THE GLOBALLY IRREDUCIBLE REPRESENTATIONS OF SYMMETRIC GROUPS

ALEXANDER KLESHCHEV AND ALEXANDER PREMET

Abstract. Let K be an algebraic number field and O be the ring of integers of K. Let G be a finite group and M be a finitely generated torsion free OG-module. We say that M is a globally irreducible O G-module if for every maximal ideal 𝔭 of O, the k_{𝔭} G-module M ⊗_{O} k_{𝔭} is irreducible, where k_{𝔭} stands for the residue field O/𝔭.

Answering a question of Pham Huu Tiep, we prove that the symmetric group Σ_{n} does not have non-trivial globally irreducible modules. More precisely we establish that if M is a globally irreducible O Σ_{n}-module, then M is an O-module of rank 1 with the trivial or sign action of Σ_{n}.

1. Introduction and generalities

The notion of global irreducibility was first introduced by J. G. Thompson in the course of constructing the sporadic group F_{3} ([9, 10]). He observed that there exists an even unimodular ℤ-lattice Λ of rank 248 with Aut (Λ) ≅ ℤ_{2} × F_{3} such that the 𝔽_{p} F_{3}-module Λ/pΛ is irreducible for every prime p. Other interesting examples of the same nature were discovered later, and different (closely related) notions of global irreducibility were introduced. A good survey on the subject is [11].

Let K be an algebraic number field and O be the ring of integers of K. Let G be a finite group and M be a finitely generated torsion free O G-module. We say that M is globally irreducible if for every maximal ideal 𝔭 of O, the k_{𝔭} G-module M ⊗_{O} k_{𝔭} is irreducible, where k_{𝔭} denotes the residue field O/𝔭. This definition seems to “include” all versions of global irreducibility introduced before.

Let Σ_{n} denote the symmetric group on n letters. The main result of this paper is the following theorem which answers a question posed by Pham Huu Tiep. Theorem A. Let M be a globally irreducible O Σ_{n}-module. Then M is an O-module of rank 1 with the trivial or sign action of Σ_{n}.

Throughout the paper:

ℕ = {1, 2, … } is the set of all natural numbers,

F is an arbitrary field of characteristic p > 0 (unless otherwise stated).

For r, s ∈ ℕ we denote:

[r, s] = {t ∈ ℕ | r ≤ t ≤ s}; (r, s] = {t ∈ ℕ | r < t ≤ s};
[r, s) = {t ∈ ℕ | r ≤ t < s}; (r, s) = {t ∈ ℕ | r < t < s}.

Mathematics Subject Classification: 20C30, 20C10.

The authors thank G. Michler and A. Zalesskii who organized a conference on representations of finite groups in Bad-Honnef where this collaboration began, and the Volkswagen foundation for a financial support. The first author was also supported by the NSF.

Given a partition λ of n, we often gather together equal parts of λ and write

λ = (λ_{1}^{a_{1}}, λ_{2}^{a_{2}}, …, λ_{k}^{a_{k}}),

where λ_{1} > λ_{2} > ⋯ > λ_{k} > 0, a_{i} > 0, and a_{1}λ_{1} + a_{2}λ_{2} + ⋯ + a_{k}λ_{k} = n. A partition λ is called p-regular if a_{i} < p for all i. We denote by λ′ the partition conjugate to λ (the Young diagram of λ′ is the transpose of the Young diagram of λ). The symbol (i, j) denotes the node in the ith row and the jth column of the Young diagram of λ, and h(i, j) = h_{λ}(i, j) stands for the length of the hook of λ corresponding to the node (i, j). Note that the notation for the node (i, j) should not be confused with the one for the interval (i, j) as defined above! In what follows we do not distinguish between partitions and their Young diagrams.

We introduce some notions from the representation theory of symmetric groups. Details can be found in [6]. Let S_{ℤ}^{λ} denote the Specht module over ℤ corresponding to λ (it carries a canonical ℤ Σ_{n}-module structure). Given a field F set S_{F}^{λ} = S_{ℤ}^{λ}⊗_{ℤ}F. It is well–known that if char F = 0 then S_{F}^{λ} is an irreducible F Σ_{n}-module. In general, the irreducible F Σ_{n}-module corresponding to a p-regular partition λ is denoted by D_{F}^{λ}. Note that D_{F}^{λ} is always a quotient of S_{F}^{λ}. Another fact which is going to be used without comments is that 𝔽_{p} is the splitting field for the group Σ_{n}. So the question of the irreducibility of S_{F}^{λ} depends on the characteristic of F only, not on F itself.

We shall deduce Theorem A from the following Theorem B. The Specht module S_{ℤ}^{λ} is globally irreducible if and only if λ = (n) or (1^{n}).

We now explain why Theorem B implies Theorem A.

Let M be a globally irreducible O Σ_{n}-module, and M_{K} = M ⊗_{O} K. Then M_{K} is a direct sum of Specht modules over K (see [6, Theorem 4.12]). It follows that there exists a ℤ Σ_{n}-module M_{ℤ} such that M_{K} ≅ M_{ℤ} ⊗_{ℤ} K as K Σ_{n}-modules. Moreover, M_{ℤ} can be chosen to be a direct sum of Specht modules over ℤ. By a theorem of Brauer–Nesbitt, the k_{𝔭} Σ_{n}-modules M ⊗_{O} k_{𝔭} and M_{ℤ}⊗_{ℤ}k_{𝔭} have the same composition factors (see, e.g., [2, Theorem 82.1]). We claim that M_{ℤ} is a globally irreducible ℤ Σ_{n}-module isomorphic to a Specht module S_{ℤ}^{λ} for some partition λ of n. Indeed, if \overline{M} = M_{ℤ} ⊗_{ℤ} 𝔽_{p} is reducible for some prime p, then, for each maximal ideal 𝔮 of O such that 𝔮 ∩ ℤ = p ℤ, the k_{𝔮} Σ_{n}-module M_{ℤ} ⊗_{ℤ} k_{𝔮} ≅ \overline{M} ⊗_{𝔽_{p}} k_{𝔮} is reducible. Also, if M_{ℤ} is a direct sum of more than one Specht module over ℤ, then, for each maximal ideal 𝔭 of O, the k_{𝔭} Σ_{n}-module M_{ℤ} ⊗_{ℤ} k_{𝔭} is reducible again. This establishes the claim in view of the global irreducibility of M and the result of Brauer–Nesbitt mentioned above.

Now suppose we have proved Theorem B. Then the ℤ-rank of M_{ℤ} is 1. So dim M_{K} = 1, i.e. M is an O-module of rank 1. Hence the subgroup A_{n} = Σ′_{n} acts trivially on M, and Theorem A follows.

2. Some representation theory and some number theory

The following result known as Carter’s conjecture was proved by James and Murphy [5, 7]. Recall that the p-exponent of an integer n is the largest power of the prime p dividing n.

Proposition 2.1. Let λ be a p-regular partition of n. Then S_{F}^{λ} is irreducible if and only if the p-exponents of the hook lengths of the Young diagram of λ are constant along the columns.

Our first lemma is a straightforward consequence of Proposition 2.1.

Lemma 2.2. Assume that both λ and λ′ are p-regular partitions of n. If there is a hook in the Young diagram of λ whose length is divisible by p, then S_{F}^{λ} is reducible.

Proof. Suppose that S_{F}^{λ} is irreducible and some h_{λ}(i, j) is divisible by p. In view of Proposition 2.1 all hook lengths in the jth column of the Young diagram of λ should be divisible by p. But this is impossible for the bottom hook in the jth column, since λ′ is p-regular.

From now on we fix an arbitrary partition λ of n. Let

λ = (λ_{1}^{a_{1}}, λ_{2}^{a_{2}}, …, λ_{k}^{a_{k}}), λ′ = (μ_{1}^{b_{k}}, μ_{2}^{b_{k−1}}, …, μ_{k}^{b_{1}}).

[figure: a_{1}; b_{1}; b_{k}; a_{k}]

Put

A_{s} = (∑_{i=1}^{s} a_{i}, λ_{s}), s = 1, 2, …, k;

B_{r} = (∑_{i=1}^{r−1} a_{i} + 1, λ_{r} + 1), r = 1, 2, …, k + 1

(λ_{k+1} is interpreted as 0). Then A_{1}, …, A_{k} and B_{1}, …, B_{k+1} are all the removable and the addable nodes for λ, respectively. We denote by λ(r, s) the partition of n whose Young diagram is obtained from that of λ by removing A_{s} and adding B_{r}. This partition is not well defined exactly in the following cases:

(a) r = s, a_{r} = 1;

(b) r = s + 1, b_{s} = 1. So when we write λ(r, s) we assume that neither (a) nor (b) occurs. Clearly, λ(r, s) ▷ λ if r ≤ s and λ(r, s) ◁ λ if r > s.

We make use of the so called p-ladders introduced by James in [4]. Those are, roughly speaking, straight lines with slope (p − 1). More precisely, the u-th p-ladder is defined as the following set of nodes:

L_{u} := {(i, j) | i, j ∈ ℕ, i + (p − 1)(j − 1) = u}.

If we slide the nodes of λ along the p-ladders as far up as they can go, then we obtain a new partition λ^{R}, called the p-regularization of λ. It is proved in [4] that λ^{R} is well defined and that λ is p-regular if and only if λ = λ^{R}.

Proposition 2.3 ([4]). D_{F}^{λ^{R}} occurs in S_{F}^{λ} with multiplicity 1, and if D_{F}^{μ} is a composition factor of S_{F}^{λ}, then μ ⊵ λ^{R}.

Lemma 2.4. Suppose that λ and λ(r, s) are p-regular. Let A_{s} belong to the v-th p-ladder, and B_{r} belong to the u-th p-ladder. Then λ(r, s) ▷ λ if and only if u > v.

Proof. It was mentioned above that λ(r, s) ▷ λ if and only if r ≤ s. By definition,

v = ∑_{i=1}^{s} a_{i} + (p − 1)(λ_{s} − 1) and u = ∑_{i=1}^{r−1} a_{i} + 1 + (p − 1)λ_{r}.

Now u > v is equivalent to

∑_{i=1}^{s} a_{i} − ∑_{i=1}^{r−1} a_{i} < (p − 1)(λ_{r} − λ_{s}) + p.  (1)

Suppose r > s and u > v. Then

∑_{i=s+1}^{r−1} a_{i} > (p − 1)(λ_{s} − λ_{r}) − p.  (2)

Note that (p − 1)(λ_{s} − λ_{r}) ≥ (p − 1)(r − s), and

∑_{i=s+1}^{r−1} a_{i} ≤ (r − 1 − s)(p − 1),

since λ is p-regular. Therefore (2) holds if and only if a_{i} = p−1 for s+1 ≤ i ≤ r −1 and λ_{s} − λ_{r} = r − s. If r = s + 1 this gives b_{s} = λ_{s} − λ_{s+1} = 1, so λ(r, s) is not defined. If r > s+1, we get a_{r−1} = p−1, and b_{r−1} = 1. Thus λ(r, s) is not p-regular, giving a contradiction. Hence r > s implies u ≤ v.

Let r ≤ s. Then

∑_{i=r}^{s} a_{i} ≤ (s − r + 1)(p − 1) < (s − r)(p − 1) + p ≤ (λ_{r} − λ_{s})(p − 1) + p,

which in combination with (1) forces u > v. Thus we have proved that r > s implies u ≤ v and r ≤ s implies u > v, as desired.

Lemma 2.5. Suppose A_{s} belongs to the v-th p-ladder and B_{r} belongs to the u-th p-ladder.

(i) If u = v, then λ(r, s)^{R} = λ^{R}. (ii) If u ≠ v, then λ(r, s)^{R} is obtained from λ^{R} by moving the bottom removable

node of λ^{R} on the p-ladder L_{v} to the top addable node for λ^{R} on the p-ladder L_{u} (and such nodes exist).

Proof. This follows from the definition of p-regularization.

Denote

A_{r,s} = ∑_{i=r}^{s} a_{i}, B_{r,s} = ∑_{j=r}^{s} b_{j}.

(The sums are interpreted as 0 if r > s.)

Lemma 2.6. Let r ≤ s. Then λ(r, s)^{R} ▷ λ^{R} if and only if A_{r,s} < p + (p − 1)B_{r,s−1}.

Proof. Let A_{s} belong to the v-th p-ladder and B_{r} belong to the u-th p-ladder. By Lemma 2.5, λ(r, s)^{R} is obtained from λ^{R} by moving a removable node on the v-th p-ladder to an addable node on the u-th p-ladder. Since both λ^{R} and λ(r, s)^{R} are pregular, Lemma 2.4 implies that λ(r, s)^{R} ▷λ^{R} if and only if u > v, which is equivalent to

A_{1,s} + (p − 1)(λ_{s} − 1) < A_{1,r−1} + 1 + (p − 1)λ_{r}.

Now the result follows from the equality λ_{r} − λ_{s} = ∑_{j=r}^{s−1} b_{j}.

The following proposition is a special case of the main result of Carter–Payne [1].

Proposition 2.7 ([1]).

(i) Let 1 ≤ r ≤ s ≤ k. If p divides A_{r,s} + B_{r,s−1}, then Hom_{F Σ_{n}} (S_{F}^{λ(r,s)}, S_{F}^{λ}) ≠ 0. (ii) Assume that k ≥ 2 , b_{2} ≥ 2, and p is odd. Let

μ = (λ_{1} + 2, λ_{1}^{a_{1}−1}, λ_{2}^{a_{2}−1}, λ_{2} − 2, λ_{3}^{a_{3}}, …, λ_{k}^{a_{k}}).

If p divides a_{1} + a_{2} + b_{1} + 1 then Hom_{F Σ_{n}} (S_{F}^{μ}, S_{F}^{λ}) ≠ 0.

Our further arguments rely heavily on the following consequence of Proposition 2.7.

Lemma 2.8. Let λ be a partition of n.

(i) Suppose there exist r, s ∈ ℕ with 1 ≤ r < s ≤ k such that p divides A_{r,s} +B_{r,s−1} and A_{r,s} < (p − 1)B_{r,s−1} + p. Then S_{F}^{λ} is reducible.

(ii) Suppose that k ≥ 2, b_{2} ≥ 2, a_{1} + a_{2} < (p − 1)(b_{1} + 2) + 1, and p is odd. If p divides a_{1} + a_{2} + b_{1} + 1 then S_{F}^{λ} is reducible.

Proof. (i) By Proposition 2.7(i),

Hom_{F Σ_{n}} (S_{F}^{λ(r,s)}, S_{F}^{λ}) ≠ 0.

By Lemma 2.6, our assumption guarantees that λ(r, s)^{R} ▷ λ^{R}. If S_{F}^{λ} is irreducible, then it is isomorphic to D_{F}^{λ^{R}}, see Proposition 2.3. On the other hand, by the same proposition, each composition factor D_{F}^{μ} of S_{F}^{λ(r,s)} satisfies μ ⊵ λ(r, s)^{R} ▷ λ^{R}, hence the Hom–space above should be zero. This contradiction proves (i).

For (ii), let μ be as in Proposition 2.7(ii). Then the Young diagram of μ can be obtained from that of λ by moving two nodes from the (a_{1} + a_{2})th row of the Young diagram of λ to the first row. Observe that μ is well defined as b_{2} ≥ 2. By Proposition 2.7(ii),

Hom_{F Σ_{n}} (S_{F}^{μ}, S_{F}^{λ}) ≠ 0.

Let C = (a_{1} +a_{2}, λ_{2}), D = (a_{1} +a_{2}, λ_{2} −1), X = (1, λ_{1} +1), and Y = (1, λ_{1} +2). The Young diagram of μ can be obtained from that of λ in two steps. First we remove C and add X to get a partition ν. Then we remove D from the Young diagram of ν and add Y. By our assumption,

a_{1} + a_{2} + b_{1} + 1 < p(b_{1} + 1) + p.

As p divides a_{1} + a_{2} + b_{1} + 1 we must have

a_{1} + a_{2} + b_{1} + 1 ≤ p(b_{1} + 1)

whence a_{1} + a_{2} < (p − 1)b_{1} + p. So ν^{R} ▷ λ^{R} by Lemma 2.6 (one should keep in mind that ν = λ(1, 2)). Applying Lemma 2.6 for the second time we obtain μ^{R} ▷ ν^{R} ▷ λ^{R}. Now we finish the proof as in part (i).

The next three lemmas are versions (and generalizations) of the well–known Chebyshev’s results on prime numbers.

Lemma 2.9. Let n ∈ ℕ. Then the interval (n, 2n + 1] contains at least two primes.

Proof. It is known [3, p. 21] that for n ≥ 21, the number of primes on the interval (n, 2n + 1] exceeds 3n/(5 ln n). Clearly, 3n/(5 ln n) is greater than 2 if n ≥ 21. Now check the result for n = 1, 2, …, 20.

Lemma 2.10. Let n be an integer greater than 1. Then (n, 2n) contains a prime.

Proof. This is the famous “Bertrand’s postulate” proved by Chebyshev (see, e.g., [3, p. 21]).

Lemma 2.11. Let n ∈ ℕ. Then (n, \frac{3}{2}n + 2] contains a prime.

Proof. For n ≥ 25, the interval (n, \frac{6}{5}n) contains a prime (see [8, p. 192]). Now check the result for n = 1, 2, …, 24.

Remark. It is easy to see that Lemma 2.9 also follows from the fact that the interval (n, \frac{6}{5}n) contains a prime ([8, p. 192]).

3. Proof of Theorem B

Given a partition λ of n set

A = max_{1≤i≤k} {a_{i}}, B = max_{1≤j≤k} {b_{j}}

(the parameters a_{i}, b_{j}, and k are defined in Section 2). We wish to prove that S_{ℤ}^{λ} is globally irreducible if and only if λ ∈ {(n), (1^{n})}. It is well–known that (S_{F}^{λ})^{∗} ≅ (S_{F}^{λ′}) ⊗ sgn. Therefore, S_{ℤ}^{λ} is globally irreducible if and only if S_{ℤ}^{λ′} is. So no generality is lost by assuming that B ≥ A. Let f ∈ {1, 2, …, k} be such that b_{f} = B.

Lemma 3.1. Suppose that S_{ℤ}^{λ} is globally irreducible, and B ≥ A. Then any integer in the interval [1, A_{1,f} + B_{1,f} − 1] is a hook length of λ.

Proof. We prove by the downward induction on d = f, f − 1, …, 1 that any integer in the interval [1, A_{d,f} + B_{d,f} − 1] is a hook length of λ.

If d = f we have A_{d,f} + B_{d,f} − 1 = a_{f} + b_{f} − 1. In this case, it suffices to note that h(A_{1,f}, B_{f,k} − v) = v + 1, v = 0, 1, …, b_{f} − 1, and h(A_{1,f} − w, B_{f+1,k} + 1) = b_{f} + w, w = 0, 1, …, a_{f} − 1.

Now suppose that 1 ≤ d < f. By the inductive hypothesis, the hook lengths of λ cover the interval [1, A_{d+1,f} + B_{d+1,f} − 1]. Note that

h(A_{1,d}, B_{f,k} − v) = A_{d+1,f} + B_{d,f−1} + 1 + v, v = 0, 1, …, b_{f} − 1,

and

h(A_{1,d} − w, B_{f+1,k} + 1) = A_{d+1,f} + B_{d,f} + w, w = 0, 1, …, a_{d} − 1.

Thus each number in the union

[1, A_{d+1,f} + B_{d+1,f} − 1] ∪ [A_{d+1,f} + B_{d,f−1} + 1, A_{d,f} + B_{d,f} − 1]  (3)

is a hook length of λ. If A_{d+1,f} + B_{d+1,f} > A_{d+1,f} + B_{d,f−1}, the union (3) equals the interval [1, A_{d,f} + B_{d,f} − 1], as desired. Otherwise we have

A_{d+1,f} + B_{d+1,f} ≤ A_{d+1,f} + B_{d,f−1}

which is equivalent to b_{f} ≤ b_{d}. Since b_{f} = B, we conclude that b_{d} = b_{f} = B yielding

A_{d+1,f} + B_{d+1,f} = A_{d+1,f} + B_{d,f−1} =: x,

that is the union (3) equals [1, A_{d,f} + B_{d,f} − 1] \ {x}. Note that

A_{d,f} + B_{d,f} ≥ a_{d} + a_{f} + b_{d} + b_{f} ≥ 2 + 2B.

Hence [1, A_{d,f} + B_{d,f} − 1] contains (B, 2B + 1]. By Lemma 2.9, there is a prime p > B which is a hook length of λ. Now Lemma 2.2 implies that S_{ℤ}^{λ} is not globally irreducible. This accomplishes the induction step.

Lemma 3.2. Suppose S_{ℤ}^{λ} is globally irreducible. Then either n = 1 or B ≥ 2.

Proof. Suppose B = 1 and n ≥ 2. Then λ is the “staircase partition”, and k ≥ 2. Hence 3 is a hook length for λ. Now, by Lemma 2.2, S_{F}^{λ} is reducible in characteristic 3.

In view of Corollary 3.2 we may (and will) assume that our globally irreducible Specht module S_{ℤ}^{λ} satisfies the condition B ≥ max{2, A}. Let q denote the maximal prime in the interval [2, B].

Lemma 3.3. Assume that S_{ℤ}^{λ} is globally irreducible, and B ≥ max{2, A}. If f = k, then λ = (n).

Proof. By Lemma 3.1, the hook lengths of λ cover the interval [1, A_{1,k} + B_{1,k} − 1]. We claim that a_{i} < q for all i ≤ k. Indeed, suppose a_{i} ≥ q for some i. Then

A_{1,k} + B_{1,k} − 1 ≥ b_{k} + a_{i} − 1 ≥ 2q − 1.

So the hook lengths of λ cover (q, 2q). By Lemma 2.10, there is a prime p > q such that p is a hook length of λ. Now Lemma 2.2 leads us to a contradiction (it applies as p > B). Thus a_{i} < q for all i, that is λ is q-regular. But then λ should satisfy Carter’s condition (see Proposition 2.1).

Observe that

h(A_{1,k}, b_{k} − (q − 1)) = q.

Therefore, a_{k} = 1 since otherwise

h(A_{1,k} − 1, b_{k} − (q − 1)) = q + 1

and λ does not satisfy Carter’s condition. If k = 1 this implies that λ = (n). So suppose k > 1. Then

h(A_{1,k} − 1, b_{k} − (q − 1)) = q + 1 + b_{k−1}.

Now Carter’s condition forces b_{k−1} ≡ −1(mod q). In particular, b_{k−1} ≥ q − 1. This gives A_{1,k} + B_{1,k} − 1 ≥ 2 + b_{k} + b_{k−1} − 1 ≥ 2q, so that [1, A_{1,k} + B_{1,k} − 1] ⊇ (q, 2q). Again Lemmas 2.10 and 2.2 lead us to a contradiction completing the proof.

Now we are in a position to finish the proof of Theorem B.

Theorem 3.4. Suppose S_{ℤ}^{λ} is globally irreducible and B ≥ max{2, A}. Then λ = (n).

Proof. By Lemma 3.3, we may assume that f < k. Applying Lemma 2.8(i) with r = f, s = f + 1 shows that

a_{f} + a_{f+1} ≥ (p − 1)b_{f} + p  (4)

for any prime p dividing a_{f} + a_{f+1} + b_{f}.

If an odd prime divides a_{f} + a_{f+1} + b_{f}, then (4) implies that a_{f} + a_{f+1} ≥ 2b_{f} + 3. In this case, either a_{f} > b_{f} or a_{f+1} > b_{f} contrary to the choice of b_{f}. Thus we must have

a_{f} + a_{f+1} + b_{f} = 2^{m}.  (5)

Combining this with (4) we get

a_{f} + a_{f+1} ≥ b_{f} + 2.  (6)

Assume f > 1. Applying Lemma 2.8(i) with r = f − 1, s = f + 1 gives that

a_{f−1} + a_{f} + a_{f+1} ≥ (p − 1)(b_{f−1} + b_{f}) + p  (7)

for any prime p dividing a_{f−1} + a_{f} + a_{f+1} + b_{f} + b_{f−1}. If an odd prime divides this number, then (7) yields

a_{f−1} + a_{f} + a_{f+1} ≥ 2b_{f−1} + 2b_{f} + 3.  (8)

By Lemma 3.1, the hook lengths of λ cover [1, a_{f} + a_{f−1} + b_{f} + b_{f−1} − 1]. If

a_{f} + a_{f−1} + b_{f} + b_{f−1} − 1 ≥ 2b_{f} − 1,

then [1, a_{f} + a_{f−1} + b_{f} + b_{f−1} − 1] ⊇ (q, 2q). So again Lemma 2.10 ensures that there exists a prime p_{1} > q which is a hook length of λ, and again Lemma 2.2 shows that this contradicts the global irreducibility of S_{ℤ}^{λ}. Thus

a_{f} + a_{f−1} + b_{f} + b_{f−1} − 1 < 2b_{f} − 1

which is equivalent to

2b_{f} − b_{f−1} > a_{f} + a_{f−1} + b_{f}.

However, this violates (8) as b_{f} ≥ a_{f+1}. So we conclude that

a_{f−1} + a_{f} + a_{f+1} + b_{f} + b_{f−1} = 2^{M}  (9)

for some M > m. From (5) and (9) we have

a_{f−1} + b_{f−1} = 2^{M} − 2^{m} = 2^{m}(2^{M−m} − 1) ≥ 2^{m}.

Therefore, either a_{f−1} ≥ 2^{m−1} or b_{f−1} ≥ 2^{m−1}. Now (6) says that

b_{f} < \frac{1}{2}(a_{f} + a_{f+1} + b_{f}) = 2^{m−1}.

So b_{f} < a_{f−1} or b_{f} < b_{f−1} contrary to the choice of f.

Thus we may assume that f = 1. If b_{2} ≥ 2, then applying Lemma 2.8(ii) gives that there exists an odd prime p for which

a_{1} + a_{2} ≥ (p − 1)(b_{1} + 2) + 1  (10)

(indeed, a_{1} + a_{2} + b_{1} + 1 is odd by (5)). As p ≥ 3, (10) implies that a_{1} + a_{2} ≥ 2b_{1} + 5, whence either a_{1} > b_{1} or a_{2} > b_{1}. Since this is impossible we must have b_{2} = 1.

Applying Lemma 2.8(i) to λ′ instead of λ we obtain that

1 + b_{1} ≥ (p − 1)a_{2} + p

for any prime p dividing a_{2} + b_{1} + 1. If there is an odd prime with this property, then 1 + b_{1} ≥ 2a_{2} + 3. Combining this with (6) (which is a_{1} + a_{2} ≥ b_{1} + 2 as f = 1) we get a_{1} ≥ a_{2} + 4. Therefore, a_{1} + a_{1} − 4 ≥ a_{1} + a_{2} ≥ b_{1} + 2 yielding

a_{1} ≥ \frac{b_{1}}{2} + 3.

So [1, a_{1} +b_{1} −1] ⊃ (b_{1}, \frac{3}{2}b_{1} +2]. Due to Lemmas 2.11 and 3.1 one of the hooks of the Young diagram of λ has length equal to a prime greater than q. By Lemma 2.2, this is impossible. Thus a_{2} + b_{1} + 1 = 2^{s}. In view of (5) we also have a_{1} + a_{2} + b_{1} = 2^{m}. So

a_{1} − 1 = 2^{s}(2^{m−s} − 1),

and m ≥ s. If m > s we get b_{1} < 2^{s} ≤ a_{1} − 1 which is false as 1 = f. If m = s we conclude that a_{1} = 1. But then a_{1} + a_{2} ≥ b_{1} + 2 forces a_{2} > b_{1} leading us to a contradiction with the choice of b_{f}. This contradiction proves the theorem.

References

[1] R. W. Carter and M. T. J. Payne, On homomorphisms between Weyl modules and Specht modules, Math. Proc. Camb. Phil. Soc. 87 (1980), 419–425.

[2] C. W. Curtis and I. Reiner, Representation theory of finite groups and associative algebras, John Wiley and Sons (Interscience), New York, 1962.

[3] W. and F. Ellison, Prime numbers, John Wiley and Sons, Hermann, 1985.

[4] G. D. James, On the decomposition matrices of the symmetric groups, II, J. Algebra 43 (1976), 45–54.

[5] G. D. James, On a conjecture of Carter concerning irreducible Specht modules. Math. Proc. Camb. Phil. Soc. 83 (1978), 11–17.

[6] G. D. James, The representation theory of the symmetric groups, Springer Lecture Notes 682, Berlin, Heidelberg, New York, 1978.

[7] G. D. James and G. E. Murphy, The determinant of the Gram matrix for a Specht module. J. Algebra 59 (1979), 222–235.

[8] P. Ribbenboim, The book of prime number records, Springer–Verlag, New York, Berlin, Heidelberg, 1988.

[9] J. G. Thompson, Finite groups and even lattices. J. Algebra 38 (1976), 523–524.

[10] J. G. Thompson, A simple subgroup of E_{8}(3). In: N. Iwahori (ed.), Finite groups symposium, Japan Soc. for promotion of Science, 1976, pp. 113–116.

[11] Pham Huu Tiep, Globally irreducible representations of finite groups and integral lattices, Geometriae Dedicata 64 (1997), 85–123.

Department of Mathematics, University of Oregon, Eugene, OR 97403, USA

E-mail address: klesh@math.uoregon.edu

Department of Mathematics, University of Manchester, Oxford Road, M13 9PL, UK

E-mail address: sashap@ma.man.ac.uk