PARTITIONS WHICH ARE p- AND q-CORE

J.-C. Puchta

Mathematical Institute, 24-29 St. Giles’, Oxford OX1 3LB, United Kingdom

                          puchta@maths.ox.ac.uk

Received: 6/19/01, Revised: 9/1/01, Accepted: 9/6/01, Published: 9/12/01

Abstract

Let p and q be distinct primes, n an integer with n > p^{2}q^{2}. Then there is no partition of n which is at the same time p- and q-core. Hence there is no irreducible representation of S_{n} which is of p- and q-defect zero at the same time.

Let n be an integer. Then there is a natural bijection between the set of partitions of n and the irreducible representations of the symmetric group on n letters S_{n}. A representation of a finite group G with character χ is called of p-defect zero, if |G|_{p}|χ(1). In the case of the symmetric group this is known to be equivalent to the statement that the corresponding partition has no hook-number divisible by p, in this case the partition is called a p-core partition. Granville and Ono [2] proved that for any t ≥ 7 and any n there is a t-core partition of n, thus for every p ≥ 7 there is an irreducible representation of S_{n} with p-defect zero, an easier proof was given by Kiming [4].

In a recent paper Navarro and Willems [5] asked for relations between the p- and the q-blocks of representations. In this note we will show that the property of having defect zero exclude each other, if n is large enough compared to p and q. More precisely we will prove the following theorem.

Theorem 1. Let p and q be primes, n an integer with n > p^{2}q^{2}. Then there is no irreducible representation of S_{n} with p- and q-defect zero.

By the correspondence between irreducible representations of the S_{n} and partitions of n this will follow from the following statement.

Theorem 2. Let s and t be relatively prime integers, n an integer with n > s^{2}t^{2}. Then there is no partition of n which is at the same time s- and t-core.

Especially, the number of partitions which are simultaneously s- and t-core is finite. J. Kohles Anderson [3] proved a more precise version of this statement: The number of

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 1 (2001), #A06 2

partitions with this property is in fact equal to \frac{1}{s+t} \binom{s+t}{t}. However, the proof we give here seems to be simpler then the one given by her.

I would like to thank the referee for making me aware of [3].

The proof will use the description of t-core partitions introduced by Garvan, Kim and Stanton [1].

For the sequel we choose an arbitrary partition n = λ_{1} + … + λ_{k} of n and assume that it is t-core and s-core at the same time. We thus have to show that n < s^{2}t^{2}.

Consider the diagram of the partition, i.e. the set of cells whose first row consists of λ_{1} cells (1, 1), (1, 2), …, (1, λ_{1}), the second of λ_{2} cells and so on. Label a cell (i, j) with j − i (mod st), cells in column 0 are labeled in the same way. A cell at the end of a row is called exposed. Now divide the diagram into regions S_{k}, such that a cell belongs to S_{k} if and only if s(k − 1) ≤ j − i < sk, in the same way T_{k} denotes the cells with t(k − 1) ≤ j − i < tk. Now by [1], paragraph 2, we know that if the partition is s-core, and there is an exposed cell labeled with i in the region S_{k}, then there is an exposed cell labeled with ĩ ≡ i (mod s) in every region S_{l} with l ≤ k. Especially, there is some sequence k_{ν}, 0 ≤ ν ≤ l, k_{0} = 1, such that λ_{k_{ν}} ≡ λ_{1} − (k_{ν} − 1) (mod s), (k_{ν+1} − k_{ν}) < λ_{k_{ν}} − λ_{k_{ν+1}} < 2s − (k_{ν+1} − k_{ν}) and λ_{k_{l}} < s, i.e. λ_{k_{ν}} = λ_{1} − νs + k_{ν}. Assume that l < t. Since λ_{k_{ν}} ≤ λ_{k_{ν+1}}, we have k_{ν+1} ≤ k_{ν} + s, thus the partition under consideration consists of at most ls < st summands, each being st at most, thus we have n ≤ s^{2}t^{2}.

Now if l > t, then the labels of the exposed cells in the rows k_{ν} run through a complete remainder system (mod t), since s and t are coprime, the remainders of λ_{k_{ν}} − k_{ν} = λ_{1} − νs, 0 ≤ ν < t are therefore all different. However, by [1] we know that if the partition is t-core, and there is an exposed cell in region T_{k} with the label i, then there is no exposed cell with a label \overline{i} ≡ t−i−1 (mod t) in any region T_{l} with l ≥ 1−k. If λ_{1} is in region T_{k}, then λ_{k_{t−1}} is in region T_{l} with l ≥ k − s, thus k − s < 1 − k, i.e. k ≤ s/2. By the definition of T_{k} we have λ_{1} < t(s/2 + 1) ≤ st.

Since the property of being a t-core partition is unchanged under conjugation, by the same reasoning we get that there are less than st summands, thus we obtain n < s^{2}t^{2} again.

Thus in any case the assumption that our partition is at the same time s-core and t-core leads to the estimate n < s^{2}t^{2} which proves our theorem.

References

[1] F. Garvan, D. Kim, D. Stanton, Cranks and t-cores, Invent. Math. 101, No.1, 1-17 (1990)

[2] A. Granville, K. Ono, Defect zero p-block for finite simple groups, Trans. Am. Math. Soc. 348, No.1, 331-347 (1996)

[3] J. Kohles Anderson, Partitions which are simultaneously t_{1}- and t_{2}-core, to appear in Discrete Mathematics

[4] I. Kiming, A note on a theorem of A. Granville and K. Ono, J. Number Theory 60, No.1, 97-102 (1996)

[5] G. Navarro, W. Willems, When is a p-block a q-block?, Proc. Am. Math. Soc. 125, No.6, 1589-1591 (1997)

Mathematics Subject Classification: 05A17, 11P83, 20C30