LABELLING THE CHARACTER TABLES OF SYMMETRIC AND ALTERNATING GROUPS

Let X be a character table of the symmetric group S_n. It isshown that unless n = 4 or n = 6, there is a unique way to assignpartitions of n to the rows and columns of X so that for all and , X is equal to (), the value of the irreducible characterof S_n labelled by on elements of cycle type . Analogous resultsare proved for alternating groups, and for the Brauer charactertables of symmetric and alternating groups.     

A Proof of the Murnaghan–Nakayama Rule Using Specht Modules and Tableau Combinatorics

Annals of Combinatorics - The Murnaghan–Nakayama rule is a combinatorial rule for the character values of symmetric groups. We give a new combinatorial proof by explicitly finding the trace...     

A combinatorial method for calculating the moments of Lévy area

We present a new way to compute the moments of the Lévy area of a two-dimensional Brownian motion. Our approach uses iterated integrals and combinatorial arguments involving the shuffle product.

     

Counting partitions on the abacus

In 2003, Maróti showed that one could use the machinery of ℓ-cores and ℓ-quotients of partitions to establish lower bounds for p(n), the number of partitions of n. In this paper we explore these ideas in the case ℓ=2, using them to give a largely combinatorial proof of an effective upper bound on p(n), and to prove asymptotic formulae for the number of self-conjugate partitions, and the number of partitions with distinct parts. In a further application we give a combinatorial proof of an identity originally due to Gauss. Dedicated to the memory of Dr. Manfred Schocker (1970–2006)     

High valency subgraphs of cubes

Let G be a subgraph of the 1-skeleton of the unit cube in R”, in which each vertex is of degree at least k. Our main result is that each connected component of G has at least 2^k vertices, and so G has at most 2^n?k components.     

Knights, spies, games and ballot sequences

This paper solves the Knights and Spies Problem: In a room there are n people, each labelled with a unique number between 1 and n. A person may either be a knight or a spy. Knights always tell the truth, while spies may lie or tell the truth as they see fit. Each person in the room knows the identity of everyone else. Apart from this, all that is known is that strictly more knights than spies are present. Asking only questions of the form: ‘Person i, what is the identity of person j?’, what is the least number of questions that will guarantee to find the true identities of all n people? We present a questioning strategy that uses slightly less than 3n/2 questions, and prove that it is optimal by solving a related two-player game. The performance of this strategy is analysed using methods from the famous ballot-counting problem. We end by discussing two questions suggested by generalisations of the original problem.     

Vertices of Specht modules and blocks of the symmetric group

This paper studies the vertices, in the sense defined by J.A. Green, of Specht modules for symmetric groups. The main theorem gives, for each indecomposable non-projective Specht module, a large subgroup contained in one of its vertices. A corollary of this theorem is a new way to determine the defect groups of symmetric groups. The main theorem is also used to find the Green correspondents of a particular family of simple Specht modules; as a corollary, this gives a new proof of the Brauer correspondence for blocks of the symmetric group. The proof of the main theorem uses the Brauer homomorphism on modules, as developed by M. Broué, together with combinatorial arguments using Young tableaux.