Prime power degree representations of the double covers of the symmetric and alternating groups

Christine Bessenrodt, Jørn B. Olsson

September 29, 2000

1 Introduction

In 1998, Zalesskii proposed to classify all instances of irreducible characters of quasi-simple groups which are of prime power degree. In joint work, Malle and Zalesskii then dealt with all quasi-simple groups with the exception of the alternating groups and their double covers [5]. In an earlier article [1] we have classified all the irreducible characters of S_{n} of prime power degree and have deduced from this also the corresponding classification for the alternating groups. In the present article we complete Zalesskii’s programme by dealing with the final case left open in [5], the double covers of the alternating groups. We derive this result from a corresponding result on the double covers of the symmetric groups. If one is only interested in spin characters, one easily sees that only 2-powers can occur as prime power degrees. But from a combinatorial point of view it is natural to ask more generally: when is the number of shifted standard tableaux of a given shape a prime power? Since our method is independent of the prime, we will answer this question, showing that apart from a few accidental cases for small n only the ’obvious’ partitions satisfy the prime power condition. Thus in turn for the spin characters, apart from exceptions for small n, the ’obvious’ spin characters of 2-power degree are indeed the only ones (this confirms the conjecture stated in [5]).

The paper is organized as follows. In section 2, some results on bar lengths in strict partitions are proved. These results lead to an algorithm which allows us to show that every strict partition of n whose shifted degree is a prime power has a large bar. In section 3, we obtain a new result concerning the degree polynomial of spin characters resp. the shifted tableaux count (Theorem 3.1). In section 4 we prove Theorem 2.3, the main result. To do so, we combine the algorithm above with Theorem 3.1 and some analysis of special ’small’ cases. Some numerical work was done by the aid of a computer. We then derive the classification of irreducible spin characters of the double covers of S_{n} and A_{n} of prime power degree.

Acknowledgements. The authors are grateful to the Danish Natural Science Foundation for the support of their cooperation on this work.

2 An algorithm for bar lengths

We refer to [3], [4], [6], [8] for details about partitions, Young diagrams, shifted diagrams, tableaux, hooks and bars. Consider a partition λ = (λ_{1}, λ_{2}, …, λ_{m}) of the integer n into distinct parts. Thus λ_{1} > λ_{2} > ⋯ > λ_{m} > 0 and λ_{1} + λ_{2} + … + λ_{m} = n. We call the λ_{i}’s the parts of λ and m the length l(λ) of λ. The shifted diagram of λ consists of n nodes (boxes) with λ_{i} nodes in the ith row, indented along the diagonal (see the example below). We refer to the nodes in matrix notation, i.e. the (i, j)-node is the jth node in the ith row. Thus

is the shifted diagram of λ = (5, 4, 1). A node (i, λ_{i}) is called a removable node if λ_{i} > λ_{i+1} +1, so that after removing the node the resulting partition is again a partition into distinct parts. The length b_{ij} of the (i, j)-bar is the length of the (i, j +i)-hook in the shiftsymmetric diagram to λ, which is obtained from λ by a reflection along the diagonal. The shift-symmetric diagram to λ = (5, 4, 1) and the bar lengths filled into the corresponding nodes of λ are depicted by

. 9 6 5 3 2
. . 5 4 2 1
… 1
. .
. .

We put b_{i} = b_{1i} for the first row bar lengths, abbreviated by frb. Moreover, we let B(λ) = ∏_{i,j} b_{ij} be the product of all bar lengths in λ. Then the number of shifted standard tableaux g_{λ} of λ is given by the Bar formula:

g_{λ} = \frac{n!}{B(λ)} .

We will call g_{λ} the shifted degree of the partition λ. It is known that the degree of the complex irreducible spin character of the double cover S̃_{n} of the symmetric group S_{n} labelled by λ is then obtained as

⟨λ⟩(1) = 2^{[ \frac{n−l(λ)}{2} ]} g_{λ} .

(see [8]). In particular, for λ = (n), g_{(n)} = 1 and the (basic) spin character ⟨n⟩ is of 2-power degree 2^{[ \frac{n−1}{2} ]}.

We first have to deal with the case of two-part partitions.

Lemma 2.1 Let λ = (n−k, k) be a partition of n with 0 < k < n−k. Then

g_{λ} = \frac{n − 2k}{k} \binom{n − 1}{k − 1} .

This is a power of the prime p if and only if either λ = (n − 1, 1) and n = 2 + p^{a} for some r, or λ = (3, 2), (4, 2), (5, 2), (4, 3), (8, 2) with p = 2, 5, 3, 5, 3 respectively.

Proof. Using the Bar formula we obtain immediately

\begin{aligned}g_{λ} & = & \frac{n!}{n · (n − k)(n − k − 1) ⋯ (n − 2k + 1) · (n − 2k − 1)! · k!} \\ & = & \frac{n − 2k}{k} \binom{n − 1}{k − 1} . \end{aligned}

One easily checks that for all the partitions stated in the assertion the shifted degree is indeed a power of the given prime p, as described in the Lemma. In particular, the assertion of the Lemma holds for n ≤ 6. So we may now assume that n ≥ 7. For the converse part of the Lemma we now consider a partition λ = (n − k, k) with 0 < k < n − k which satisfies g_{λ} = p^{a} for a prime p and an integer a ∈ ℕ. Since k < n − k, k ≤ \frac{n−1}{2} and hence, since n ≥ 7, k ≤ \frac{(n−2)(n−3)}{6}. Since \binom{n−1}{k−1} is a divisor of p^{a}k, and since any prime power dividing this binomial coefficient is bounded by n − 1 [9], we obtain

\binom{n − 1}{k − 1} ≤ (n − 1)k ≤ \frac{(n − 1)(n − 2)(n − 3)}{6} = \binom{n − 1}{3} .

Since n ≥ 7, the monotonicity of the binomial coefficients then yields k ≤ 4. We now discuss all these cases in turn. For k = 1, the partition has the form λ = (n − 1, 1) with g_{λ} = n − 2, and hence n = 2 + p^{a} as stated in the assertion. For k = 2, the equation p^{a} = \frac{1}{2}(n − 4)(n − 1) leads immediately to p = 3. If n − 4 = 3^{j} for some j ∈ ℕ, then n − 1 = 3^{j} + 3 = 2 · 3^{a−j}. But then a − j = 1 and hence n = 7. This corresponds to the partition (5, 2) in our list, which is of shifted degree 9. In the second case, when n − 1 = 3^{j} for some j ∈ ℕ, j > 1, we have n − 4 = 3^{j} − 3 = 2 · 3^{a−j}, and thus again a − j = 1 and then n = 10. This corresponds to the partition (8, 2) in our list, which is of shifted degree 27. For k = 3, the condition p^{a} = \frac{1}{6}(n − 6)(n − 1)(n − 2) leads to (n − 6)(n − 1)(n − 2) = 6 p^{a}; since n − 1, n − 2 cannot both be p-powers, this implies n − 1 = 6 or n − 2 = 6. The first case gives the example (4, 3) of shifted degree 5, while the second case leads to a contradiction. For k = 4, we obtain similarly as above (n − 8)(n − 1)(n − 2)(n − 3) = 24 p^{a}. Using again that at least one of the consecutive factors is not a p-power, one easily sees that this case cannot occur. ⋄

The following result includes in particular a ”spin analogue” of a theorem of Burnside, describing the minimal shifted degrees for n sufficiently large:

Proposition 2.2 Let λ be a partition of n into distinct parts.

(i) For λ = (n), g_{λ} = 1.

(ii) For n ≥ 3 and λ = (n − 1, 1), g_{λ} = n − 2.

(iii) For n ≥ 5 and λ = (n − 2, 2), g_{λ} = \frac{1}{2}(n − 1)(n − 4).

(iv) Some special cases for small n:

n  λ  g_{λ}
6 (3, 2, 1) 2
7 (4, 3)  5
(4, 2, 1) 7
8 (5, 3) 14
(4, 3, 1) 12
9 (5, 4) 14
(4, 3, 2) 12
10 (4, 3, 2, 1) 12

If λ is not a partition in one of the cases above (in particular, n ≥ 8), then g_{λ} > \frac{1}{2}(n − 1)(n − 4) > n.

Proof. The degrees occurring in the statement of the Proposition can all easily be computed by the Bar formula or they have already appeared in the previous Lemma.

We now prove by induction on n the main assertion that for sufficiently large n the degrees in cases (i)-(iii) are the minimal shifted degrees, using the branching property:

g_{λ} = ∑_{A} g_{λ\A}

where A runs through the removable nodes of λ.

So let λ be a partition of n into distinct parts, not occurring in one of the cases of the statement of the Proposition. We assume first that λ has at least 2 removable nodes. If λ does not arise by adjoining a node to one of the partitions in the statement of the Proposition, then we obtain by induction g_{λ} > (n − 2)(n − 5), which is greater than \frac{1}{2}(n − 1)(n − 4), as n ≥ 7. Now we discuss the situation where λ is an extension of one of the partitions in the list by a node. If λ = (n − 3, 3), then g_{λ} = \frac{1}{6}(n − 1)(n − 2)(n − 6), which is greater than \frac{1}{2}(n − 1)(n − 4) for n > 8 (the case n = 8 gives the example (5, 3) in case (iv) of the Proposition). If λ = (n − 3, 2, 1), then for n ≥ 9 we obtain by induction:

g_{λ} = \frac{1}{2}(n − 2)(n − 5) + g_{(n−4,2,1)} > (n − 2)(n − 5)

which is greater than \frac{1}{2}(n − 1)(n − 4). For n = 8 we have g_{(5,2,1)} = 9 + 7 = 16 > 14. For the partitions (5, 3, 1), (5, 2, 1), (6, 4), (5, 4, 1), (5, 3, 2), (5, 3, 2, 1) one easily checks that the assertion holds.

Now we turn to the case where λ has only one removable node. First we consider the case where λ = (k, k − 1, …, r) for some r ∈ ℕ, 1 < r < k. Note that n ≥ 11. Using branching and induction (and an easy check for (6, 5)) we have:

g_{λ} = g_{(k,... ,r)} = g_{(k,... ,r+1,r−1)}
= g_{(k,... ,r+1,r−2)} + g_{(k,... ,r+2,r,r−1)}
> (n − 3)(n − 6)

For n ≥ 11, this is greater than \frac{1}{2}(n − 1)(n − 4), as claimed. We finally deal with the case where r = 1; here we know that n ≥ 15. Again, by branching and induction we obtain

g_{λ} = g_{(k,... ,2,1)} = g_{(k,... ,2)} = g_{(k,... ,3,1)}
= g_{(k,... ,3)} + g_{(k,... ,4,2,1)}
> (n − 4)(n − 7)

For n ≥ 14, this is greater than \frac{1}{2}(n − 1)(n − 4), as claimed. ⋄

Our aim is to prove:

Theorem 2.3 Let λ be a partition of n into distinct parts. Then g_{λ} = p^{a} for some prime p and a ∈ ℕ, if and only if one of the following occurs:

n = p^{a} + 2 , λ = (p^{a} + 1, 1)

or we are in one of the following exceptional cases:

n = 5 : λ = (3, 2), g_{λ} = 2
n = 6 : λ = (4, 2), g_{λ} = 5
λ = (3, 2, 1), g_{λ} = 2
n = 7 : λ = (5, 2), g_{λ} = 9
λ = (4, 3), g_{λ} = 5
λ = (4, 2, 1), g_{λ} = 7
n = 8 : λ = (5, 2, 1), g_{λ} = 16
n = 10 : λ = (8, 2), g_{λ} = 27

For the rest of the section we assume:

λ is a partition of n into distinct parts, of length m > 2.

First we state some elementary results about bar lengths.

Lemma 2.4 Let a, b be any two different bar lengths in λ. Then

a + b ≤ λ_{1} + n

resp. equivalently:

n − λ_{1} ≤ 2n − a − b .

Proof. Suppose a > b. We have a ≤ b_{1} = λ_{1} + λ_{2} and b ≤ b_{2} = λ_{1} + λ_{3}. Now

λ_{1} + n = (λ_{1} + λ_{2}) + (λ_{1} + λ_{3}) + (λ_{4} + ⋯ + λ_{m}) ≥ b_{1} + b_{2} ≥ a + b

⋄

In addition to our previous assumption on λ, we also assume now: g_{λ} is a prime power.

Proposition 2.5 Let n ≥ 8. If q is a prime for which n − λ_{1} < q ≤ n, then

q, 2q, …, [\frac{n}{q}] q

are all frb’s of λ.

Proof. Put w = [\frac{n}{q}], so n = wq + r, with 0 ≤ r < q. Note that q ≥ 5, since otherwise n − λ_{1} ≤ 2, contradicting m > 2. We have

n − λ_{1} = λ_{2} + … + λ_{m} < q

so that bars of length divisible by q are all frb’s. Now

λ_{1} > n − q = (w − 1)q + r ≥ (w − 1)q .

If a mixed bar λ_{1} + λ_{i} has length divisible by q, then

q | λ_{1} + λ_{i} > λ_{1} ≥ (w − 1)q

whence λ_{1} + λ_{i} = wq. In this case, the result follows from [8], (4.3). We now assume that only unmixed bars have length divisible by q. The unmixed frb’s are in the set

{1, 2, …, λ_{1}} \ {λ_{1} − λ_{i} | i = 2, …, m}

If for some 1 ≤ w_{1} ≤ w − 1, w_{1}q is not an frb, then we obtain λ_{1} − λ_{i} = w_{1}q for some i > 1. Then

q > λ_{i} = λ_{1} − w_{1}q > (w − 1 − w_{1})q ≥ 0

yielding w_{1} = w − 1. Thus only (w − 1)q and wq may not be frb’s. If one or both of them is not an frb, then g_{λ} is a q-power. If only one of them is not an frb, then

g_{λ} | ((w − 1)q)_{q} or g_{λ} | (wq)_{q}

and hence g_{λ} ≤ n, a contradiction in view of Proposition 2.2. If both are not frb’s, then

g_{λ} | ((w − 1)qwq)_{q} ≤ qn .

If w ≥ 3, then q ≤ \frac{n}{3}, and hence g_{λ} ≤ \frac{n^{2}}{3} . But this contradicts Proposition 2.2 since for n ≤ 15, \frac{n^{2}}{3} ≤ \frac{1}{2}(n − 1)(n − 4). In the remaining case w = 2 we have g_{λ} | q^{2} ≤ \frac{n^{2}}{4} , and this again contradicts Proposition 2.2. Hence also both (w − 1)q and wq are frb’s, as claimed. ⋄

As the classification result is easily checked for n ≤ 28, we assume from now on also that n > 28.

Lemma 2.6 If q is a prime with \frac{n}{2} < q ≤ n, then q is an frb of λ.

Proof. As n > 28, we may choose different primes p_{1}, p_{2} with \frac{3}{4}n < p_{1}, p_{2} ≤ n (see [2]). Then λ has to contain bars of length p_{1} and p_{2} since otherwise g_{λ} = p_{i} ≤ n, a contradiction because of Proposition 2.2. By Lemma 2.4

λ_{1} ≥ p_{1} + p_{2} − n > \frac{n}{2}

so that n − λ_{1} < \frac{n}{2}. If q > \frac{n}{2}, then q > n − λ_{1}. Now apply Proposition 2.5. ⋄

We now have the exact analogue of the crucial result in [1]:

Proposition 2.7 Suppose we have sequences of integers s_{1} < s_{2} < ⋯ < s_{r} ≤ n, t_{1} < t_{2} < ⋯ < t_{r} ≤ n satisfying

(i) s_{i} < t_{i} for all i;

(ii) s_{1} and t_{1} are primes > \frac{n}{2};

(iii) For 1 ≤ i ≤ r − 1, s_{i+1} and t_{i+1} contain prime factors exceeding 2n − s_{i} − t_{i}.

Then s_{1}, …, s_{r}, t_{1}, …, t_{r} are all frb’s of λ.

Proof. Induction on i to show that s_{i} and t_{i} are frb’s for λ; use Lemma 2.6, Lemma 2.4 and Proposition 2.5. ⋄

We get an algorithm from Proposition 2.7 which shows that b_{1} is large and thus λ is “almost” a two-part partition: start with two large primes s_{1} < t_{1} close to n. Then 2n − s_{1} − t_{1} is small. Choose if possible two integers s_{2} and t_{2} with s_{2} < t_{2}, s_{1} < s_{2} ≤ n, t_{1} < t_{2} ≤ n each having a prime divisor exceeding 2n − s_{1} − t_{1}. Then 2n − s_{2} − t_{2} < 2n − s_{1} − t_{1}. Choose if possible two integers s_{3} and t_{3} with s_{3} < t_{3}, s_{2} < s_{3} ≤ n, t_{2} < t_{3} ≤ n each having a prime divisor exceeding 2n − s_{2} − t_{2} and so on. If this process reaches s_{r}, t_{r}, then t_{r} ≤ b_{1} = λ_{1} + λ_{2} by Proposition 2.7. ¿From [1] we quote a result that shows that for large n the algorithm always terminates close to n. First we need some notation. Suppose that n ≥ 3 is a positive integer. Consider two finite increasing sequences of integers {A_{i}} and {B_{i}} which satisfy the following properties:

(i) A_{1} < B_{1} ≤ n are two “large” primes not exceeding n.

(ii) For every i, we have that

A_{i} < B_{i} ≤ n.

(iii) If B_{i} < n, then A_{i+1} < B_{i+1} are integers not exceeding n each with a prime factor exceeding 2n − A_{i} − B_{i}.

Then denote by A(n) (resp. B(n)) the largest integer in such a sequence {A_{i}} (resp. {B_{i}}).

Theorem 2.8 [1] If n > 3.06 · 10^{8}, then there is a pair of sequences {A_{i}} and {B_{i}} as above for which

n − B(n) ≤ 225.

Corollary 2.9 Let λ be a partition of n into distinct parts with largest bar length b_{1}. If g_{λ} is a prime power and n > 3.06 · 10^{8}, then n − b_{1} ≤ 225.

3 Degree polynomials

Let μ = (μ_{1}, …, μ_{m}) be a partition of k into distinct parts. For n > k + μ_{1}, set λ = (n − k, μ_{1}, …, μ_{m}). Denote by μ̂ = (d_{1}, …, d_{k}) the shift-symmetric partition of 2k associated to μ, possibly extended by zero parts. With this notation we have

Theorem 3.1 The number of standard shifted tableaux of shape λ is given by

g_{λ} = \frac{1}{B(μ)} ∏_{i=1}^{k}(n − d_{i} − k + i) .

Thus, viewing the degree as a function g_{μ}(n) of n, this is a polynomial of degree k in n.

Before we embark on the proof, we illustrate this by an example.

Example. Let k = 5, μ = (4, 1). Thus μ̂ = (5, 3, 1, 1, 0) and B(μ) = 5 · 4 · 2 · 1 · 1. Choose n > 9, so λ = λ(n) = (n − 5, 4, 1) is a partition of n into distinct parts. Then

g_{λ} = \frac{1}{40} n(n − 2)(n − 3)(n − 6)(n − 9) .

Proof of Theorem 3.1. We have already mentioned before that the bar lengths in a partition are the hook lengths in the upper half of its associated shift-symmetric partition; let λ̂ be the shift-symmetric partition associated with λ, then we have for i < j:

(∗) h_{ij}(λ̂) = b_{i j−i}(λ)

The partition μ̂ has largest part μ_{1} +1 and length μ_{1}. Let Let ν be obtained by adding a largest part n − k to μ̂. If X is the set of first column hook lengths for μ̂, then Y = X∪{n−k+μ_{1}} is the set of first column hook lengths for ν. Obviously, ν is just λ̂ with the first column removed. Therefore the set H of first row hook lengths of ν is

H = {h_{1j}(λ̂) | j ≥ 2}

Using (∗) we obtain

H = {b_{1 j−1}(λ) | j ≥ 2}

the set of first row bar lengths of λ. Denote the product of the numbers in H by B_{1}. Then

(∗∗) B(λ) = B_{1}B(μ) .

In the following, we use the notation and some results from [8] resp. [7]. We extend the β-set Y for ν to a β-set Z for ν of cardinality k + 1. Thus

Z = Y^{+((k+1)−(μ_{1}+1))} = Y^{+(k−μ_{1})} .

Since Y = X ∪{n−k+μ_{1}}, we have Z = X^{+(k−μ_{1})}∪{n}. Set X̂ = X^{+(k−μ_{1})}. Since Z is a β-set for ν, we obtain

H = {1, 2, …, n} \ {n − d | d ∈ X̂}

and thus

B_{1} = \frac{n!}{∏_{d∈X̂}(n − d)}

As X̂ = {d_{i} + k − i | i = 1, …, k}, now (∗∗) implies

B(λ) = B(μ)\frac{n!}{∏_{i=1}^{k}(n − d_{i} − k + i)}

Since g_{λ} = \frac{n!}{B(λ)}, the assertion now follows. ⋄

As an application of this result we prove the classification theorem in the case where n − b_{1} ≤ 4, where as before b_{1} = λ_{1} + λ_{2} is the largest bar length in a partition λ of n into distinct parts.

Proposition 3.2 Let λ be a partition of n with n − b_{1} ≤ 4 and g_{λ} a prime power greater than 1. Then λ is one of the partitions given in Theorem 2.3.

Proof. We prove the result by a case-by-case analysis using Theorem 3.1. Set c = n − b_{1}. Let λ = (λ_{1}, …, λ_{m}) as before, and assume g_{λ} = p^{a} for a prime p and a ∈ ℕ. If c = 0, then λ is a 2-part partition and the assertion is true by Lemma 2.1.

Set k = n − λ_{1} and μ = (λ_{2}, …, λ_{m}). Denote by μ̂ = (d_{1}, …, d_{k}) the shift-symmetric partition of 2k associated to μ, possibly extended by zero parts. For n ≤ 15 it is easy to check the assertion directly, so we now assume that n > 15. If c = 1, then μ̂ = (k, 3, 1^{k−3}, 0) and hence by Theorem 3.1 we know:

\begin{matrix}p^{a} · B(μ) & = & p^{a} · k · (k − 1) · (k − 3)! \\ & = & n · (n − 2) ⋯ (n − k + 2) · (n − k − 1) · (n − 2k + 1) \end{matrix} (∗)

So

p^{a} · \binom{k}{2} = \frac{1}{2}n · \binom{n − 2}{k − 3} · (n − k − 1) · (n − 2k + 1)

Again, we use the fact that any prime power in a binomial coefficient \binom{s}{t} is at most s [9] and thus we deduce:

p^{a} ≤ \frac{1}{2}n · (n − 2) · (n − k − 1) · (n − 2k + 1) .

Inserting this into (∗) and cancelling, we obtain

(n − 3) ⋯ (n − k + 2) ≤ \frac{1}{2}k · (k − 1) · (k − 3)! .

Since 2(k − 1) ≤ n − 3 and \frac{3}{2}k ≤ n − 4 (here we use n > 12), we obtain

(n − 5) ⋯ (n − k + 2) ≤ 4 ⋯ (k − 3) .

But this implies n − 5 ≤ k − 3 and hence the contradiction n ≤ k + 2. If c = 2, then μ̂ = (k − 1, 4, 2, 1^{k−5}, 0^{2}) and hence

p^{a} · B(μ) = p^{a} · k (k − 2) (k − 3) · (k − 5)! · 2
= n (n − 1) · (n − 3) ⋯ (n − k + 3) · (n − k + 1)·
·(n − k − 2) (n − 2k + 2)

So

p^{a} · k · \binom{k − 2}{2} = \frac{1}{2}\binom{n}{2} · \binom{n − 3}{k − 5} · (n − k + 1) · (n − k − 2) · (n − 2k + 2)

and hence, similarly as before,

(n − 1) · (n − 4) ⋯ (n − k + 3) ≤ k · (k − 2) · (k − 3) · (k − 5)!

Using n − 1 ≥ 2k and n − 4 ≥ 12, we obtain

(n − 5) ⋯ (n − k + 3) ≤ 5 ⋯ (k − 5) · (k − 3) · (k − 2)

which implies n − 5 ≤ k − 2, and hence the contradiction n ≤ k + 3. Similar arguments also lead to a contradiction in the cases where c = 3 resp. c = 4 (here one has to consider two possibilities for μ in each case), so we have indeed no partition of n ≥ 16 of shifted prime power degree with 1 ≤ c ≤ 4. ⋄

4 Classification results

We start by proving our main result, Theorem 2.3, the combinatorial classification of partitions of prime power shifted degree stated in section 2.

Proof of Theorem 2.3. We have already remarked before that for n up to 28 it is easily checked using Maple that the result holds. For n in the range from 29 to 9.25 · 10^{8} it was already checked for the proof of the classification result in [1] that the following holds: Given n, let p_{1}, p_{2} be the two largest primes below n. Then there is a prime divisor q of n(n − 1)(n − 2)(n − 3)(n − 4) with p_{1} + p_{2} + q > 2n. Hence we can apply the results in section 2 and obtain n − b_{1} ≤ 4. Thus Proposition 3.2 yields the result in this case. If n is larger than 3.06 · 10^{8} we apply Theorem 2.8 resp. Corollary 2.9 to obtain c = n − b_{1} ≤ 225. Because of Proposition 3.2 we may assume in this case that c ≥ 5 (and so k ≥ 7), and we now have to show how to reach a contradiction. As before, we let λ = (λ_{1}, …, λ_{m}), k = n − λ_{1}, μ = (λ_{2}, …, λ_{m}) and μ̂ = (d_{1}, …, d_{k}) the shift-symmetric partition of 2k associated to μ, possibly extended by zero parts. We assume g_{λ} = p^{a} for a prime p and a ∈ ℕ. First we want to estimate the number of different parts in μ̂. These arise from parts of μ differing at least by 2 and their reflections in the shiftsymmetric diagram. Hence, if l ∈ ℕ is maximal with ∑_{i=0}^{l−l} i = l^{2} ≤ k then μ̂ has at most 2l + 1 different parts (including possibly a zero part). But since equal parts in μ̂ induce consecutive factors f_{i} = n − d_{i} − k + i in the product

∏_{i=1}^{k}(n − d_{i} − k + i) = g_{λ}B(μ) = p^{a} B(μ)

(see Theorem 3.1), at least [\frac{k−2l−1}{2}] factors f_{i} are not divisible by p. As the minimal factor in the product is f_{1} = n − d_{1} − k + 1, this implies

B(μ) ≥ \begin{matrix}[ & \frac{k−2l−1}{2} & ] \\ & ∏ \\ & i=1 \end{matrix}(n − d_{1} − k + i) .

¿From the proof of Theorem 2.8 (see [1]) we know not only that c = n−b_{1} = n − (λ_{1} + λ_{2}) ≤ 225, but also that n − b_{2} = n − (λ_{2} + λ_{3}) ≤ 450. Hence

k ≤ (n − λ_{1}) + (n − (λ_{1} + λ_{2} + λ_{3})) ≤ 675 .

Thus, as μ_{1} = k − c, we obtain

n − d_{1} − k + 1 = n − μ_{1} − k = n − 2k + c ≥ n − 1345 .

Hence

B(μ) ≥ (n − 1345)^{[^{\frac{k−2l−1}{2}}]} .

But on the other hand B(μ) ≤ k!, and

k! ≤ N^{[^{\frac{k−2l−1}{2}}]} = N^{[^{\frac{k−1}{2}}]−[\sqrt{k}]}

for 7 ≤ k ≤ 675 and N ≥ 4 · 10^{5} (as may easily be checked with Maple), giving the desired contradiction. ⋄

Remark. The arguments applied above are sufficiently efficient to deal also with larger bounds for n − b_{1}. Thus, using a variant of Theorem 2.8 with a larger bound for n − B(n) and a smaller lower bound for n would reduce the necessary computer checking for “mid-sized” n considerably.

¿From the combinatorial classification result we can now easily deduce the corresponding classification result for the spin characters of the double covers of the symmetric groups:

Theorem 4.1 Let n ≥ 4 and let λ be a partition of n into distinct parts. Then the spin character ⟨λ⟩ of S̃_{n} is of p-power degree for a prime p if and only if p = 2 and one of the following cases occurs:

(i) λ = (n), i.e. the character is a basic spin character, and in this case ⟨n⟩(1) = 2^{[ \frac{n−1}{2} ]}.

(ii) n = 2 + 2^{a} for some a ∈ ℕ, and λ = (n − 1, 1), and in this case ⟨n − 1, 1⟩(1) = 2^{a+2^{a−1}}.

(iii) The spin character belongs to the following list of exceptional cases:

\begin{aligned}⟨3, 2⟩(1) & = & 4 \\ ⟨3, 2, 1⟩(1) & = & 4 \\ ⟨5, 2, 1⟩(1) & = & 64 \end{aligned}

Proof. Assume that the spin character ⟨λ⟩ is of prime power degree p^{a}. Since 2^{[ \frac{n−l(λ)}{2} ]} divides the degree of ⟨λ⟩, it is clear that p = 2. Hence g_{λ} must be a 2-power as well, and then the result follows immediately from Theorem 2.3. ⋄

Applying this, we also obtain the classification result for the spin characters of the double cover Ã_{n} of the alternating groups, confirming the conjecture stated in [5]:

Theorem 4.2 Let λ be a partition of n into distinct parts. Then the spin character ⟨⟨λ⟩⟩ of Ã_{n} is of p-power degree for a prime p if and only if p = 2 and one of the following cases occurs:

(i) λ = (n), i.e. the character is a basic spin character, and in this case ⟨⟨n⟩⟩(1) = 2^{[ \frac{n−2}{2} ]}.

(ii) n = 2 + 2^{a} for some a ∈ ℕ, and λ = (n − 1, 1), and in this case ⟨⟨n − 1, 1⟩⟩(1) = 2^{a−1+2^{a−1}}.

(iii) The spin character belongs to the following list of exceptional cases:

\begin{aligned}⟨⟨3, 2⟩⟩(1) & = & 4 \\ ⟨⟨3, 2, 1⟩⟩(1) & = & 4 \\ ⟨⟨5, 2, 1⟩⟩(1) & = & 64 \end{aligned}

Proof. The degree of the spin character ⟨⟨λ⟩⟩ of Ã_{n} is related to that of the spin character ⟨λ⟩ of S̃_{n} by:

⟨⟨λ⟩⟩(1) = \{ \begin{matrix}⟨λ⟩(1) & if n − l(λ) is odd \\ \frac{1}{2}⟨λ⟩(1) & if n − l(λ) is even \end{matrix}

Hence the result follows immediately from the previous classification theorem. ⋄

References

[1] A. Balog, C. Bessenrodt, J. B. Olsson, K. Ono, Prime power degree representations of the symmetric and alternating groups, preprint 2000

[2] H. Harborth, A. Kemnitz, Calculations for Bertrand’s postulate, Math. Mag. 54 (1981) 33-34

[3] P. N. Hoffman, J. F. Humphreys, Projective representations of the symmetric groups, Clarendon Press, Oxford 1992 [4] G. James, A. Kerber, The representation theory of the symmet-

ric group. Encyclopedia of Mathematics and its Applications, 16, Addison-Wesley 1981

[5] G. Malle, A. Zalesskii, Prime power degree representations of quasisimple groups, preprint 2000

[6] A. O. Morris, A. K. Yaseen, Some combinatorial results involving Young diagrams, Math. Proc. Camb. Phil. Soc. 99 (1986) 23-31

[7] J.B. Olsson, Remarks on symbols, hooks and degrees of unipotent characters, J. Comb. Theory (A) 42 (1986) 223-238

[8] J.B. Olsson, Combinatorics and representations of finite groups, Vorlesungen aus dem FB Mathematik der Univ. Essen, Heft 20, 1993

[9] H. Scheid, Die Anzahl der Primfaktoren in \binom{n}{k}, Arch. Math. 20 (1969) 581–582

Christine Bessenrodt Fakultät für Mathematik, Otto-von-Guericke-Universität Magdeburg, D-39016 Magdeburg, Germany Email address: bessen@mathematik.uni-magdeburg.de

Jørn B. Olsson Matematisk Institut, Københavns Universitet Universitetsparken 5, 2100 Copenhagen Ø, Denmark Email address: olsson@math.ku.dk