Prime power degree representations of the symmetric and alternating groups

Antal Balog, Christine Bessenrodt, Jørn B. Olsson, Ken Ono

June 30, 2000

1 Introduction

In 1998, the second author raised the problem of classifying the irreducible characters of S_{n} of prime power degree. Zalesskii proposed the analogous problem for quasi-simple groups, and he has, in joint work with Malle, made substantial progress on this latter problem. With the exception of the alternating groups and their double covers, their work provides a complete solution. In this article we first classify all the irreducible characters of S_{n} of prime power degree (Theorem 2.4), and then we deduce the corresponding classification for the alternating groups (Theorem 5.1), thus providing the answer for one of the two remaining families in Zalesskii’s problem. This classification has another application in group theory. With it, we are able to answer, for alternating groups, a question of Huppert: Which simple groups G have the property that there is a prime p for which G has an irreducible character of p-power degree > 1 and all of the irreducible characters of G have degrees that are relatively prime to p or are powers of p?

The paper is organized as follows. In section 2, some results on hook lengths in partitions are proved. These results lead to an algorithm which allows us to show that every irreducible representation of S_{n} with prime power degree is labelled by a partition having a large hook. In section 3, we obtain a new result concerning the prime factors of consecutive integers (Theorem 3.4). In section 4 we prove Theorem 2.4, the main result. To do so, we combine the algorithm above with Theorem 3.4 and work of Rasala on minimal degrees. This implies Theorem 2.4 for large n. To complete the proof, we check that the algorithm terminates appropriately for small n (i.e. those n ≤ 9.25·10^{8}) with the aid of a computer. In the last section we derive the classification of irreducible characters of A_{n} of prime power degree, and we solve Huppert’s question for alternating groups.

Acknowledgements. The second and third author are grateful to the Danish Natural Science Foundation for the support of their cooperation on this work. The fourth author thanks the National Science Foundation, the Alfred P. Sloan Foundation, and the David and Lucile Packard Foundation for their generous support. The authors are indebted to Rhiannon Weaver for writing an efficient computer program which was vital for this work.

2 An algorithm for hook lengths

We refer to [5], [7] for details about partitions, Young diagrams and hooks. Consider a partition λ = (λ_{1}, λ_{2}, …, λ_{m}) of the integer n. 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 of λ. Moreover for i ≥ 1, m_{i} = m_{i}(λ) denotes the number of parts equal to i in λ. Thus m = ∑_{i≥1} m_{i}. The Young diagram of λ consists of n nodes (boxes) with λ_{i} nodes in the ith row. We refer to the nodes in matrix notation, i.e. the (i, j)-node is the jth node in the ith row. The (i, j)-hook consists of the nodes in the Young diagram to the right and below the (i, j)-node, and including this node. The number of nodes in this hook is its hooklength, denoted by h_{ij}. Thus

• • • •

is the Young diagram of (5^{2}, 4, 1), where we have marked the (2, 3)-hook belonging to the third node in the second row, and the corresponding hook length h_{23} is 4.

We put h_{i} = h_{i1} = λ_{i} + (m − i) for 1 ≤ i ≤ m; these are the first column hook lengths, abbreviated by fch. The degree f_{λ} of λ is

f_{λ} = \frac{n!}{Π_{i,j}h_{ij}} .

It is known that this is the degree of the complex irreducible representation of the symmetric group S_{n} labelled by λ (see [5], [7]).

Example 2.1 If λ = (n − k, 1^{k}), a partition of n with 0 < k < n then f_{λ} = \binom{n−1}{k}, a binomial coefficient.

Binomial coefficients are prime powers only in the “trivial” cases ([3], [11]):

Proposition 2.2 The binomial coefficient \binom{n}{k} is a (nontrivial) power of a prime exactly when n is a prime power and k = 1 or k = n − 1.

This immediately implies the characterization of hook partitions of (nontrivial) prime power degree:

Corollary 2.3 Suppose that λ = (n − k, 1^{k}) with 0 < k < n − 1. Then f_{λ} = p^{r} for some prime p and integer r ≥ 1 if and only if n = p^{r} + 1 and k = 1 or k = p^{r} − 1.

The following theorem characterizes those partitions (resp. irreducible characters of symmetric groups) that are of prime power degree.

Theorem 2.4 Let λ be a partition of n. Then f_{λ} = p^{r} for some prime p, r ≥ 1, if and only if one of the following occurs:

n = p^{r} + 1 , λ = (p^{r}, 1) or (2, 1^{p^{r}−1}) , f_{λ} = p^{r}

or we are in one of the following exceptional cases:

n = 4 : λ = (2^{2}),  f_{λ} = 2
n = 5 : λ = (2^{2}1) or (3, 2) , f_{λ} = 5
n = 6 : λ = (4, 2) or (2^{2}1^{2}) , f_{λ} = 3^{2}
λ = (3^{2}) or (2^{3}) ,  f_{λ} = 5
λ = (321) ,  f_{λ} = 2^{4}
n = 8 : λ = (521) or (321^{3}) , f_{λ} = 2^{6}
n = 9 : λ = (72) or (2^{2}, 1^{5}) , f_{λ} = 3^{3}

First we state some elementary results about hook lengths. In the following, λ is a partition of n, m_{1} is the multiplicity of 1 as a part of λ, and h_{ij}, h_{i} are the hook lengths as defined above.

The following lemma is elementary.

Lemma 2.5 If h_{i2} ≠ 0 (i.e. λ_{i} ≥ 2) then

h_{i2} = h_{i} − m_{1} − 1 .

Proposition 2.6 Let 1 ≤ i, j ≤ m, i ≠ j. Then

h_{i} + h_{j} − n − 1 ≤ m_{1} .

Proof. It suffices to prove this for h_{1} and h_{2}. If λ = (n − k, 1^{k}) is a hook partition, then the result is trivially true. If λ is not a hook partition, then h_{22} ≠ 0 so that h_{22} = h_{2} − m_{1} − 1 by Lemma 2.5. Since h_{11} = h_{1} and since h_{11} + h_{22} ≤ n the result follows. ⋄

Lemma 2.7 Suppose that s = h_{ik} and t = h_{jℓ} where (i, k) ≠ (j, ℓ).

(1) If i ≠ j and k ≠ ℓ, then s + t ≤ n.

(2) If s + t > n, then either i = j = 1 (both hooks in the first row) or k = ℓ = 1 (both hooks in the first column).

Proof. (1) By assumption, the two hooks have at most one node in common. If they have a node in common, none of the hooks is the (1, 1)-hook. Thus, the hooks plus possibly the (1, 1)-node comprise s + t nodes, whence s + t ≤ n.

(2) By (1), we know that i = j or k = ℓ. Assume the former so that k ≠ ℓ. If i > 1, then h_{1k} > h_{ik} whence h_{1k} + h_{jℓ} > n, contradicting (1) applied to (1, k) and (j, ℓ). The case k = ℓ is similar. ⋄

Corollary 2.8 For i ≥ 2, every hook of length t > n − h_{i} = n − h_{i1} is in the first column of λ.

Proof. Assume that t = h_{jℓ}. If (j, ℓ) = (i, 1), then the result is true. Otherwise, apply (2) of Lemma 2.7 with (i, k) = (i, 1) to get ℓ = 1. ⋄

From now on, assume that f_{λ} is a power of a prime and that λ is not a hook partition. For n ≤ 6 one easily checks Theorem 2.4 by hand (or by using the tables in [5]). So we assume from now on that n > 6. Consequently, it follows that f_{λ} ≥ n + 1 ([5], Theorem 2.4.10).

Proposition 2.9 If q is a prime for which n − m_{1} ≤ q ≤ n, then

q, 2q, …, [\frac{n}{q}] q

are all fch of λ.

Proof. Put w = [\frac{n}{q}], n = wq + r, 0 ≤ r < q. By assumption, we have that (w − 1)q ≤ (w − 1)q + r = n − q ≤ m_{1}. Since m_{1} is the multiplicity of 1 in λ, the numbers 1, 2, …, m_{1} are fch. In particular, we have that q, 2q, …, (w − 1)q are fch. If wq ≤ m_{1}, then we are done. Assume that m_{1} < wq. At most w hooks in λ are of lengths divisible by q (see e.g. [7], Proposition (3.6)). If there are only the above (w − 1) hooks in the first column of length divisible by q, then q|f_{λ} since ∏_{i=1}^{w}(iq) | n!. By assumption, f_{λ} is then a power of q. We get f_{λ} = (wq)_{q}, the q-part of wq. Thus f_{λ}|wq ≤ n, whence f_{λ} ≤ n, a contradiction. Let h_{ij} be the additional hook length divisible by q. Since λ ≠ (1^{n}), m_{1} ≤ h_{2}. If h_{2} > m_{1}, then h_{ij} + h_{21} > q + m_{1} ≥ n. By Corollary 2.8 we get j = 1. If h_{2} = m_{1}, then λ = (n − m_{1}, 1^{m_{1}}) and since m_{1} < wq there has to be a hook of length divisible by q in the first row. Since n − m_{1} ≤ q it has to be the (1, 1)-hook. Thus h_{11} = wq. ⋄

Corollary 2.10 Let 1 ≤ i < j ≤ m. If h ≤ n has a prime divisor q satisfying 2n − h_{i} − h_{j} < q, then h is a fch of λ.

Proof. By Proposition 2.6, n − m_{1} ≤ 2n + 1 − h_{i} − h_{j}. By assumption 2n + 1 − h_{i} − h_{j} ≤ q ≤ h ≤ n, whence n − m_{1} ≤ q ≤ n. By Proposition 2.9, any multiple of q less or equal to n is a fch. In particular h is a fch of λ. ⋄

Lemma 2.11 If q is a prime, \frac{n}{2} < q ≤ n, then λ has a hook of length q.

Proof. This follows immediately from the degree formula and the fact that f_{λ} ≥ n + 1. ⋄

We are now going to strengthen our assumption on n and λ slightly. According to Table 3 of [2] there are, for all n ≥ 12, at least two distinct primes p, q with \frac{n}{2} < p, q ≤ n. By Lemma 2.11 there are hooks of length p and q in λ. We will assume that such primes p, q exist for n and that p and q are fch. This is not a restriction, see Lemma 2.7(2) (if necessary we may replace λ by its conjugate partition λ^{0} as we have f_{λ} = f_{λ^{0}}). Now the above assumption forces any prime between \frac{n}{2} and n to be a fch of λ.

Proposition 2.12 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 fch of λ.

Proof. We use induction on i to show that s_{i} and t_{i} are fch for λ. For i = 1 this is true by our assumption. If s_{i} and t_{i} are fch, then Corollary 2.10 shows that s_{i+1} and t_{i+1} are fch of λ. ⋄

We get an algorithm from Proposition 2.12 which shows that h_{1} is large and thus λ is “almost” a hook: 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} ≤ h_{1} by Proposition 2.12.

Example 2.13 n = 189 = 3^{3} · 7. Choose s_{1} = 179, t_{1} = 181. Then 2n−s_{1}−t_{1} = 18. Now choose t_{2} = 188 = 4·47 and s_{2} = 186 = 2·3·31 which have prime factors exceeding 18. Then 2n − s_{2} − t_{2} = 4. Choose t_{3} = 189, s_{3} = 188. Thus if f_{λ} is a prime power then h_{1} = 189 and λ is a hook partition, contradicting Corollary 2.3. Thus none of the 1.527.273.599.625 partitions of n = 189 is of prime power degree greater than one.

3 Prime factors in consecutive integers and good sequences

In this section we show, for sufficiently large n, that there are suitable sequences as in Proposition 2.12 that end with numbers close to n. Thus, for a partition λ of prime power degree the algorithm described in the previous section shows that λ differs from a hook partition only by a small amount.

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}}). We want to show that there are such sequences with n − B(n) “small”.

More precisely, we prove the following theorem:

Theorem 3.1 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.

We note that the 225 in the theorem above can be reduced to 2 for sufficiently large n. However, this result is of no use in the present paper.

We first review some facts about the distribution of primes, and we prove a theorem on the prime divisors of a product of consecutive integers. Using this, we then prove Theorem 3.1.

Throughout this section p shall denote a prime. Now we recall three relevant functions. If X > 0, then define π(X) and θ(X) by

\begin{aligned}π(X) & := & #\{p ≤ X\}, & (1) \\ θ(X) & := & ∑_{p≤X} log p. & (2) \end{aligned}

Moreover, recall that von Mangoldt’s function Λ(n) is defined by

Λ(n) := \begin{cases}log p & if n = p^{α} with α ∈ ℤ, \\ 0 & otherwise. \end{cases} (3)

Rosser and Schoenfeld [9], [10] proved the following unconditional inequalities. These inequalities will be important in the proof of Theorem 3.1.

Theorem 3.2 (Rosser-Schoenfeld)

(1) If X > 1, then

π(X) ≤ 1.25506 · \frac{X}{log X} .

(2) If X > 1, then

∑_{1≤n≤X} \frac{Λ(n)}{n} ≤ log X.

(3) If X ≥ 1319007, then

0.998684 · X < θ(X) < 1.001102 · X.

Now we prove the following crucial result about the prime factors of consecutive integers.

Lemma 3.3 If 1 ≤ m ≤ k ≤ y < n are integers for which

(\frac{n}{k})^{k} ≥ (n + k)^{π(y)+m−1}, (4)

then at least m of the integers n + 1, n + 2, …, n + k have a prime factor exceeding y.

Proof. If L is defined by

L := ∏_{j=1}^{k}(n + j) = ∏_{p} p^{α_{p}},

then let R and S be the unique integers for which L = RS and

\begin{aligned}R & = & ∏_{p>y} p^{α_{p}}, & (5) \\ S & = & ∏_{p≤y} p^{α_{p}}. & (6) \end{aligned}

Since n + k is the largest factor defining L, if

R > (n + k)^{m−1}, (7)

then at least m of the numbers n + 1, n + 2, …, n + k have a prime factor exceeding y. Therefore, it suffices to prove (7). Since L = RS > n^{k}, we trivially have that

R > \frac{n^{k}}{S} ,

and so by (7) it suffices to prove that

\frac{n^{k}}{S} > (n + k)^{m−1}. (8)

Now we derive an upper bound for S. By definition, we have that

log S = log ∏_{p≤y} p^{α_{p}} = ∑_{p≤y} α_{p} log p = ∑_{p≤y} log p ∑_{α=1}^{∞} ( ∑_{1≤j≤k , p^{α}|n+j} 1) .

Now, the innermost sum clearly has the upper bound [\frac{k}{p^{α}} ] + 1 Moreover, since this bound equals 1 whenever p^{α} > k, by Theorem 3.2(2) we find that

log S ≤ ∑_{p^{α}≤k} log p [\frac{k}{p^{α}} ] + ∑_{p≤y} log p ∑_{p^{α}≤n+k} 1
≤ k ∑_{d=1}^{k} \frac{Λ(d)}{d} + ∑_{p≤y} log p [\frac{log(n+k)}{log p}]
≤ k log k + π(y) log(n + k).  (9)

Therefore, we have that

S ≤ k^{k}(n + k)^{π(y)},

and so

\frac{n^{k}}{S} ≥ (\frac{n}{k})^{k} (n + k)^{−π(y)}.

However, since (\frac{n}{k})^{k} ≥ (n + k)^{π(y)+m−1} by (4) we have that

\frac{n^{k}}{S} ≥ (n + k)^{m−1}

which is (8). ⋄

As a consequence of Theorem 3.2 and Lemma 3.3, we obtain the following crucial result.

Theorem 3.4 If n > 3.06 · 10^{8} is an integer and k is a positive integer satisfying

168 ≤ k ≤ \frac{n}{4},

then at least three of the integers n + 1, n + 2, …, n + k have a prime factor exceeding 4k.

Proof. By Theorem 3.2 (3), we have that

θ(n + k) − θ(n) > 0.998684(n + k) − 1.001102n = 0.998684k − 0.002418n.

So, if n/400 < k ≤ n/4 and n ≥ 1319007, then

θ(n + k) − θ(n) > 0.0000787n > 2 log(5n/4) ≥ 2 log(n + k). (10)

Since

θ(n + k) − θ(n) = ∑_{n<p≤n+k} log p ≤ (π(n + k) − π(n)) log(n + k),

(10) implies that there are at least three primes among the numbers n+1, n+ 2, …, n + k provided that n/400 < k ≤ n/4 and n ≥ 1319007. Moreover, by hypothesis these primes are ≥ n + 1 > 4k. Next we consider the cases where 100 ≤ k ≤ n/400 and n ≥ 1.8 · 10^{14}. By Lemma 3.3 it suffices to verify that

(\frac{n}{k})^{k} ≥ (n + k)^{π(4k)+2} (11)

whenever 100 ≤ k ≤ n/400 and n > 1.8 · 10^{14}. By Theorem 3.2(1), if k ≥ 100, then

π(4k) + 2 ≤ \frac{5.02024k}{log(4k)} + 2 < \frac{5.15k}{log(4k)}.

Therefore, (11) holds as soon as

(\frac{n}{k})^{k} ≥ ((1 + \frac{1}{400})n)^{\frac{5.15k}{log(4k)}} ≥ (n + k)^{\frac{4.95k}{log(4k)}} .

By taking logarithms, the first inequality is equivalent to

log(4k) log(n/k) ≥ 5.15 log ((1 + \frac{1}{400}) n) . (12)

However, for a fixed value of n the function on the left hand side of this inequality is an increasing function in k in the interval [1, \sqrt{n}/2] and is decreasing for larger k thus taking the minimal value log 400 · log(n/100) at the endpoints. It is easy to verify that (12) holds for all k in the interval [100, n/400] provided that n > 1.8 · 10^{14}. Similarly, we can show that (11) holds for all 500 ≤ k ≤ n/2000 and n ≥ 3 · 10^{8}.

To complete the proof in the remaining cases we use Maple on a PC. On one hand, one can, using the Nextprime function, check that for j = 0, …, 6 there is always a prime in any interval of type (m · 10^{4+j}, (m + 1) · 10^{4+j}] where 10^{4} ≤ m ≤ 10^{5}. This implies immediately that there are at least three primes in any interval of type (n, n + n/2000] where 10^{8} ≤ n ≤ 10^{15}. On the other hand using log(n + k) < log n + k/n, (11) follows from

log n ≥ \frac{k log k + (π(4k) + 2)k/n}{k − π(4k) − 2}.

This is easily verified (using only a table of primes below 2000) for 168 ≤ k ≤ 500 and n > 3.06 · 10^{8}. All cases are considered. ⋄

Proof of Theorem 3.1. Suppose that n > 3.06 · 10^{8} and pick two primes 0.8n < A_{1} < B_{1} ≤ 0.9n which is allowable by Theorem 3.2 (3). Note that B_{1} − A_{1} ≤ n − B_{1} by hypothesis. Now suppose that when constructing the sequences we have

B_{i} − A_{i} ≤ n − B_{i}.

Now we seek new integers A_{i+1} < B_{i+1} < n for which

\begin{matrix}B_{i+1} − A_{i+1} ≤ n − B_{i+1}, & (13) \\ B_{i} < A_{i+1} < B_{i+1} ≤ n. & (14) \end{matrix}

and each with a prime factor exceeding 3(n − B_{i}). Now we apply Theorem 3.4 with n = B_{i} and k = ⌈\frac{3}{4}(n − B_{i})⌉ (Note. ⌈x⌉ denotes the smallest integer ≥ x). Obviously, we have 4k ≥ 3(n − B_{i}) > 2n − A_{i} − B_{i}. As long as k ≥ 168 we can find three integers B_{i} < a < b < c ≤ B_{i} + k each with a prime factor > 4k. We now show that either the pair a and b or the pair b and c satisfies (13) and (14). If neither does, then

c − b > n − c and b − a > n − b

which are equivalent to

c − b ≥ n − c + 1 and b − a ≥ n − b + 1.

These imply that

a ≤ 2b − n − 1 ≤ 2(2c − n − 1) − n − 1 = 4c − 3n − 3
≤ 4(B_{i} + k) − 3n − 3
≤ 4B_{i} + 4 (\frac{3}{4}(n − B_{i}) + \frac{3}{4}) − 3n − 3 = B_{i}.

Since B_{i} < a we see that we can always choose such an A_{i+1} and B_{i+1} provided that k ≥ 168. Suppose that A_{i+1} and B_{i+1} are the last terms which are found this way, then we have n−B_{i+1} ≤ 225. This proves Theorem 3.1. ⋄

Corollary 3.5 Let λ be a partition of n with largest hook length h_{1}. If λ is of prime power degree and n > 3.06 · 10^{8}, then n − h_{1} ≤ 225.

Proof. This follows immediately from Theorem 3.1 and Proposition 2.12. ⋄

4 Proof of the classification result for S_{n}

For dealing with the situation where c = n − h_{1} is small, we provide a good upper bound for the p-powers in the character degrees for S_{n}. This is similar to the case of binomial coefficients (i.e. the case of hook partitions).

Proposition 4.1 Let λ be a partition of n, and set c = n − h_{1}. Let p be a prime, and l the integer with p^{l} ≤ n < p^{l+1}. Then

ν_{p}(f_{λ}) ≤ ν_{p}((2c + 2)!) + 2l .

In particular, a bound for the p-part of f_{λ} is given by

(f_{λ})_{p} ≤ n^{2} · ((2c + 2)!)_{p} .

Proof. Let k = m_{1} be the multiplicity of 1 in λ. By looking at the Young diagram we see that

λ_{1} − λ_{2} ≥ h_{1} − k − (c + 2) = n − 2c − k − 2 .

Let A denote the set of nodes other than the final k nodes in the leg and the final n − 2c − k − 2 nodes in the arm of the (1, 1)-hook; for a node y ∈ A let h_{y} denote the corresponding hook length. Then from the degree formula we obtain

ν_{p}(f_{λ}) = ∑_{i=1}^{l} ([\frac{n}{p^{i}} ] − [\frac{k}{p^{i}} ] − [\frac{n − 2c − k − 2}{p^{i}}]) − ∑_{y∈A} ν_{p}(h_{y})

For fixed i, the ith summand in the first sum gives a contribution of at most [\frac{2c+2}{p^{i}}] + 2, hence

ν_{p}(f_{λ}) ≤ ∑_{i=1}^{l} ([\frac{2c + 2}{p^{i}}] + 2) ≤ ν_{p}((2c + 2)!) + 2l .

The second inequality follows immediately from this. ⋄

We first consider the case of small n. Assume again that λ is a partition of n of prime power degree but not a hook.

Using the available tables (or with the aid of MAPLE), it is easy to check the main theorem for n < 43. In other words, the only partitions having prime power degree are of the form (n−1, 1) together with their conjugates, or are on the short list of exceptions given in the theorem.

For a midsized n (i.e. 43 ≤ n ≤ 9.25 · 10^{8}), we use the following number theoretic condition.

Fix a number b. Given a number n, let p_{1}, p_{2} be the two largest primes below n. Then check whether there is a prime divisor q of n(n−1)(n−2) ⋯ (n−b) with p_{1} + p_{2} + q > 2n.

A computer program (written in C++ using the LiDIA number theory package, run on a super computer with 32 nodes, running time 2 hours) was used to check that this condition is satisfied for all n from 29 to 9.25·10^{8} for b = 4.

We now want to use this in the situation where λ is a partition of n of prime power degree, 43 ≤ n ≤ 9.25 · 10^{8}. Since the two largest primes p_{1}, p_{2} are fch of λ, if q is a prime divisor as in the condition above (with b = 4), then by Corollary 2.10 one of the numbers n, n − 1, … n − 4 is a fch of λ. Hence n − h_{1} ≤ 4.

From Proposition 4.1 we know that if λ is of prime power degree, then

f_{λ} ≤ o(c) · n^{2}

where c = n − h_{1} and o(c) = max{(2c + 2)!_{p} | p prime }. In our situation we have c ≤ 4, so o(c) ≤ o(4) = 256, and hence we know

f_{λ} ≤ 256 · n^{2}

Now we use information on the minimal degrees of S_{n}-representations. Burnside’s theorem on the minimal degree > 1 for S_{n} was greatly generalized by Rasala [8], giving in a suitable sense the list of the minimal degrees for sufficiently large n (depending on the requested length of the list). We use the notation from [8].

For any k, the list of minimal degrees for S_{n} starts with the degrees of partitions of n coming from partitions of numbers d ≤ k by adding on a part n − d, if n ≥ B_{k}, a bound which is provided explicitly in [8]. The degree polynomial φ_{μ}(n) for any partition μ of k is also given explicitly.

For k = 5, one has B_{5} = 43. One easily checks that for n ≥ 43

256 · n^{2} ≤ φ_{5}(n) = \frac{1}{5!}(n − 9) ∏_{i=0}^{3}(n − i)

Hence f_{λ} is among the minimal degrees for S_{n}. But the list of minimal degrees is easily computed [8], and none of these is a prime power except possibly n − 1 which occurs only for (n − 1, 1) and its conjugate (use that the degree formula in [8], Theorem A, gives a factorized expression in which consecutive numbers appear). Hence we do not get any further partitions of prime power degree.

Now we deal with the case of large n (i.e. n > 9.25 · 10^{8}). By Corollary 3.5 we know that n − h_{1} ≤ 225. We want to use the bound for the degree and the minimal degree argument again in this situation. From [8] we get the numbers B_{k} from which on the minimal degrees all come from partitions of weight at most k.

Since B_{18} = 310.390.100 < n, we thus know a long list of minimal degrees. The maximal entry on this list comes from the partition μ = (64321^{3}) which is of maximal degree 16.336.320 among the partitions of 18. For c = 225 we have o(c) = 2^{448} and one checks (for example, using MAPLE) that for n ≥ 9.25 · 10^{8} one has

f_{λ} ≤ 2^{448} · n^{2} ≤ φ_{μ}(n) = \frac{[μ](1)}{18!} ∏_{i=1}^{18}(n − μ_{i} − 18 + i)

where μ = (μ_{1}, …, μ_{18}), extending μ by parts 0 if necessary.

Hence f_{λ} is among the minimal degrees, but as before, one can check that all the minimal degrees on this list are not prime powers, except possibly n − 1. This completes the proof. ⋄

5 Alternating groups and a question of Huppert

The purpose of this section is to prove the analogue of Theorem 2.4 for the alternating groups. Also we answer a question of B. Huppert about character degrees in alternating groups. If λ is a partition of n, then the irreducible representation of S_{n} labelled by λ remains irreducible when restricted to A_{n} if and only if λ ≠ λ^{0}, the conjugate (associated) partition of λ. If λ = λ^{0} the restriction is a sum of two irreducible representations of the same degree. This leads us to the following definition providing the character degrees in A_{n} ([5], 2.5). Let λ be a partition of n. Then

f̃_{λ} := \begin{cases}f_{λ} & if λ ≠ λ^{0} \\ \frac{1}{2}f_{λ} & if λ = λ^{0}. \end{cases}

Theorem 5.1 Let λ be a partition of n. Then f̃_{λ} = p^{r} for some prime p, r ≥ 1, if and only if one of the following occurs:

n = p^{r} + 1 > 3 , λ = (p^{r}, 1) or (2, 1^{p^{r}−1}) , f̃_{λ} = p^{r}

or we are in one of the following exceptional cases

n = 5 : λ = (2^{2}1) or (3, 2) , f̃_{λ} = 5
λ = (31^{2}) , f̃_{λ} = 3
n = 6 : λ = (4, 2) or (2^{2}1^{2}) , f̃_{λ} = 3^{2}
λ = (3^{2}) or (2^{3}) ,  f̃_{λ} = 5
λ = (321) ,  f̃_{λ} = 2^{3}
n = 8 : λ = (521) or (321^{3}) , f̃_{λ} = 2^{6}
n = 9 : λ = (72) or (2^{2}, 1^{5}) , f̃_{λ} = 3^{3}

Proof. If λ ≠ λ^{0} we may apply Theorem 2.4. Suppose λ = λ^{0}. The only self conjugate partitions occurring in the list of Theorem 2.4 are (2, 1) and (2^{2}) and here f̃_{λ} = 1. Using the character tables of [1] there are no further occurrences of self conjugate partitions with f̃_{λ} = p^{r} for n ≤ 13. When n > 13 we may always find two primes p_{1}, p_{2} satisfying \frac{n+1}{2} < p_{1} < p_{2} ≤ n, (using [2], Table 3). Then 2p_{1} and 2p_{2} are not character degrees of S_{n}. If λ = λ^{0} and f_{λ} = 2p^{r} then p ≠ p_{1}. Thus λ has to contain hooks of lengths p_{1} and p_{2}. Since λ contains only one hook of length p_{1} and p_{2} respectively, both of them have to be in the diagonal, i.e. p_{1} = h_{ii}, p_{2} = h_{jj} for some i, j.

This contradicts Lemma 2.7. ⋄

In recent work on the characterization of the finite simple groups P SL(n, q) by character degree properties [4], B. Huppert needs the following:

Corollary 5.2 Suppose that for some simple alternating group A_{n} there is a prime p, such that all irreducible character degrees of A_{n} are either prime to p or powers of p. Assume that some power of p is a degree for A_{n}. Then n = 5 and p = 2, 3 or 5, or n = 6 and p = 3.

Proof. Using the Atlas [1] we may assume n > 13. By Theorem 5.1, A_{n} only has a prime power character degree p^{r} when n = p^{r} + 1. But then

f̃_{(n−2,1^{2})} = \frac{(n − 1)(n − 2)}{2} = p^{r}(p^{r} − 1)/2

is a character degree which is divisible by p and not a power of p.

References

[1] J.H. Conway, R.T. Curtis, S.P. Norton, R.A. Parker, R.A. Wilson, Atlas of Finite Groups. Clarendon Press, Oxford, 1985

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

[3] F. Hering, Eine Beziehung zwischen Binomialkoeffizienten und Primzahlpotenzen, Arch. Math. 19 (1968) 411-412

[4] B. Huppert, Some simple groups which are determined by the set of their character degrees, II, preprint 2000

[5] G. James, A. Kerber: The representation theory of the symmetric group. Encyclopedia of Mathematics and its Applications, 16, Addison-Wesley 1981

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

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

[8] R. Rasala, On the minimal degrees of characters of S_{n}, J. Algebra, 45(1977), 132–181

[9] J. B. Rosser, L. Schoenfeld, Approximate formulas for some functions of prime numbers, Illinois J. Math. 6 (1962) 64-94 [10] J. B. Rosser, L. Schoenfeld, Sharper bounds for the Chebyshev func-

tions θ(x) and Ψ(x), Math. Comp. 29 (1975) 243-269

[11] W. Stahl, Bemerkung zu einer Arbeit von Hering, Arch. Math. 20 (1969) 580

Antal Balog Mathematical Institute of the Hungarian Academy of Sciences P. O. Box 127, Budapest 1364, Hungary Email address: balog@hexagon.math-inst.hu

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

Ken Ono Department of Mathematics, University of Wisconsin, Madison, Wisconsin 53706, USA Email address: ono@math.wisc.edu