of type I if r = 1. In this case, removing the (1, 1)-rim hook from λ leaves a self-conjugate p-core ξ with ξ 1 h.
2. λ is a R-partition of type II if the (r, r)-hook length of λ equals p. In this case, removing the (r, r)-rim hook from λ leaves a self-conjugate JM-partition.
Note that the R-partitions of type II include all self-conjugate partitions of p-weight 1. It was shown in [F3] that every R-partition is either of type I or type II; however, the two types are not mutually exclusive.

The alternating group and the main result
In this section we introduce the alternating group and give our main result, which we then re-cast in the symmetric group setting.
Suppose throughout this section that F is a splitting field for A n , and that p 2. We also assume that n 2 (so that A n has index 2 in S n ), the case n = 1 being trivial. If M is an FS n -module and H S n , let res H M denote the restriction of M to H. If M is irreducible, then by basic Clifford theory res A n M is irreducible if M ⊗ sgn M, and otherwise res A n M splits as the direct sum M + ⊕ M − of two irreducible modules. Furthermore, all irreducible FA n -modules arise in this way.
If p = ∞ then, by Theorem 2.2 and the fact that simple FS n -modules are self-dual, we have S λ ⊗ sgn S λ ′ , so the character χ λ of S λ restricts to an irreducible character ψ λ of A n if λ λ ′ (and in this case χ λ ′ also restricts to ψ λ ), while if λ = λ ′ then χ λ restricts to the sum of two irreducible characters ψ λ+ , ψ λ− . With this notation, we can give our main result.
Theorem 3.1. Suppose F is a splitting field for A n of characteristic p, and ψ = ψ λ or ψ λ± is an ordinary irreducible character of A n .

If p 5, then ψ is irreducible over F if and only if λ is a p-JM-partition or an R-partition.
Most of Theorem 3.1 has already been proved in [F3], beginning with the following reduction of the problem to the representation theory of S n .
Theorem 3.2 [F3, Theorem 4.1]. Suppose F is a splitting field for A n of odd characteristic p, and ψ = ψ λ or ψ λ± is an irreducible character of the alternating group A n . Then ψ is irreducible over F if and only if one of the following holds.
2. λ = λ ′ and the Specht module S λ has exactly two composition factors.
So from now on we can restrict attention entirely to the symmetric group, and prove the following.
Theorem 3.3. Suppose F has characteristic p, and λ is a self-conjugate partition. Then S λ has exactly two composition factors if and only if one of the following holds.
• λ is an R-partition of type I.
• p 5 and λ is an R-partition of type II.
We have already proved the 'if' part of Theorem 3.3 in [F3, §5.1]. So the remainder of this paper is dedicated to proving the 'only if' part. We do this in Section 7, after we have recalled some more background and developed further tools.

Restriction functors
In this section, we describe some results on restriction functors, which will be our main tool. The definition of these functors goes back to Robinson [Ro1], though our main reference here is the survey of Brundan and Kleshchev [BK]. We translate the partition combinatorics from [BK] to the p-restricted convention.

The restriction functors e i
Assume throughout this section that p < ∞. We shall feel free to identify Z/pZ with the set {0, . . . , p − 1}. In [BK,§2.2 Now we recall some results describing the effect of these operators on Specht modules and simple modules. Given a partition λ and i ∈ Z/pZ, let rem i (λ) denote the number of removable i-nodes of λ and λ ▽i the partition obtained by removing all the removable i-nodes from λ. Example. Suppose p = 3 and λ = (4, 3, 1 2 ). The residues of the removable nodes of λ are as follows: 0 1 0 .
Representations of the alternating group which remain irreducible in characteristic p 9 Hence we have ǫ 0 S λ = 2, e (max) 0 S λ S (3 2 ,1) , The corresponding result for the simple modules D λ is more complicated. If λ is a prestricted partition, define the i-signature of λ to be the list of + and − signs obtained by examining the addable and removable i-nodes of λ from top to bottom, writing a + for each addable i-node and a − for each removable i-node. Construct the reduced i-signature from the i-signature by successively deleting adjacent pairs −+. The removable i-nodes corresponding to the − signs in the reduced i-signature are the normal i-nodes of λ. Let nor i (λ) denote the number of normal i-nodes of λ, and λ i the partition obtained by removing all the normal i-nodes.

Lemma 4.2. Suppose λ is a p-restricted partition and i
Example. Suppose p = 3 and λ = (6, 5, 3, 2, 1 3 ). The Young diagram of λ, with the residues of addable and removable nodes marked, is as follows. We see that the 0-signature of λ is + − − + −. So the reduced 0-signature is + − −, and the normal 0-nodes are (2, 5) and (7, 1). Hence ǫ 0 D λ = 2, and e (2) 0 D λ D (6,4,3,2,1 2 ) . These results will be very helpful in finding lower bounds for the number of composition factors of a Specht module S λ : since e (r) i is an exact functor, we have ǫ i T ǫ i S λ for any composition factor T of S λ ; furthermore, the number of composition factors T (with multiplicity) for which equality holds must equal the composition length of S λ ▽i . Hence we have the following result. Analogously to the restriction functors e i , one can define induction functors f i : FS n -mod → FS n+1 -mod and obtain similar combinatorial results. In particular, we have the following analogue of Lemma 4.3, which will occasionally be helpful.
Lemma 4.4. Suppose λ is a partition and i ∈ Z/pZ, and let λ △i denote the partition obtained from λ by adding all the addable i-nodes. Then the composition length of S λ is at least the composition length of S λ △i .

Restriction and the Mullineux map
It follows fairly easily from the definition of the e i that they behave well with respect to the functor − ⊗ sgn. In fact, the following lemma is the basis for Kleshchev's combinatorial algorithm for computing the Mullineux map.
As a consequence of this lemma and Lemma 4.2, we have nor i (λ) = nor −i (m p (λ)) for any p-restricted λ.

Restriction and restrictisation
In this section we show how to extract more information from the preceding discussion on restriction functors using Theorem 2.1. Given i ∈ Z/pZ, we write i (λ) if e rem i (λ) i D λ rest = 0; that is, if there is a power of e i which kills D λ rest but not S λ . By Lemma 4.2, this is just the combinatorial condition nor i (λ rest ) < rem i (λ). More generally, given residues i 1 , . . . , i r , we which is the same as saying that for some 1 l r nor i l ((λ rest ) i 1 ...i l−1 ) < rem i l (λ ▽i 1 ...i l−1 ). Now the following is an immediate consequence of Theorem 2.1 and Lemma 4.3.
Lemma 4.6. Suppose λ is a partition and that i (λ) for some i ∈ Z/pZ. Then S λ is reducible.
We now prove a more complicated version of this lemma for self-conjugate partitions. We assume for the moment that p is odd, and write p = 2h + 1. Given 0 i h and a self-conjugate partition λ, we define λ ▽±i to be the partition obtained by repeatedly removing all removable nodes of residue i and −i; then λ ▽±i is self-conjugate, and Now we have the following, which will be our main inductive tool for proving Theorem 3.3.
Proposition 4.7. Suppose p = 2h + 1 is an odd prime and λ is a self-conjugate partition, and that one of the following occurs.
1. There is some 0 i h such that S λ ▽±i has at least three composition factors.
2. There is some 0 i < h such that S λ ▽±i is irreducible.
3. There is some 0 i h such that S λ ▽±i is reducible and either i (λ) or −i (λ).
Then the composition length of S λ is not 2.

Proof.
1. This follows by applying Lemma 4.3 one, two or three times.
2. Assume i > 0; a similar but simpler argument applies in the case i = 0. Since i < h, there is no i-node adjacent to a (−i)-node, and so rem i (λ ▽−i ) = rem i (λ). So If S λ has exactly two composition factors, then by Lemma 2.3 these are of the form S and S ⊗ sgn. Since S λ ▽±i is irreducible and the restriction functors are exact, exactly one of e (S ⊗ sgn) must be non-zero. But by Lemma 4.5 and the fact that (since i < h) e i and e −i commute, we have 3. Suppose i (λ). Then e rem i (λ) i D λ rest = 0, so that by Lemma 4.3 S λ ▽i has strictly fewer composition factors than S λ . Hence S λ ▽±i has strictly fewer composition factors than S λ . Since S λ ▽±i is reducible, this means that S λ has at least three composition factors.
The case where −i (λ) is proved in the same way.
4. If S λ has exactly two composition factors, then by Lemma 2.3 and Theorem 2.1 these are D λ rest and D λ rest ⊗ sgn. If h(−h)h (λ) and (−h)h(−h) (λ), then by Lemma 4.2 we have and the second equality together with Lemma 4.5 and the fact that λ is self-conjugate gives so that by exactness In order to use Proposition 4.7, it will be very helpful to be able to test the condition nor i (λ rest ) < rem i (λ) for a given partition λ without having to construct λ rest . We now prove a new result which will enable us to do this.
Recall the definition of the ramps used in the definition of λ rest , and write rmp l (λ) for the number of nodes of λ in ramp l, rmp + l (λ) for the number of addable nodes of λ in ramp l, and rmp − l (λ) for the number of removable nodes of λ in ramp l, setting all of these numbers to be zero when l < 0.
Lemma 4.8. For any λ and any l, we have Hence if λ and µ are partitions with λ rest = µ rest , then rmp + Proof. We assume p > 2; a modification to the argument is required when p = 2, and since this case is not relevant to the main results in this paper, we feel content to leave this case to the reader. We also assume l > 0, with the case l = 0 being trivial.
Suppose (r, c) is a node in ramp l. Assuming first that r, c > 1, we have nodes (r, c − 1), (r − 1, c) and (r − 1, c − 1) in ramps l − 1, l − p + 1 and l − p respectively. By checking the possible cases for which of these four nodes are nodes of λ, we easily find that the formula holds when restricted just to these four nodes. If r = 1, then the same argument applies looking at just the two nodes (1, c) and (1, c − 1), and a similar statement applies when c = 1. Summing over all (r, c) in ramp l gives the result.
The second sentence of the lemma now follows, since if λ rest = µ rest , then rmp l (λ) = rmp l (µ) for all µ. Furthermore, if equality occurs then

Remarks.
1. Of course, the inequality nor i (λ rest ) rem i (λ) follows from Theorem 2.1 and Lemmas 4.1 and 4.2. But it naturally comes out of the argument below, and it is interesting to have a purely combinatorial proof.
2. The final statement of the proposition (together with Lemma 4.2) shows that given a partition λ and residues i 1 , . . . , i r , we have i 1 ...i r (λ) if and only if for some 1 l r This will be very useful in the proof of our main theorem.
So λ satisfies the hypothesis in Proposition 4.9. We have for all a, and this is non-negative for a 1 and non-positive for a 2. The 0-signature of µ, with the signs labelled according to the ramps containing the corresponding nodes, is So µ has three normal nodes, in ramps 6, 9 and 18. Since the removable 0-nodes of λ lie in ramps 6, 9 and 18, we have (λ rest ) 0 = (λ ▽0 ) rest .

The abacus
In this section we describe the abacus notation for partitions, which we shall use in the proof of our main theorem. We also discuss p-quotient-separated partitions, which label Specht modules whose composition factors are well understood.

The abacus display for a partition
We assume throughout this section that p < ∞, and fix an abacus with p infinite vertical runners, which we number 0, . . . , p − 1 from left to right. We mark positions . . . , −2, −1, 0, 1, 2, . . . on the runners, reading from left to right along successive rows, with runner i containing the positions congruent to i modulo p. We say that position m is later than position l (or position l is earlier than position m) if m > l. For example, if p = 5 then the positions are marked as follows. Now given a partition λ, we place a bead on the abacus in position λ i − i for each i ∈ N. The resulting configuration is called the abacus display for λ. We call a position occupied if there is a bead at that position, and vacant otherwise; we may also say that there is a space in position l if that position is vacant.
Abacus combinatorics are well-established, so we quote some facts without justification. Taking an abacus display for λ and sliding all the beads up their runners as far as possible, we obtain an abacus display for the p-core of λ. The p-weight of λ is the number of pairs l < m such that l ≡ m (mod p), position l is vacant and position m is occupied. Now we consider p-quotients. Given an abacus display for a partition λ, we define a partition λ (i) by examining runner i in isolation as a 1-runner abacus display and reading off the corresponding partition. In other words, λ (i) j equals the number of vacant positions above the jth lowest bead on runner i. The p-tuple λ (0) , . . . , λ (p−1) is called the p-quotient of λ, and from the last paragraph the sum |λ (0) | + · · · + |λ (p−1) | equals the p-weight of λ.
Since we shall be dealing with self-conjugate partitions, we record the following fact: if λ is a self-conjugate partition, then for any l there is a bead in position l in the abacus display for λ if and only if there is a space in position −l − 1. Moreover, the p-quotient λ (0) , . . . , λ (p−1) satisfies λ (i) = λ (p−1−i) ′ for each i.

Addable and removable nodes and ramps
Suppose λ is a partition and i ∈ Z/pZ. The removable i-nodes of λ correspond to the occupied positions l on runner i in the abacus display for λ for which position l − 1 is vacant. Removing such a node corresponds to moving the bead from position l to position l − 1. Similarly, addable i-nodes correspond to occupied positions l − 1 on the abacus display with position l vacant, and adding such a node corresponds to moving the bead from position l − 1 to position l. The order of the addable and removable i-nodes from top to bottom of the Young diagram for λ is the same as the order of the corresponding positions on runner i from bottom to top.
We record a small lemma that we shall use later.
Lemma 5.1. Suppose λ is a partition, and that for some i ∈ Z/pZ λ has at least one addable i-node and at least one removable i-node. Then Proof. If λ (i−1) = λ (i) = ∅, then there are integers a, b such that in the abacus display for λ: • position kp + i − 1 is occupied if and only if k a; • position lp + i is occupied if and only if l b.
If a b, then λ has no removable i-nodes, while if a b, then λ has no addable i-nodes; contradiction. Now we use the description of addable and removable nodes on the abacus to enable us to apply Proposition 4.9 using the abacus.
Proof. From the above discussion, the occupied position lp + i with a space in position lp + i − 1 corresponds to a removable i-node (c, λ c ), while the vacant position kp + i with a bead in position kp + i − 1 corresponds to an addable i-node (d, λ d + 1), with c < d. In fact, the definition of the abacus display means that The definition of the abacus display gives λ c − c = lp + i and λ d − d = kp + i − 1, and hence the removable node (c, λ c ) lies in ramp (c + l − 1)p + i, and the addable node (d, λ d + 1) lies in ramp (d + k − 1)p + i. The difference between these ramp numbers is which is strictly positive by assumption. Hence λ has a removable i-node and an addable i-node in a later ladder, and so i (λ) by Proposition 4.9.

Rouquier partitions
We now describe a class of partitions for which the corresponding decomposition numbers are very well understood. In order to do this, it will be helpful to impose a new ordering on the runners of the abacus; this approach was first taken by Richards [Ri].
summing over all p-restricted partitions µ with the same p-core and p-weight as λ.

p-quotient-separated partitions
In this section we generalise the notion of a Rouquier partition, to define a class of Specht modules whose composition length can be deduced from Lemmas 4.3 and 4.4 and Theorem 5.3. This material will be familiar to many experts, although it does not seem to have appeared in this form before.
Suppose λ is a partition and construct the abacus display for λ. Say that λ is p-quotientseparated if the following property holds: there do not exist runners i j such that the first space on runner i is earlier than the last bead on runner j and the first space on runner j is earlier than the last bead on runner i. This terminology was first used by James and Mathas [JM1] in the case p = 2.
We assemble some facts about p-quotient-separated partitions. Given a partition λ, let q i = q i (λ) for i ∈ Z/pZ, and π = π λ . Proposition 5.4. Suppose λ is a partition, and i, j ∈ Z/pZ.
1. If λ is p-quotient-separated and q i < q j , then the first space on runner j is later than the last bead on runner i.
2. If λ is p-quotient-separated and has at least two addable i-nodes, then q i−1 > q i .
4. λ is Rouquier if and only if every partition with the same p-core and p-weight as λ is p-quotientseparated.

Proof.
1. By construction, the first space on runner i is in position q i − pλ (i) ′ 1 , and the last bead on runner j is in position q j − p + pλ ( j) 1 . So either the first space on runner i is earlier than the last bead on runner j (in which case the first space on runner j is later than the last bead on runner i by the p-quotient-separated property), or q j − q i < p and λ (i) = λ ( j) = ∅. But in this case the first space on runner j is as position q j and the last bead on runner i is in position q i − p, so the required result still holds.
2. Since λ has at least two addable nodes, there are integers k < l such that positions kp + i − 1 and lp + i − 1 are occupied while positions kp + i and lp + i are vacant. Then the first space on runner i is earlier than the last bead on runner i − 1, so q i−1 > q i by part (1).
3. We assume λ △i is not p-quotient-separated, and show that λ is not either. Since λ ▽±i is not p-quotient-separated, there are k l such that in the abacus display for λ △i the first space on runner k occurs before the last bead on runner l, and vice versa. Clearly if {k, l} ∩ {i − 1, i} = ∅, then λ is not p-quotient-separated, so assume otherwise.
Suppose k = i − 1 and l = i, and that in the abacus display for λ △i the first space on runner i is in position ap + i, and the last bead on runner i − 1 in position bp + i − 1; then by assumption a < b. But now the definition of λ △i means that in the abacus display for λ positions ap + i − 1 and ap + i are both vacant, and positions bp + i − 1 and bp + i are both occupied. So the first space on runner i − 1 is earlier than the last bead on runner i and vice versa, so λ is not p-quotient-separated.
Next consider the case where k = i − 1 and l i. In the abacus display for λ △i the first space on runner l is earlier than the last bead on runner i − 1, which is in position bp + i − 1, say. In the abacus display for λ, positions bp + i − 1 and bp + i are both occupied, so the first space on runner l is earlier than the last bead on runner i − 1 and the last bead on runner i. Now suppose that in the abacus display for λ △i the first space on runner i − 1 is in position ap + i − 1; by assumption this is earlier than the last bead on runner l. In the abacus display for λ at least one of the positions ap + i − 1 and ap + i is vacant, so the last bead on runner l is later than either the first space on runner i − 1 or the first space on runner i. Either way, λ fails to be p-quotient-separated.
The case where k = i and l i − 1 works in a very similar way.
4. First we show that if λ is Rouquier, then λ is p-quotient-separated. Take i, j ∈ Z/pZ, and suppose without loss that q i < q j . As in the proof of (1), the first space on runner j is in position q j − pλ ( j) ′ 1 , and the last bead on runner i is in position where w is the p-weight of λ. Since q j − q i > (w − 1)p, the last bead on runner i is earlier than the first space on runner j. So λ is p-quotient-separated. Since every partition with the same p-core and p-weight as λ is also Rouquier, it follows that every such partition is also p-quotient-separated.
If λ is not Rouquier, take i, j such that q i < q j and q j − q i < (w − 1)p. Then from the last paragraph is it clear how to construct a partition with the same p-core and p-weight as λ which is not p-quotient-separated.
Our aim is to show that the composition length of a Specht module labelled by a p-quotientseparated partition is the same as that of a Specht module labelled by a Rouquier partition with the same ordered p-quotient. The following proposition gives us the inductive step.
Proposition 5.5. Suppose λ is a p-quotient-separated partition and i ∈ Z/pZ, with q i−1 > q i . Then: 1. λ has at least one addable i-node and has no removable i-nodes; 2. λ △i is p-quotient-separated with the same ordered p-quotient as λ; 3. S λ and S λ △i have the same composition length.

Proof.
1. λ cannot have a removable i-node, since then in the abacus display for λ there would be a space on runner i − 1 with a bead in a later position on runner i, contradicting Proposition 5.4(1). If λ does not have an addable i-node either, then for every integer k, position kp + i − 1 is occupied if and only if position kp + i is occupied. Clearly then the same is true in the abacus display for the p-core of λ, which gives q i−1 = q i − 1, a contradiction.
2. The fact that λ △i is p-quotient-separated has already been proved in Proposition 5.4(3), so we just check that λ and λ △i have the same ordered p-quotient. Since λ has no removable i-nodes, the abacus display for λ △i is obtained from that for λ by 'swapping runners i − 1 and i'; that is, position kp + i is occupied in the abacus display for λ △i if and only if position kp + i − 1 is occupied in the abacus display for λ, and vice versa. Hence we have and the latter statement gives π λ △i = (i − 1, i) • π λ . Hence λ △i has the same ordered p-quotient as λ.
3. Since λ has no removable i-nodes, this follows from Lemmas 4.3 and 4.4. Now we can prove the main result of this section.
Proposition 5.6. Suppose λ is a p-quotient-separated partition. Then there is a Rouquier partition µ with the same ordered p-quotient as λ, and S λ and S µ have the same composition length.

Proof.
Define We shall proceed by induction on and for fixed d 1 , . . . , d p−1 we use downwards induction on N := i q 2 i . Since i q i is constant, N is bounded for fixed (d 1 , . . . , d p−1 ); so it suffices to show that we can replace λ with a partition which yields a smaller value of M, or the same values of d 1 , . . . , d p−1 ' and a larger value of N.
Suppose that for some i, k ∈ Z/pZ we have q i−1 > q k > q i , and consider the partition λ △i . By Proposition 5.5, λ △i has the same ordered p-quotient as λ, and the Specht modules S λ and S λ △i have the same composition length. So we can replace λ with λ △i . From the proof of Proposition 5.5 we see that making this replacement does not change any of the integers d j and strictly increases N.
Alternatively, assume there are no such i, k. The only way this can happen is if there is some l ∈ Z/pZ such that q l > q l+1 > · · · > q l−1 . But now if λ is not Rouquier, take a j 1 such that d j < w − 1, and let i = π(j − 1). Then π(j) = i − 1, and again we consider the partition λ △i . Again, we can replace λ with λ △i by Proposition 5.5, and now the replacement increases d j by 1 while fixing all the other d k , and in particular decreases M.

JM-partitions and R-partitions on the abacus
Since we shall be using the abacus extensively in the proof of our main theorem, it will be useful to have characterisations of JM-partitions and R-partitions in terms of their abacus displays. We take these from [F2, F3]. • λ is p-quotient-separated.
We observe an important consequence of Proposition 5.8 for the proof of our main theorem. Proposition 5.10. Suppose p = 2h + 1 is an odd prime and 0 i < h, and that λ is a self-conjugate partition such that λ ▽±i is an R-partition of type I. Then λ is an R-partition of type I.
Proof. In general, the abacus display for λ is obtained from the abacus display for λ ▽±i by moving beads from runner i − 1 to the adjacent positions on runner i, and moving beads from runner −i − 1 to the adjacent positions on runner −i. Since λ ▽±i is an R-partition of type I, Proposition 5.8 shows that the only possible beads that can be moved are in positions i − 1 and −i − 1. But even if beads are moved from these positions to positions i and −i, then the resulting abacus display still satisfies the conditions in Proposition 5.8, so λ is an R-partition of type I.

Proof of Theorem 3.3 for p-quotient-separated partitions
We are now ready to prove Theorem 3.3 in the case where λ is p-quotient-separated. Throughout this section we assume that p is an odd prime, and write p = 2h + 1.
First we make an observation about the ordered p-quotient of a self-conjugate partition. Suppose λ is self-conjugate; then the p-core of λ is also self-conjugate, and together with the last paragraph of Section 5.1 this implies that q i + q p−1−i = p − 1 for every i. Hence the permutation π = π λ satisfies π(p − 1 − i) = p − 1 − π(i) for every i. As a consequence, the ordered p-quotient λ [0] , . . . , λ [p−1] has the same symmetry as the p-quotient, namely that λ [p−1−i] = λ [i] ′ for every i. Proposition 5.11. Suppose λ is a self-conjugate p-quotient-separated partition and that S λ has exactly two composition factors. Then λ is an R-partition of type II, and if p = 3 then λ has 3-weight 1.

Proof.
To begin with, we assume λ is a Rouquier partition, and use Theorem 5.3. Let It is a well-known fact that if α and β are both non-empty partitions, then there are at least two partitions γ for which c γ αβ > 0. For each pair of partitions α, β, we fix an arbitrary choice of a partition α • β such that c α•β αβ > 0, and if α ∅ β, then we fix an arbitrary choice of a partition α * β α • β such that c α * β αβ > 0. Suppose first that there is some 1 i h − 1 such that λ [i] ∅. Consider the three partitions µ 1 , µ 2 , µ 3 with the same p-core as λ, and with ordered p-quotients Since these ordered p-quotients are distinct, the partitions µ 1 , µ 2 , µ 3 are distinct. By Theorem 5.3(1), µ 1 , µ 2 , µ 3 are p-restricted, with d λµ k > 0, and hence (since a µµ = 1 for any µ) [S λ : D µ k ] > 0 for k = 1, 2, 3. So S λ has at least three composition factors; contradiction.
So we have λ [i] = ∅ for 1 i h − 1, and hence also for h + 1 i p − 2. Now suppose that λ [h] = ∅. Then the only p-restricted partition µ for which d λµ > 0 is the partition with the same p-core as λ and with ordered p-quotient for which we have d λµ = 1. Since λ [0] = λ [p−1] ′ , the sum ν a µν is a perfect square (namely, the square of the composition length of the Weyl module ∆ λ [0] ). So the composition length of S λ is a perfect square; contradiction.
So we have λ [h] ∅. Now we look at the case p = 3, and observe that in this case λ [0] = λ [2] = ∅; for if not, then the partitions µ with the same p-core as λ and with ordered p-quotients all give d λµ > 0 and hence [S λ : D µ ] > 0.
Next we suppose (with p arbitrary again) that |λ [h] | > 1. In this case, we can find (thanks to another well-known fact about Littlewood-Richardson coefficients) three pairs of partitions α, β such that c λ [h] αβ > 0. For each such pair, we construct the partition µ with the same p-core as λ and with ordered p-quotient (where the α ′ , β occur in positions h − 1, h), and we have d λµ > 0 and hence [S λ : D µ ] > 0. So again S λ has at least three composition factors, a contradiction.
So we have λ [h] = (1). This completes the analysis in the case p = 3. In the case p 5, the only p-restricted µ for which d λµ > 0 are those with ordered p-quotient where the (1) occurs either in position h − 1 or in position h. For each of these two partitions we have d λµ = 1, while ν a µν is again equal to the square of the composition length of ∆ λ [0] . So ∆ λ [0] must be simple, i.e. λ [0] is p-restricted JM-partition. So by Proposition 5.9 λ is an R-partition of type II.

Homomorphisms between Specht modules
In this section, we review some results on homomorphisms between Specht modules and prove a result that we shall need later. This material is discussed at length elsewhere, so in the interests of brevity we specialise as much as possible.

Tableau homomorphisms
We begin with some combinatorics. Throughout this section let λ and µ be fixed compositions of n. A λ-tableau of type µ is a function T from the Young diagram of λ to N with the property that exactly µ i nodes are mapped to i, for each i. We write T r,c for the image of the node (r, c) under T, and we illustrate T by drawing the Young diagram and filling the (r, c)-box with T r,c , for each (r, c). A tableau is row-standard if its entries weakly increase from left to right along the rows, and semistandard if it is row-standard and its entries strictly increase down the columns.
Recall that M λ denotes the Young permutation module associated with λ. For each rowstandard λ-tableau T of type µ, there is an FS n -homomorphism Θ T : M λ → M µ defined in [J2, §13]. If λ is a partition, then the Specht module S λ is a submodule of M λ , and the restriction of Θ T to S λ is denotedΘ T . If T is semistandard, we refer toΘ T as a semistandard homomorphism. Now we have the following result.

If λ is a partition, the set
Θ T T a semistandard λ-tableau of type µ is an F-basis for the space of all FS n -homomorphisms S λ → M µ which can be extended to M λ .

If λ is a partition and p 3, then every FS n -homomorphism S λ → M µ can be extended to M λ .
In view of this theorem, a natural way to construct a homomorphism S λ → S µ when p 3 is to find a linear combination of semistandard homomorphisms S λ → M µ whose image lies in S µ ; the latter condition can be checked using James's Kernel Intersection Theorem [J2, Corollary 17.18] and the author's results from [F4], so that we now have a reasonably fast algorithm [F4] for computing Hom FS n (S λ , S µ ). Even when p = 2, this method can often be used to construct homomorphisms between Specht modules (including the homomorphisms described in this paper), though it will not in general find all homomorphisms.
In this section we want to show the existence of a non-zero homomorphism S λ → S µ in a certain case, and we construct this as the composition of two known homomorphisms between Specht modules. But we have some work to do in showing that this composition is non-zero. In order to do this, we need to discuss dominance. If T is a row-standard λ-tableau of type µ, let T[l, r] denote the total number of entries less than or equal to l in rows 1, . . . , r of T. If U is another row-standard λ-tableau of type µ, we say that T

dominates U (and write T U) if T[l, r] U[l, r] for all l, r.
Example. The dominance order on the set of row-standard (3, 2)-tableaux of type (2 2 , 1) may be represented by the following Hasse diagram. The following lemma will be very useful.

Lemma 6.2. Suppose T is a row-standard λ-tableau, and writeΘ T as a linear combination S t SΘS of semistandard homomorphisms. Then S T for each S with t S 0.
Proof. This follows directly from the algorithm given in [F4, §5.2] for semistandardisingΘ T (and in fact is quite easy to see from [J2, §13] where the homomorphismsΘ T are introduced).
Example. Continuing from the last example, if we let T = 1 2 2 1 3 , then by Lemma 6.2Θ T should be a linear combination of homomorphismsΘ S with S T. From the diagram above we see that the only such S is S = 1 1 2 2 3 . And indeed (as can easily be shown using the results below)Θ T = −Θ S . Now we describe two particular constructions of homomorphisms that we shall use. Throughout this section we assume that p is finite.

One-node Carter-Payne homomorphisms
Suppose λ is a partition of n with a removable node (a, b) and an addable node (c, d) of the same residue, with c > a. Let µ be the partition obtained by removing (a, b) and adding (c, d). Then there is a non-zero FS n -homomorphism S λ → S µ ; this is a special case of the Carter-Payne Theorem [CP], and an explicit formula for this homomorphism may be found in the paper of Lyle [L2]. For simplicity, we concentrate on a special case.

Restrictisation homomorphisms
In this section we consider the homomorphisms arising in the following theorem, which may be regarded as a homomorphism-space analogue of Theorem 2.1.

Theorem 6.4 [FLM, Theorem 1.5]. Suppose λ is a partition of n. Then
A non-zero homomorphism S λ → S λ rest is constructed in [FLM] as a tableau homomorphism Θ T , but for a tableau T which is not necessarily semistandard. This leads to an additional problem (which creates a large part of the work in [FLM]) of showing thatΘ T is non-zero. In order to prove our result on composition of homomorphisms, we shall find the least dominant tableau occurring whenΘ T is expressed as a linear combination of semistandard homomorphisms; we note that this gives a new proof thatΘ T is non-zero.
First we must describe the tableau T. In fact, there is a range of possibilities for T, yielding homomorphismsΘ T which agree up to sign. These tableaux are called magic tableaux in [FLM], and we define them here using one of the recursive characterisations given in [FLM], which we re-phrase for our own purposes.
Let λ be a partition of n. Write full(λ) for the number of 'full' ramps in λ (i.e. ramps in which every node is a node of λ); then it is easy to see that full(λ) = λ rest 1 . If we look at the first ramp which is not full, we can find a node of this ramp, in row m say, which is not a node of λ. Then we have λ m + (m − 1)(p − 1) = full(λ), while for any 1 l m we have λ l + (l − 1)(p − 1) full(λ). Call such a value of m a nice value for λ. Having chosen a nice value m, we define a partition λ • by It is easy to see that (λ • ) rest = (λ rest 2 , λ rest 3 , . . . ); in particular, full(λ • ) = λ rest 2 . Now given any λ • -tableau U, define a λ-tableau U + by Now we can define magic tableaux recursively: the unique tableau for the empty partition is magic, and if λ ∅, a magic tableau for λ is any tableau of the form U + , where U is a magic λ • -tableau for some nice value m.

Theorem 6.5 [FLM, Theorem 2.2 & Lemma 4.2]. Suppose λ is a partition, and T is a magic λ-
tableau. ThenΘ T defines a non-zero homomorphism from S λ to S λ rest . If U is any other magic λ-tableau, thenΘ U = ±Θ T .
Now we consider expressing a 'magic homomorphism' as a linear combination of semistandard homomorphisms. Given a partition λ, define a λ-tableau Re(λ) as follows: let Re(λ) x,y equal x plus the number of nodes below x in the same ramp which are not nodes of λ. Informally, we construct Re(λ) by filling each box with the number of the row that box moves to when we construct λ rest from λ. Hence Re(λ) has type λ rest .
Our aim in this section is to prove the following statement.
Proposition 6.6. Suppose λ is a partition and T a magic λ-tableau, and writeΘ T as a linear combination S a SΘS of semistandard homomorphisms. Then a Re(λ) = ±1, and for any S with a S 0 we have S Re(λ).
We begin with a lemma about Re(λ).

Lemma 6.7. Suppose λ is a partition, and (a, b) and (c, d) are nodes of λ with
In particular, Re(λ) is semistandard.
Proof. For this proof, say that a node is missing if it not a node of λ.
1. The conditions imply that b d. For every node (x, y) in the same ramp as (a, b) which is not a node of λ, there is a node (x, y + d − b) in the same ramp as (a, d) which is not a node of λ. Hence the number of missing nodes below (a, d) in the same ramp is at least the number of missing nodes below (a, b) in the same ramp.
2. Arguing as in the previous case, the number of missing nodes below row a and in the same ramp as (a, b) is at least the number of missing nodes below row a and in the same ramp as (c, d). Of the latter nodes, at most c − a − 1 lie between rows a and c; so the number of missing nodes below row c and in the same ramp as (c, d) is strictly greater than the number of missing nodes below row a and in the same ramp as (a, b) plus a − c, and this gives the result.
Now we describe the relations in [FM,F4] used to 'semistandardise' homomorphisms. For these and subsequent results, we need some notation for multisets of positive integers, which we collect here.
• Given a multiset X, let X i denote the multiplicity of i as an element of X.
• Given two multisets X, Y, let X ⊔ Y denote the multiset with (X ⊔ Y) i = X i + Y i for all i.
• Given a multiset X, let X + 1 denote the multiset obtained from X by increasing each element by 1.
• Given any l, n ∈ N let {l} n denote the multiset with n elements all equal to l.
• Given a tableau T, let T i denote the multiset of entries in the ith row of T.
With this notation in place, we can state a useful result for manipulating tableau homomorphisms. This is a combination of [F4,Theorem 3.1] and [FM,Lemma 4].
Proposition 6.8. Suppose λ is a partition, A is a row-standard λ-tableau and 1 h < k. Suppose R, S, T are multisets of positive integers with R ⊔ S ⊔ T = A h ⊔ A k and |S| > λ h . Let S be the set of all pairs (U, V) of multisets such that S = U ⊔ V and |R| .
Remark. It is shown in [F4, §5.2] that the relations obtained from Proposition 6.8 are sufficient to express a tableau homomorphismΘ T as a linear combination of semistandard homomorphisms; in fact, only the case k = h + 1 is required. Since the semistandard homomorphisms are linearly independent, this means that any linear relation between tableau homomorphisms is a consequence of the relations obtained from Proposition 6.8.
Before we proceed, we note a simple corollary. Now we show that the operation U → U + preserves relations between tableau homomorphisms.

Corollary 6.9. Suppose λ is a partition, A is a row-standard λ-tableau and
Proposition 6.10. Suppose λ is a partition and m is a nice value for λ, and let λ • be as above. Suppose U is a set of row-standard λ • -tableaux of type (λ • ) rest , and that coefficients (c U ) U∈U are chosen such that U∈U c UΘU = 0. Then U∈U c UΘU + = 0.
Proof. From the remark following Proposition 6.8, it suffices to consider only relations of the form given in that proposition. So suppose A, h, k, R, S, T are as in Proposition 6.8 (with λ • in place of λ). Then we have and we want to show We consider three cases.

Case 1: k < m.
In this case we have ( ThenR,Ŝ,T satisfy the hypotheses of Proposition 6.8 (with A + in place of A), so (with the obvious definition ofŜ) we have Any term withV 1 > 0 can be neglected, since p−1+v p−1 ≡ 0 (mod p) for any 0 < v < p. When V 1 = 0, we haveÛ 1 = p − 1, and so Case 2: h < m k.

In this case (A
NowR,Ŝ,T satisfy the hypotheses of Proposition 6.8 (with A + in place of A, and k + 1 in place of k); proceeding as in Case 1, we have For any pair (Û,V) withV 1 > 0 we haveΘ A + [Û,V] = 0 by Corollary 6.9 (with l = 1 and with m and k + 1 in place of h and k). So now we need only consider pairs (U, V) witĥ U 1 = p − 1 andV 1 = 0, and we can proceed as in Case 1.
Next we introduce a result which gives relations between tableau homomorphisms which allow us to move all the 1s in a tableau up to the top row. We use the following variation on Proposition 6.8; this was actually proved before Proposition 6.8, although it follows from the latter fairly easily by induction.
Proposition 6.11 [FM,Lemma 7]. Suppose λ is a partition, B is a row-standard λ-tableau and h, r ∈ N. Let V denote the set of submultisets V of B h such that r V and |V| = B h+1 Using this, we can prove the following.
Proposition 6.12. Suppose λ is a partition and m is a nice value for λ, and define λ • as above.

Proof.
1. Let R = Re(λ • ). We use Proposition 6.11 to re-writeΘ R + by moving all the 1s up to the top row. We apply Proposition 6.11 m − 1 times, each time with r = 1, taking h = m − 1, m − 2, . . . , 1 in turn. At a given step, we move all the 1s from row h + 1 to row h, and move a multiset of entries greater than 1 from row h to row h + 1. Since R is semistandard, the entries in row h of R are all at least h; hence the entries not equal to 1 in row h of R + are all at least h + 1. Furthermore, there are at least full(λ so we see that the first full(λ) − h(p − 1) entries in row h of R + are equal to h + 1. So each time we apply Proposition 6.11, one of the terms we obtain involves moving only entries equal to h + 1 down to row h + 1. Taking this term at every stage we obtain the tableau Re(λ), and the coefficient ofΘ Re(λ) obtained is ±1; indeed, the binomial coefficients in Proposition 6.11 are always trivial, since all the entries in row h + 1 of R + (other than the 1s) are strictly greater than h + 1.
Any other term we obtain from our repeated applications of Proposition 6.11 involves moving all the 1s up to row 1, and moving full(λ) − h(p − 1) entries greater than or equal to h + 1 down from row h to row h + 1 for each 1 h < m, with a strict inequality at some point. Hence the resulting tableau will strictly dominate Re(λ).
2. This case is similar to the previous one: when we apply Proposition 6.11 repeatedly, we move all the 1s in T up to row 1, and move full(λ) − h(p − 1) entries greater than or equal to h + 1 down from row h to row h + 1. Since T ⊲ Re(λ • ) + , any tableau resulting from this process will strictly dominate Re(λ).
Applying Proposition 6.11 twice to move the 1s up to the top row, we get the following possibilities.
Proof of Proposition 6.6. We use induction on |λ|, with the case λ = ∅ being trivial. Suppose λ ∅, and m be the nice value chosen in the construction of T. Then T = U + for a magic λ • -tableau U, and by induction we can assume that when we writeΘ U as a linear combination V u VΘV of semistandard homomorphisms, we have u Re(λ • ) = ±1 while u V = 0 for any V Re(λ • ).
By Proposition 6.10 we haveΘ by Proposition 6.12 this equals ±Θ Re(λ) plus a linear combination of homomorphismsΘ S for S ⊲ Re(λ). By Lemma 6.2 eachΘ S can be written as a linear combination of semistandard homomorphismsΘ R for R S ⊲ T, and the result follows.

Composition of homomorphisms
Our aim in this section is to show that the composition of the homomorphisms from the two previous sections is non-zero, given a certain additional condition. Specifically, we prove the following result.
Proposition 6.13. Suppose λ is a partition of n with a removable node (a, b) and an addable node (c, d) of the same residue with c > a and (p − 1)c + d (p − 1)a + b, and that λ has no removable nodes in rows a + 1, . . . , c − 1. Let µ be the partition obtained from λ by removing (a, b) and adding (c, d). Then there is a non-zero FS n -homomorphism S λ → S µ rest .
In order to prove Proposition 6.13, we need to describe how to compose tableau homomorphisms. Recall that if S is a tableau, then S j denotes the multiset of entries in row j of S, and in particular S j i denotes the number of entries equal to i in row j of S. If x 1 , x 2 , . . . are non-negative integers with finite sum x, we write (x 1 , x 2 , . . . )! for the multinomial coefficient (x 1 + x 2 + · · · )! x 1 !x 2 ! . . .
. Proposition 6.14 [DF, Proposition 4.7]. Suppose λ, µ, ν are compositions of n, S is a λ-tableau of type µ and T is a µ-tableau of type ν. Let X be the set of all collections X = (X ij ) i, j 1 of multisets such that For X ∈ X, let U X denote the row-standard λ-tableau with (U X ) j = i 1 X ij . Then Our aim is to use Proposition 6.14 to show that the composition of the homomorphisms S λ → S µ → S µ rest from the last two sections is non-zero. Rather than attempting to give an explicit expression for this composition, we use Propositions 6.3 and 6.6 to find the least dominant tableau occurring when the composition is expressed in terms of semistandard homomorphisms.
Given λ and µ as above, let V be the λ-tableau obtained from Re(µ) by moving the (c, d)-entry up to position (a, b). Lemma 6.15. Suppose λ and µ are as above, and T is a semistandard µ-tableau of type µ rest with T ⊲ Re(µ). If W is a row-standard tableau obtained from T by moving an entry from row c up to row a, then W ⊳ V.
Proof. Suppose to the contrary that V ⊲ W. Since V agrees with Re(µ) and W agrees with T in rows 1, . . . , a − 1, and we have V ⊲ W while Re(µ) ⊳ T, all four tableaux must agree on rows 1, . . . , a − 1. Similarly, all four tableaux agree on rows c + 1, c + 2, . . . . V is obtained from Re(µ) by moving an entry t = Re(µ) c,d from row c up to row a. By Lemma 6.7, t is the largest entry in rows a, . . . , c of Re(µ). By the previous paragraph (and since T and Re(µ) have the same type) the largest entry in rows a, . . . , c of T is also t. So W is obtained from T by moving an entry less than or equal to t from row c up to row a. Since T ⊲ Re(µ), this gives W ⊲ V, a contradiction.
Proposition 6.16. Suppose a < r c and T is a semistandard µ-tableau of type µ rest with T Re(µ).

If r = c and T
Proof. Taking S = CP λ µ (r) and ν = µ rest , Proposition 6.14 simplifies considerably in our situation. It says that Θ T • Θ S is a linear combination of row-standard tableaux U obtained from T by moving an entry from row r to row a and moving an entry from row j + 1 to row j for each r j < c. If we let U(T, r) be the particular tableau obtained by moving the largest possible entry at each stage, then we have U U(T, r) for any other such U. So it suffices to consider only the tableaux U(T, r).
Since µ a+1 = · · · = µ c−1 and T is semistandard, the largest entries in rows a + 1, . . . , c − 1 are the entries in column d, and these are strictly increasing. Hence for a < r < c we have U(T, r) ⊲ U(T, r + 1). So it suffices to consider only the tableau U(T, c). But by Lemma 6.15 It remains to observe that the coefficient of V in Θ Re(µ) • Θ CP λ µ (c) is 1; but this follows from Lemma 6.7: since (c, d) lies in the same ramp as or a later ramp than (a, b), we have Re(µ) c,d > Re(µ) a,b−1 , so all the multinomial coefficients in the coefficient of Θ V equal 1.
The composition β • α may be computed using Proposition 6.14. By Proposition 6.16, we get where each U is a row-standard tableau with V U. Hence when we write such a U as a linear combination of semistandard homomorphisms, the coefficient ofΘ V is zero, by Lemma 6.2. So when we write β • α as a linear combination of semistandard homomorphisms, the coefficient ofΘ V is ±1, and in particular β • α 0.
Remark. For simplicity, we have concentrated on quite a special case. It is possible to weaken the assumptions on λ and µ and use the same argument: we can allow removable nodes in λ between rows a and c, as long as these none of these removable nodes has the same residue as (a, b), and as long as µ i − µ i+1 < p for all a < i < c. We leave the reader to check the details (referring to [L2] for the formula for a homomorphism S λ → S µ ). However, it is not generally the case that the composition of a one-node Carter-Payne homomorphism and a restrictisation homomorphism is non-zero. For example, take p = 3, λ = (6) and µ = (5, 1), so that µ rest = (3, 2, 1). S (6) is isomorphic to the simple module D (2 3 ) (which happens to be the trivial FS 6 -module); since soc(S (3,2,1) ) is a different simple module D (3,2,1) , there is no non-zero homomorphism S (6) → S (3,2,1) .

Proof of the main theorem
We now come to the proof of our main theorem. We proceed by induction, with our main tool being Proposition 4.7. As we shall see, this deals with all cases except for two families of partitions which we deal with using the other techniques described above.

Notation and assumptions
Throughout this section, p = 2h + 1 is an odd prime. J denotes the set of p-JM-partitions, and A denotes the set of self-conjugate partitions that (according to Theorem 3.3) label Specht modules with exactly two composition factors. So if p 5 then A is the set of R-partitions, while if p = 3 then A comprises the R-partitions of type I, the self-conjugate partitions of 3-weight 1 and the partition (3 3 ).
It will be helpful to make some assumptions that will remain in force for the next few sections. Essentially, these say that λ is a partition which cannot be dealt with by Proposition 4.7(1,2) or Proposition 5.11. Recall that if λ is a partition and 0 i h, then λ ▽±i denotes the partition obtained by repeatedly removing all removable nodes of residue i or −i. Similarly, we define λ △±i to be the partition obtained from λ by repeatedly adding addable nodes of residue ±i.

Assumptions and notation in force for Sections 7.1-7.5
λ is a self-conjugate partition which is not p-quotient-separated and is not in A or J. λ ▽±i ∈ A ∪ {λ} for each 0 < i < h, while λ ▽±h ∈ A ∪ J ∪ {λ}.
We observe some immediate consequences of these assumptions. Suppose 0 i < h and λ ▽±i λ. Then by Proposition 5.10 λ ▽±i is not an R-partition of type I, since λ is not. Also, when p = 3 λ ▽±i cannot equal (3 3 ), since this partition has one removable node and two addable nodes, all of residue 0. So λ ▽±i must be an R-partition of type II, and if in addition p = 3 then λ ▽0 has 3-weight 1; in particular, λ ▽±i is p-quotient-separated.
λ is obtained from λ ▽±i by adding equal numbers of addable i-and (−i)-nodes. But we cannot have λ = (λ ▽±i ) △±i , since then by Proposition 5.4(3) λ would be p-quotient-separated, contradicting our assumptions. So λ is obtained from λ ▽±i by adding some but not all of the addable i-nodes, and some but not all of the addable (−i)-nodes. In particular, λ ▽±i has at least two addable i-nodes.
A similar discussion applies statement applies when λ ▽±h λ. In this case λ ▽±h can be an R-partition of type I, but only of p-weight 1, since an R-partition of type I with p-weight greater than 1 has a removable h-node. An R-partition of type I with p-weight 1 is also an R-partition of type II, and so in fact we can assume λ ▽±h is an R-partition of type II or a JM-partition, and in particular we can assume λ ▽±h is p-quotient-separated; as in the last paragraph, this means that λ must have at least one addable h-node and at least one addable (−h)-node.

First case
In this section we consider the case where λ ▽0 λ.
The abacus display for λ is obtained by moving some but not all of the beads in positions rp − 1, . . . , −rp − 1 to the adjacent positions on runner 0; moreover, since λ is self-conjugate, the bead in position ip − 1 is moved if and only if the bead in position −ip − 1 is moved. This means that there must be some −r i < j r with i 0 such that λ has beads in positions jp and ip − 1 and spaces in positions jp − 1 and ip. If j > 0, then the number of beads in positions ip + 1, . . . , jp − 2 is at least j − i, since for each i < k < j there is a bead in position kp or kp − 1, and there is also a bead in position h; hence by Lemma 5.2 we have 0 (λ). If i −2 then again the number of intervening beads is at least j − i, since there is a bead in position h − 2p. In the case where j = 0, i = −1 and p 5, there must be a bead in position h − p − 1 or h − p + 1, since if both of these positions are vacant, then (by self-conjugacy) then positions h − 1 and h + 1 are occupied, so that λ ▽0 is not p-quotient-separated, a contradiction. So again we have at least j − i beads between positions ip and jp − 1, and again 0 (λ).
The remaining case is where p = 3, and the only pair i < j such that positions 3j and 3i − 1 are occupied while positions 3j − 1 and 3i are vacant is (−1, 0). In this case we must have r = 1, and hence λ has abacus display . So λ = (3 3 ), contrary to assumption.
To help us deal with the case where (λ ▽0 ) (0) ∅, we use the following lemma.
Suppose the lemma is false, and take 1 < i h such that λ ▽±i λ. Then either λ ▽±i is an R-partition of type II, or i = h and λ ▽±i is a JM-partition. In either case, runner 0 must be either the largest or smallest runner of λ ▽±i by Propositions 5.7 and 5.9. But since λ (and hence λ ▽±i ) has an addable 0-node, runner 0 cannot be the smallest runner, by Proposition 5.4(1). So runner 0 is the largest runner of λ ▽±i , and hence (in the abacus displays for both λ ▽±i and λ) the first space on runner 0 occurs after the last bead on any other runner. But λ ▽0 is an R-partition of type II, so the abacus display for λ ▽0 (and hence for λ) has a bead in position h; so the first space on runner 0 in the abacus display for λ is at position p or later. Since λ is self-conjugate, this means that the last bead on runner p − 1 is no later than position −1 − p. But now λ has no addable 0-nodes, contrary to assumption. Proposition 7.3. Suppose λ ▽0 λ and (λ ▽0 ) (0) ∅. Then p 5, and either 0 (λ) or p−1 (λ).
Proof. The fact that p 5 follows from our standing assumptions (in particular, the definition of A when p = 3). λ ▽0 is an R-partition of type II with (λ ▽0 ) (0) ∅, and in particular (λ ▽±h ) (h) = (1) and (λ ▽±h ) (i) = ∅ for i 0, h, p − 1; the same applies to the p-quotient of λ. By Proposition 5.9 runners 0 and p − 1 must be the smallest and largest runners of λ ▽0 in some order, and in fact by Proposition 5.4(2) runner 0 is the smallest (since λ ▽0 has at least two addable 0-nodes). So in the abacus display for λ ▽0 , every position before the first space on runner 0 is occupied, and every position after the last bead on runner p − 1 is vacant.
λ ▽0 is a self-conjugate partition, which means that position ip − 1 in the abacus display is occupied if and only if position −ip is vacant. Together with the statements in the last paragraph, this implies that there is a finite set I of positive integers such that: The fact that λ ▽0 has at least two addable 0-nodes means that I is non-empty. λ is obtained from λ ▽0 by moving some but not all of the beads in positions ip − 1 (for i ∈ ±I ∪ {0}) to the right; the fact that λ is self-conjugate means that the bead i position ip − 1 is moved if and only if the bead in position −ip − 1 is moved.
Suppose first that for some i < j 0 the bead in position jp − 1 is moved but the bead in position ip − 1 is not. Then λ has beads in positions jp and ip − 1, and spaces in positions jp − 1 and ip, and we claim that the number of beads in between these two positions is at least j − i. Indeed, there is a bead in position kp − 1 or kp (or possibly both) for each i < k < j, and additionally there is a bead in position jp − h or jp − h − 2 (since λ is self-conjugate and λ (h−1) = λ (h+1) = ∅). So 0 (λ), by Lemma 5.2. So suppose there are no such i and j. Then in particular the bead in position −1 is not moved in constructing λ from λ ▽0 , but the beads in positions ±lp − 1 are moved, where l = max I. By Lemma 7.2 λ ▽±i = λ for i 0, 1, and since position h − p is vacant, this means that positions h − p + 1, h − p + 2, . . . , −2 are vacant. On the other hand, the first paragraph of this proof shows that position −lp − 2 must be occupied (in λ ▽0 , and hence in λ). So λ has beads in positions −1 and −lp − 2, and spaces in positions −2 and −lp − 1; the number of intervening beads is at least l, since for each 1 k l at east one of positions −kp − 1, −kp is occupied, and so p−1 (λ).

Second case
Here we consider the case where there is 0 < i < h such that λ ▽±i λ while λ ▽±j = λ for all 0 j < i. Proposition 7.4. Suppose 0 < i < h, and that λ ▽±i λ. Then either i (λ) or p−i (λ), or runner i − 1 is the largest runner in the abacus display for λ ▽±i , or runner i is the smallest runner in the abacus display for λ ▽±i .
Proof. Suppose first that (λ ▽±i ) (i−1) ∅. Since λ ▽±i is an R-partition of type II, this means that runner i − 1 is either the largest or the smallest runner in λ ▽±i ; but λ ▽±i has at least two addable i-nodes, so by Proposition 5.4(2) runner i − 1 must be the largest runner. Similarly, if (λ ▽±i ) (i) ∅, then runner i is the smallest runner in λ ▽±i .
So assume (λ ▽±i ) (i−1) = (λ ▽±i ) (i) = ∅. This means that there are integers b < c such that in the abacus display for λ: • for k < b, both of the positions kp + i − 1 and kp + i are occupied; • for b k < c, exactly one of the positions kp + i − 1 and kp + i is occupied; • for c k, neither of the positions kp + i − 1 and kp + i is occupied.
Suppose that for some k, positions kp + i and (k − 1)p + i − 1 are occupied while positions kp + i − 1 and (k − 1)p + i are vacant. If there is at least one bead in a position between (k − 1)p + i and kp + i − 1, then i (λ), by Lemma 5.2. On the other hand, if all of these positions are vacant, then in the abacus display for λ ▽±i every runner except runner i − 1 has a space before the bead in position kp + i − 1; so by Proposition 5.4(1) runner i − 1 is the largest runner of λ ▽±i .
Assume instead that there is no such k. Since λ is obtained from λ ▽±i by adding some but not all of the addable i-nodes (and some but not all of the addable (−i)-nodes), there must be k such that in the abacus display for λ positions kp + i − 1 and (k − 1)p + i are occupied while positions kp + i and (k − 1)p + i − 1 are vacant. Since λ is self-conjugate, this means that positions −kp + p − i + 1 and −(k + 1)p + p − i are occupied while positions −kp + p − i and (−k − 1)p + p − i + 1 are vacant. Repeating the argument from the last paragraph (with i replaced by p − i and k replaced by −k) we find that either p−i (λ) or runner p − i is the largest runner in λ ▽±i , i.e. runner i is the smallest.
We now consider the possibilities in Proposition 7.4 in more detail. We need to treat the case i > 1 and i = 1 separately.
Proposition 7.5. Suppose 1 < i < h, and that λ ▽±i λ while λ ▽±j = λ for all 0 j < i. Suppose furthermore that runner i − 1 is the largest runner in the abacus display for λ ▽±i . Then i (λ).
Proof. Since λ ▽±i is an R-partition of type II and runner i − 1 is its largest runner, we have (λ ▽±i ) ( j) = ∅ for j i − 1, h, p − i. In particular, (λ ▽±i ) (0) = ∅. Since λ (and hence λ ▽±i ) has no removable 0-nodes, position 0 in the abacus is vacant (for both partitions), and hence position lp is vacant for all l 0. The assumption that λ ▽±j = λ for all j < i then means that position lp + i − 1 is vacant (in λ) for all l 0.
Let dp + i − 1 be the first vacant position on runner i − 1 in λ ▽±i ; since λ ▽±i is p-quotientseparated and there is a bead in position h, we must have d 1. The fact that dp + i − 1 is the first vacant position on runner i − 1 means that for each c < d at least one of the positions cp + i − 1 and cp + i is occupied in λ; in fact, since (λ ▽±i ) (i) = ∅, there is some b d such that • for c = b, . . . , d − 1 exactly one of the positions cp + i − 1 and cp + i is occupied, and • for c < b both of the positions cp + i − 1 and cp + i are occupied.
By assumption λ has at least one addable i-node, so there is some k such that position kp + i − 1 is occupied while position kp + i is vacant. From the first paragraph of the proof we must have k < 0, and in particular k < d − 1. Taking a maximal such k, we therefore have positions kp + i − 1 and (k + 1)p + i occupied, while positions kp + i and (k + 1)p + i − 1 are vacant. There is at least one bead among positions kp + i + 1, . . . , (k + 1)p + i − 2: if k = −1 then we can take the bead in position −i, while if k < −1 we can take the bead in position h − 2p. Now we consider the case i = 1.
Proof. Since λ ▽0 = λ, position 0 in the abacus display for λ is vacant. If we let dp be the first vacant position on runner 0 in λ ▽±1 , then, arguing as in the proof of Proposition 7.5, we must have d 1. In particular, position 0 is occupied in the abacus display for λ ▽±1 , and hence position 1 is occupied in the abacus display for λ. Since λ is self-conjugate, position −1 is occupied and position −2 is vacant.
λ has an addable 1-node, so there is some k such that position kp is occupied and position kp + 1 is vacant. If k < 0, then taking a maximal negative such k and copying the last part of the proof of Proposition 7.5, we get 1 (λ). So suppose k > 0. Then by self-conjugacy position −kp − 2 is occupied and position −kp − 1 is vacant. For every 0 > l > −k at least one of the positions lp − 2 and lp − 1 is occupied (since otherwise we would have (λ ▽±1 ) (p−2) ∅); furthermore, at least one of positions −p and 1 − p is occupied (since d > 0); hence the number of beads between positions −kp − 1 and −2 is at least k, and so p−1 (λ). Now we consider the situation where runner i is the smallest in the abacus display for λ ▽±i . Again, we have to consider the cases i > 1 and i = 1 separately.
Proposition 7.7. Suppose 1 < i < h, and that λ ▽±i λ while λ ▽±j = λ for all 0 j < i. Suppose furthermore that runner i is the smallest runner in the abacus display for λ ▽±i . Then i (λ).
Proof. The fact that λ ▽±i is an R-partition of type II with smallest runner i means that (λ ▽±i ) ( j) = ∅ for all j i, h, p − i − 1. Suppose that for some l < k positions kp + i and lp + i − 1 are occupied in the abacus display for λ, while positions kp + i − 1 and lp + i are vacant; choose such a pair (k, l) with k − l as small as possible. Arguing as in the first paragraph of the proof of Proposition 7.5, we must have l < 0. Now for every k > m > l, positions mp + i and mp + i − 1 must both be occupied: if both were vacant then we would have (λ ▽±i ) (i−1) ∅, while if only one were vacant then the choice of k, l would be contradicted. So if k − l > 1 then the number of beads between positions lp + i and kp + i − 1 is at least k − l, and so i (λ), as required. If k − l = 1 and k < 0, then there is a bead in position lp + h, so again i (λ). Finally if k = 0 and l = −1 then there is a bead in position −i, so again i (λ).
So we can assume that there are no such k and l. This means that the last bead on runner i in the abacus display for λ is either earlier than or adjacent to the last bead on runner i − 1, and hence occurs no later than position −p + i. Furthermore, since λ has at least one addable i-node and there are no beads on runner i − 1 after position −p + i − 1, there must be a space on runner i no later than position −p + i. Now we claim that for every i < j < h, there is a space on runner j no later than position −p + j. If this is not true, take the first j for which it fails. Then there is a space on runner j − 1 with an adjacent bead on runner j, so λ ▽±j λ. Hence λ has at least one addable j-node; since there is no space on runner j before position j, there must be a bead on runner j − 1 in position j − 1 or later. This means that j > i + 1 and that λ ( j−1) ∅, and hence (λ ▽±i ) ( j−1) ∅, contradicting the first sentence of the proof.
So our claim holds. In particular, if i < h − 1 there is a space on runner h − 1 no later than position −p + h − 1. So (because λ (h−1) = ∅) there is no bead on runner h − 1 after position −2p + h − 1. But there is a bead in position h and a bead in position cp + h for all c −2, so λ has removable h-nodes but no addable h-nodes; contradiction. The same argument applies when i = h − 1 and there is a space in position −p + h − 1.
The only remaining case is where i = h − 1 and there is a bead in position −p + h − 1 and a space on runner h − 1 earlier than this bead. From our assumptions so far, there is also a bead in position −p + h − 2, and a space on runner h − 2 earlier than this bead. Hence λ (h−2) ∅. Now we consider the partition λ ▽±h . Since (λ ▽±h ) (h−2) = λ (h−2) ∅, λ ▽±h is either a JM-partition or an R-partition of type II, and runner h − 2 is either the largest or the smallest runner of λ ▽±h . The first space on runner h − 2 occurs before the last bead on runner h − 1 (which is no earlier than position h − 1), so in fact runner h − 2 is the smallest runner of λ ▽±h . However, λ has a space on runner h − 1 no later than position −2p + h − 1, which means that λ ▽±h has a space on runner h + 1 no later than position −2p + h + 1, i.e. earlier than the last bead on runner h − 2. So by Proposition 5.4(1) λ ▽±h is not p-quotient-separated, a contradiction. Now we consider the case i = 1.
Proof. Since runner 1 is the smallest in the abacus display for λ ▽±1 , the last bead on this runner occurs before the space in position h − p, i.e. no later than position 1 − p. Hence for every l 0 at most one of the positions lp and lp + 1 is occupied in the abacus display for λ. If none of the positions lp for l 0 is occupied, then we can just copy the proof of Proposition 7.7 to deduce that 1 (λ). So assume that position kp is occupied for some k 0. In fact we must have k > 0, since if position 0 were occupied then λ would have a removable 0-node. By the above analysis position kp + 1 is vacant, and position 1 is occupied, since otherwise we would have (λ ▽±1 ) (0) ∅.
Since λ is self-conjugate, this means that positions −1 and −kp − 2 are occupied, while positions −2 and −kp − 1 are vacant. Furthermore, for each 0 > l > −k at least one of the positions lp − 2 and lp − 1 is occupied, and in addition at least one of the positions −p and 1 − p is occupied (again since (λ ▽±1 ) (0) = ∅). So the number of beads between positions −kp − 1 and −2 is at least k, and so p−1 (λ).
Remark. It turns out that the results in this subsection are to some extent redundant, because (given our standing assumptions) if 1 < i < h − 1 and λ ▽±j = λ for 0 j < i, then λ ▽±i = λ also. However, proving this statement appears to be just as difficult as the results we have proved in this section, which in any case are still needed for the cases i = 1, h − 1.

Third case
We are left with the situation where λ ▽±i = λ for 0 i < h. We now have two possibilities to consider, since λ ▽±h could be an R-partition of type II or a JM-partition.
In the next few results we analyse various possibilities for the configuration of the middle three runners of the abacus display for λ. To help us, we introduce some notation in which we give portions of the abacus diagram focusing on the middle three positions in a given row. For example, given an integer k we may write k to mean that positions kp + h − 1 and kp + h + 1 are occupied in the abacus display for λ, while position kp + h is vacant.
Proof. The fact that λ ▽±j = λ for all j < h means that positions kp + h − 2, kp + h − 3, . . . , (k − 1)p + h + 1 are occupied. In particular, this means that k cannot equal 0, since then by self-conjugacy we would have −1 , a contradiction. So k 1, and hence k − 1 (since at most one of the positions kp + h − 1, kp + h and kp + h + 1 is occupied).
Exactly the same analysis applies to the partition λ ▽h ; the only essential difference is that the bead in position k − p + h has moved to position −kp + h − 1, but this makes no difference to the count of intervening beads, so −h (λ ▽h ), which gives h(−h) (λ).
Lemma 7.12. Suppose there is no k for which k . Then k for some k, and hence h (λ).
Proof. If there is no k for which k , then for every k 0 we have either k or k , and so for every k < 0 we have either k or k . But now λ has no addable h-nodes; contradiction.
So k for some k. So (by self-conjugacy) there are beads in positions kp + h and −(k + 1)p + h − 1, and spaces in positions kp + h − 1 and −(k + 1)p + h. By a similar argument to that used in the proof of Lemma 7.11, the number of intervening beads is at least 2k + 1, and so h (λ).
Lemma 7.14. Suppose that for some k < l we have k and l , and that positions (k − 1)p + h + 1 and (l + 1)p + h are vacant, and define µ to be the partition obtained from λ by repeatedly adding all addable nodes of residue not equal to h or −h. Then h (µ) and h+1 (µ), and hence S λ has at least three composition factors.
So we have h(−h)h (µ) and (−h)h(−h) (µ), so by Lemma 4.6 and Proposition 4.7 S µ has at least three composition factors. So by Lemma 4.4 S λ does too.
Lemma 7.15. Suppose that 0 . Then there is some l 1 such that either l or l .
Proof. Suppose not, i.e. for every l 1 either l or l . We will show that λ is pquotient-separated, contrary to assumption. Since λ ▽0 = λ, position 0 is vacant in the abacus display for λ. Since λ ▽±j = λ for all j < h, this then means that positions 1, . . . , h − 2 are also vacant. In addition, the fact that position h + 1 is vacant means that positions h + 2, . . . , p + h − 2 are also vacant. Symmetrically, positions −2p + h + 2, . . . , −p + h − 2 and −p + h + 2, . . . , −1 are all occupied. Now we claim that there is at least one value of l for which l . If not, then position lp + h + 1 is vacant in λ for all l 0, and hence (since λ ▽±j = λ for j < h) so are positions lp + h + 2, . . . , (l + 1)p + h − 2. In other words, all positions k 0 apart from position h are vacant, and all positions k < 0 apart from position −p + h are occupied. Hence λ is the partition = (h + 1, 1 h ); but this has p-weight 1, contrary to assumption.
By assumption λ ▽±h is a JM-partition, and in particular is p-quotient-separated. We claim that runner h − 1 is the largest runner in λ ▽±h . Choose l such that l . Then the abacus display for λ ▽±h has a bead in position lp + h − 1 p + h − 1, and every runner apart from runner h − 1 has a space before this position. So by Proposition 5.4(1) runner h − 1 is largest. Symmetrically, runner h + 1 is smallest, and hence for every i h − 1, h, h + 1 we have λ (i) = (λ ▽±h ) (i) = ∅. So the abacus display for λ has the following properties.
• For any i h − 1, h, h + 1, the last bead on runner i occurs in position −p + i, and the first space in position i.
• On runner h − 1, the last bead occurs in position −p + h − 1.
• On runner h, the last bead occurs in position h and the first space in position −p + h.
• On runner h + 1, the first space occurs in position h + 1.
It follows that λ is p-quotient-separated; contradiction.
Lemma 7.16. Suppose that for every k 0 either k or k . Then there is a p-restricted partition ν {λ rest , m p (λ rest )} such that Hom FS n (S λ , S ν ) 0. Hence S λ has at least three composition factors.
Proof. Since position lp + h + 1 is vacant in the abacus display for λ for every l 0 and λ ▽±i = λ for all 0 i < h, positions lp + h + 2, . . . , (l + 1)p + h − 2 are also vacant. In addition, position 0 is vacant, and so positions 1, . . . , h − 2 are vacant as well. So every position k 0 not on runner h is vacant, and symmetrically every position k < 0 not on runner h is occupied. part. In other words, we must prove that if λ is a self-conjugate partition not in A or J, then S λ has at least three composition factors. If λ is p-quotient-separated, then Proposition 5.11 gives the result, so we can assume this is not the case. We proceed by induction on |λ|, using the partitions λ ▽±i . If there is any i for which λ ▽±i λ and λ ▽±i A ∪ J, then by Theorem 2.5 and by induction S λ ▽±i has at least three composition factors, so by Proposition 4.7(1) S λ does too. In addition, if i < h and S λ ▽±i is irreducible then by Proposition 4.7(2) S λ has at least three composition factors. So we can assume that λ ▽±i ∈ A ∪ {λ} for each 0 i < h, while λ ▽±h ∈ A ∪ J ∪ {λ}. So the assumptions listed in Section 7.1 apply. λ must have at least one removable node, so for some i we have λ ▽±i i. Let 0 i < h be minimal with this property, and consider the possibilities for i.
If 1 i < h, then by Propositions 7.4-7.8 we have either i (λ) or p−i (λ), and so again we are done by Proposition 4.7(3).
If i = h, then from the discussion in Section 7.1, λ ▽±h is either an R-partition of type II or a JM-partition. In the first case, Proposition 7.9 gives either h (λ) or −h (λ), or p = 3 and λ = (4 3 , 3). If h (λ) or −h (λ), then we are done by Proposition 4.7(3), while if p = 3 and λ = (4 3 , 3), then we have 121 (λ) and 212 (λ), so Proposition 4.7(4) gives the result (or we can just use the readily-available decomposition numbers for S 15 ). Finally, we have the case where i = h and λ ▽±h is a JM-partition. This is dealt with in Proposition 7.17. Theorem 3.3 now follows by induction.

Irreducible representations which remain irreducible modulo every prime
We conclude this paper with a corollary of our main theorem, in which we classify the irreducible representations of A n that remain irreducible modulo every prime. The result is unsurprising.
Theorem 8.1. Suppose ψ is an ordinary irreducible character of the alternating group A n and that ψ remains irreducible modulo every prime. Then ψ is one-dimensional.
The same result for the symmetric groups was proved by Kleshchev and Premet in [KP], though an easier proof [JM3] follows from the classification of irreducible Specht modules in characteristic 2.
Proof of Theorem 8.1. Assume n 2, with the case n = 1 being trivial, and use the notation of Section 3.
For the reader's convenience we conclude with an index of the notation we use in this paper. We provide references to the relevant subsections.