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REPRESENTATIONS OF THE SYMMETRIC GROUPS AND

COMBINATORICS OF THE FROBENIUS-YOUNG CORRESPONDENCE

By

MATTHEW JOHN HALL

A Thesis Submitted to The Honors College

In Partial Fulfillment of the Bachelor’s degreeWith Honors in

Mathematics

THE UNIVERSITY OF ARIZONA

MAY 2008

Approved by:

Christopher Ryan VinrootDepartment of Mathematics

STATEMENT BY AUTHOR

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SIGNED:

Abstract

After introducing the concepts of partitions, Young diagrams, represen-tation theory, and characters of representations, the 2006 paper by A. M.Vershik entitled “A New Approach to the Representation Theory of theSymmetric Group, III: Induced Representations and the Frobenius-YoungCorrespondence” is discussed. In tracing through Vershik’s line of reason-ing, a flaw emerges in his attempt to prove one of the key lemmas. Theattempted proof is an induction argument which, if valid, would lead to apurely combinatorial proof of the Frobenius-Young correspondence. Vershikasserts that two statements are equivalent, while in fact the implication istrue in only one direction. Counterexamples are then given to Vershik’scombinatorial statement and are then translated into counterexamples inrepresentation theory. Finally, possible directions for further study arenoted.

CONTENTS 4

Contents

1 Introduction 5

2 Partitions 62.1 Conjugate Partitions . . . . . . . . . . . . . . . . . . . . . . . . . . 62.2 Diagrams and Tableaux . . . . . . . . . . . . . . . . . . . . . . . . 62.3 Orderings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.4 Dominance Ordering on Conjugate Partitions . . . . . . . . . . . . 9

3 Representation Theory 113.1 Group Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113.2 CG-modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123.3 Irreducible Representations . . . . . . . . . . . . . . . . . . . . . . 123.4 Decomposition of Representations . . . . . . . . . . . . . . . . . . . 133.5 Restricted and Induced Representations . . . . . . . . . . . . . . . 133.6 Characters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143.7 Inner Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153.8 Frobenius Reciprocity . . . . . . . . . . . . . . . . . . . . . . . . . . 163.9 The Number of Irreducible Representations . . . . . . . . . . . . . . 163.10 Double Cosets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173.11 Mackey’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

4 Representations of the Symmetric Groups 184.1 The Irreducible Representations of Sn . . . . . . . . . . . . . . . . . 194.2 Branching Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194.3 Young Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194.4 Frobenius-Young Correspondence . . . . . . . . . . . . . . . . . . . 214.5 Mackey’s Theorem Applied to Sn . . . . . . . . . . . . . . . . . . . 214.6 The Snapper Conjecture . . . . . . . . . . . . . . . . . . . . . . . . 23

4.6.1 Combinatorial Statement . . . . . . . . . . . . . . . . . . . . 234.6.2 Combinatorial Counterexamples . . . . . . . . . . . . . . . . 254.6.3 Representation Theory Counterexamples . . . . . . . . . . . 26

4.7 Frobenius-Young Correspondence, Revisited . . . . . . . . . . . . . 28

5 Further Questions 28

1 INTRODUCTION 5

Prelude

The theory of representations of finite groups focuses on homomorphisms from afinite group to the group of invertible linear transformations of a vector space.After spending a semester actively studying these areas with the help of the excel-lent text Representations and Characters of Groups by Gordon James and MartinLiebeck, I was encouraged by my thesis adviser, Professor Ryan Vinroot, to ex-plore a recent journal article by the mathematician A. M. Vershik entitled “ANew Approach to the Representation Theory of the Symmetric Groups, III: In-duced Representations and the Frobenius-Young Correspondence,” since this arti-cle served to illustrate and apply many of the concepts I had been learning. Whilemaking my way through this dense work, I attempted to prove for myself someof the theorems Vershik gave, so as to better understand his reasoning. However,after a rather lengthy struggle to prove a central lemma of the paper, I realizedthat there were in fact counterexamples, and hence the proof of the lemma wasnot valid. After discussing the problem with Ryan, who agreed that my coun-terexamples held, I began to investigate the ramifications of this error, whetheror not the paper’s main argument could still stand, and whether the error couldbe corrected. This thesis is the culmination of these efforts; through it, I hope togive to the reader familiar with the basic concepts of group theory a mathemat-ical background sufficient to understand Vershik’s paper, and then to review hisargument and the counterexamples.

1 Introduction

This paper begins with an overview of some broad notions from combinatorics, inparticular the concepts of partitions of positive integers, Young diagrams, Youngtableaux, and orderings on the set of Young diagrams. These tools are foundationalto understand Vershik’s argument in [10], because he attempts a purely combi-natorial proof of the Frobenius-Young correspondence (see Section 4.4). Next,some basic concepts of the representation theory of finite groups are discussed,including Maschke’s Theorem and its consequences, induced representations, theFrobenius Reciprocity Theorem, the number of non-isomorphic irreducible rep-resentations, and Mackey’s Theorem. These tools are necessary to understandwhat the Frobenius-Young correspondence states, as well as to translate com-binatorial counterexamples to one of Vershik’s statements into their equivalentrepresentation theoretic counterexamples. Finally, the representation theory ofthe symmetric group Sn is explored, using both methods from both combinatorics

2 PARTITIONS 6

and representation theory of general finite groups. After a formal statement ofthe Frobenius-Young correspondence, facts such as the branching rules and thenotion of Young subgroups are used to trace through Vershik’s argument. Anin-depth analysis of Vershik’s third lemma (see section 4.6) reveals that whereVershik claims an equivalence between statements, there is in fact only an impli-cation in one direction. This error is explored, an infinite family of combinatorialcounterexamples is given, and the equivalent counterexamples in representationtheory are also given. Ultimately, though Vershik’s attempted purely combina-torial proof collapses, it is possible that there may be another method by whichto successfully complete such a proof, and this question is proposed for furtherinvestigation.

2 Partitions

Partitions are an important tool in group theory, particularly in the theory of thesymmetric group. A partition λ of the integer n, written λ ` n, is a sequence ofpositive integers λ = (λ1, λ2, . . . , λk) such that λi ≥ λi+1 for all integers 1 ≤ i ≤(k − 1) and

∑ki=1 λi = n. Each element λi of the partition is called a part of the

partition.

2.1 Conjugate Partitions

For each partition λ of n, there is a conjugate partition of n, denoted λ′ =(λ′1, λ

′2, . . . , λ

′l), where

λ′i = max{j | λj ≥ i}.

The fact that λ′ is actually a partition of n may not be immediately apparent butwill become clearer with the introduction of Young diagrams.

2.2 Diagrams and Tableaux

One important way of visualizing partitions is through the use of Young diagrams.For each partition λ = (λ1, λ2, . . . , λk) the corresponding Young diagram, [λ],is a left-justified diagram in which the ith row from the top contains λi boxes, orcells. For example, the Young diagram of the partition λ = (6, 4, 1, 1) of n = 12

2 PARTITIONS 7

is:

[λ] =

The conjugate partition, as defined above, yields a conjugate Young dia-gram. In relation to the original Young diagram [λ], the conjugate Young diagram[λ′] turns out to be the diagram which interchanges the rows and columns of [λ];that is, the diagram obtained by reflecting [λ] about the main diagonal. This helpsto justify the fact that the conjugate partition is in fact another partition of n,since by considering the Young diagrams, one may realize that reflection aboutthe main diagonal will result in a new diagram with the same number of boxes,namely n. In particular, if λ = (6, 4, 1, 1) as above, the conjugate partition isλ′ = (4, 2, 2, 2, 1, 1) and the Young diagram associated with this partition is:

[λ′] =

Proposition 2.1. (λ′)′ = λ

Proof. Passing to the conjugate is equivalent to making the rows of the originaldiagram into the columns of the conjugate diagram and vice versa. Thus, takingthe conjugate again will switch the rows and columns back, once again giving theoriginal diagram.

Sometimes, such as when working with the certain aspects of the symmetricgroup Sn, it may be desirable to place the integers 1 through n, once each, inthe boxes of the Young diagram; the result of this is a Young tableau. Usingthe aforementioned partition λ = (6, 4, 1, 1) of n = 12, one Young tableau of thispartition is given as follows:

1 2 3 4 5 67 8 9 101112

One other important term is the shape of a Young tableau, which refers to thepartition that corresponds to the underlying Young diagram. The Young tableau

2 PARTITIONS 8

above has shape λ = (6, 4, 1, 1), for example.

2.3 Orderings

There are two important orderings on Young diagrams, and thus also on partitionsof n. The dominance ordering is a partial ordering on partitions of n. Givenany two partitions µ = (µ1, µ2, . . . , µl) and λ = (λ1, λ2, . . . , λk) of n, µ dominatesλ, denoted µ D λ or λ E µ, if and only if

∑mi=1 µi ≥

∑mj=1 λj for all m between 1

and max{k, l}, inclusive, defining µi = 0 if i > l and λj = 0 if j > k. In terms ofYoung diagrams, [µ] D [λ] if and only if for all m between 1 and max{k, l}, thereare more boxes in the first m rows of [µ] than there are in the first m rows of [λ].For instance, given the partitions µ = (6, 3, 2, 1) and λ = (5, 3, 3, 1) of n = 12, onecan see that µ D λ, that is:

D

For n ≤ 5, the dominance ordering is more than a partial ordering; it is in facta total ordering. Consider, for instance, the first two values of n. When n = 1,the only Young diagram is the unique diagram with one block, that is:

[λ1] =

Moving up to n = 2, there are two distinct Young diagrams: the one obtainedby adding a second block to the one row of [λ1] and the one obtained by addinga second row of one block to [λ1]. It is clear that this is total ordering, since

D

By constructing the tree of Young diagrams which partition successively greatern, it can be checked that for n ≤ 5, any two diagrams partitioning n can be com-pared by the dominance ordering; hence, for these n, the dominance ordering isa total ordering. However, when n = 6, the partitions and corresponding dia-grams of λ = (3, 1, 1, 1) and µ = (2, 2, 2) cannot be compared with the dominanceordering, and so the ordering becomes only a partial ordering.

The other important ordering is used to relate two Young diagrams whichcorrespond to partitions of integers n and n + 1. If γ ` n and λ ` (n + 1), then

2 PARTITIONS 9

one writes that γ ≺ λ or λ � γ if and only if the diagram [λ] can be obtainedby adding one block to [γ] in such a way that the result is still a valid diagram,or equivalently, [γ] can be obtained by removing a block from [λ] in such a waythat the result is still a valid diagram. Using the partition λ = (6, 3, 2, 1) of 12,and defining a partition γ = (5, 3, 2, 1) of 11, one can see by looking at the Youngdiagrams that γ ≺ λ:

≺

2.4 Dominance Ordering on Conjugate Partitions

In Vershik’s attempt to develop a purely combinatorial proof of the Frobenius-Young correspondence (see Section 4.4), he seeks to build a connection betweentwo collections of Young diagrams. His goal in doing so is to eventually show theuniqueness of a diagram that he claims is common to both collections. The proofof Vershik’s second lemma is a corollary of another lemma, which follows.

Lemma 2.2. Let λ and µ be two partitions of n. Then µ D λ if and only ifλ′ D µ′.

Proof. (Following the proof of (1.11) in [4].) First, note that proving one implica-tion also proves the other, because if (µ D λ) ⇒ (λ′ D µ′), then Proposition 2.1implies that (λ′ D µ′) ⇒ (µ′′ D λ′′) ⇒ (µ D λ).

Define partitions λ = (λ1, λ2, . . . , λa) and µ = (µ1, µ2, . . . , µb) of the integer n.Suppose λ′ 4 µ′ and µ′ 4 λ′ (this second condition is necessary; without it, onecould simply switch the roles of µ and λ). Then there exists an j ∈ N such that

p∑i=1

λ′i ≥p∑

i=1

µ′i

for all 1 ≤ p ≤ j − 1, andj∑

i=1

λ′i <

j∑i=1

µ′i. (1)

Taken together, these two equations imply that µ′j = m is greater than λ′j = l.Also, since the sum of all the parts of both µ′ and λ′ must equal n, equation (1)

2 PARTITIONS 10

implies thata∑

i=(j+1)

λ′i >

b∑i=(j+1)

µ′i. (2)

Now consider the Young diagrams [λ] and [µ] corresponding to λ and µ, re-spectively. The sum

∑ai=(j+1) λ

′i gives the number of boxes of [λ] which are to the

right of column j, and similarly the sum∑b

i=(j+1) µ′i gives the number of boxes of

[µ] which are to the right of column j. Hence,

a∑i=(j+1)

λ′i =l∑

k=1

(λk − j) andb∑

i=(j+1)

µ′i =m∑

k=1

(µk − j).

Thus, applying equation (2) and observing that m > l and µk ≥ j whenever1 ≤ k ≤ m, one may see that

l∑k=1

(λk − j) >m∑

k=1

(µk − j) ≥l∑

k=1

(µk − j). (3)

Adding lj to both sides of the inequality of the leftmost and rightmost sum inequation (3) yields

l∑k=1

λk >l∑

k=1

µk

and thus, µ 4 λ.Therefore, (λ′ 4 µ′) ⇒ (µ 4 λ), and the contrapositive of this statement shows

that (µ D λ) ⇒ (λ′ D µ′), which by the earlier argument also demonstrates that(µ D λ) ⇔ (λ′ D µ′).

Corollary 2.3. (Lemma 2 in [10]) Let [λ] be an arbitrary diagram which partitionsn, and let [λ′] be its conjugate diagram. Define two collections of diagrams: A ={[α] | α ` n, α D λ} and B = {[β′] | β′ ` n, β D λ′}. Then A ∩B = {[λ]}.

Proof. By the above lemma, B is equivalent to the set of all diagrams [β′] forwhich λ D β′ (since λ′′ = λ by Proposition 2.1). So for a diagram [φ] to be in theintersection of A and B, it must be true that both φ D λ (to be in A) and λ D φ(to be in B). Since D is a partial ordering, this can only occur if φ = λ. Thus, [λ]is the only diagram common to both sets.

3 REPRESENTATION THEORY 11

3 Representation Theory

Let G be a finite group. A representation of G over F is a homomorphism φfrom a group G to GL(n,F) for some field F. In this paper, the field F will betaken to be the field of complex numbers, C, so a representation will be a mapφ : G → GL(n,C) such that φ(gh) = φ(g)φ(h) for all g, h ∈ G. The value of n iscalled the dimension of the representation φ. All groups G have a representationφ : G→ GL(1,C) where

φ(g) = 1

for all g ∈ G. This representation is called the trivial representation of G and isdenoted 1. For the groupG = Sn, one other important representation of dimension1 is the sign representation, denoted sgn, for which

sgn(g) =

{1 if g is an even permutation,

−1 if g is an odd permutation.

Two representations ρ : G → GL(m,C) and σ : G → GL(n,C) are said to beequivalent if and only if m = n and there exists a matrix T ∈ GL(n,C) suchthat σ(g) = T−1(ρ(g))T for all g ∈ G. In this case, one writes ρ ∼= σ.

3.1 Group Algebras

Given a finite group G and the field C, one useful vector space is the groupalgebra of G over C, denoted CG. This vector space is the set of all formalC-linear combinations of elements in G, that is,

CG =

{∑g∈G

αgg | αg ∈ C

}.

The group algebra CG has addition defined by(∑g∈G

αgg

)+

(∑g∈G

βgg

)=∑g∈G

(αg + βg)g

and multiplication defined by(∑g∈G

αgg

)(∑h∈G

βhh

)=∑

g,h∈G

αgβh(gh).


3.2 CG-modules

Given a group G and a finite-dimensional vector space V over the field C, V is aCG-module if there exists a multiplication gv which satisfies the following fiveproperties for all g, h ∈ G, u, v ∈ V, c ∈ C:(1) gv ∈ V(2) (gh)v = g(hv)(3) 1v = v(4) g(cv) = c(gv)(5) g(u+ v) = gu+ gv

Theorem 3.1. CG-modules correspond exactly to representations of G over C.

Proof. Given a group G, a representation ρ : G → GL(n,C), and V = Cn, aCG-module can be obtained by defining multiplication gv as

gv = (ρ(g))v for all g ∈ G, v ∈ V .

One can check that the five axioms hold for this definition, and thus the represen-tation gives rise to a CG-module.

To prove the correspondence in the other direction, consider a CG-module Vwith basis B = {v1, v2, . . . , vn}. For each g ∈ G, define by [g]B the matrix corre-sponding to the map f : V → V such that f(v) = gv, relative to the basis B. Thensince (gh)v = g(hv) for all g, h ∈ G and all v ∈ B, [gh]B = [g]B[h]B. Furthermore,since this implies that [1]B = [g]B[g−1]B, each matrix [g]B is invertible. Thus, themap φ for which φ(g) = [g]B is a homomorphism from G to an invertible matrix,that is, φ : G → GL(n,C) (where n = dim V ), and so φ is a representation of Gover C.

For a given CG-module V , a CG-submodule of V is a subspace W of V whichis also a CG-module under the same multiplication. Note here that for any CG-module V , both the zero subspace {0} and the whole space V are CG-submodules.

Now define V to be the vector space CG, with the multiplication of CG onV defined in the natural way. Then, since the five axioms are satisfied, V is aCG-module, called the regular CG-module.

3.3 Irreducible Representations

A CG-module V is called irreducible if V 6= {0} and the only CG-submodules ofV are {0} and V ; otherwise, V is said to be reducible. Likewise, a representation


ρ : G→ GL(n,C) is irreducible if and only if the corresponding CG-module Cn

given bygv = (ρ(g))v for all g ∈ G, v ∈ Cn

is irreducible. Any representation which is not irreducible is also said to be re-ducible.

3.4 Decomposition of Representations

One of the most important theorems in representation theory (see [2], Chapter 8)is Maschke’s Theorem, which implies that any representation can be decomposedinto a direct sum of irreducible representations.

Theorem 3.2. (Maschke’s Theorem [over C]) Let G be a finite group and V bethe corresponding CG-module. Then if U is any CG-submodule of V , there existsa CG-submodule W of V such that V = U ⊕W .

A CG-module V is said to be completely reducible if

V =k⊕

i=1

Ui

where each Ui is an irreducible CG-submodule of V . Using Maschke’s Theoremand induction on dim(V ), one can prove the following important fact: if G is afinite group, then every nonzero CG-module is completely reducible.

3.5 Restricted and Induced Representations

If ρ is a representation of a group G and H is a subgroup of G, then the re-stricted representation, ResHρ, is a representation of H obtained by defining(ResHρ)(h) = ρ(h) for all h ∈ H.

In the same sense that restricted representations yield information about rep-resentations of a smaller group from information about representations of a largergroup, one may also consider induced representations, which use information abouta subgroup to give information about representations of the whole group. How-ever, before defining an induced representation, new notation must be established.Given a finite group G, consider a subset X of CG and define a subspace X(CG)of CG by

X(CG) = spanC{gx : g ∈ G, x ∈ X}.


ThenX(CG) is a CG-submodule of CG. Now letH be a subgroupG. If U is a CH-submodule of CH, then the induced module from H to G is the CG-moduleU(CG). This CG-module corresponds to a representation called the inducedrepresentation, which is denoted by IndG

Hρ where ρ is the representation of Hcorresponding to U .

3.6 Characters

For each representation φ of a group G, define the character χ of the represen-tation to be the function χ : G → C such that χ(g) = tr(φ(g)) for all g ∈ G.A central theorem of representation theory, the proof of which may be found inany standard text on the subject (for example, Proposition 13.5(1) and Theorem14.21 of [2]), is the following surprising and useful fact:

Theorem 3.3. Suppose ρ and σ are two representations of a group G over Cwith characters χ and ψ, respectively. Then ρ is equivalent to σ if and only ifχ(g) = ψ(g) for all g ∈ G.

Given a finite group G which has a representation ρ with character χ and asubgroup H of G which has a representation σ with character ψ, the characterof the representation ResHρ obtained by evaluating χ on only the elements ofH is called the restriction of χ to H and is written ResHχ. Similarly, thecharacter of IndG

Hσ is called the induction of ψ to G, written IndGHψ. Let 1 denote

both the trivial representation of a group as well as the character of the trivialrepresentation, which are essentially the same, since the trivial representation hasdimension one. As proved by Proposition 21.19 of [2], this character has valuesgiven by the equation

(IndGHψ)(g) =

1

|H|∑y∈G

ψ̇(y−1gy)

for all g ∈ G, where the value of ψ̇ is given as follows:

ψ̇(g) =

{ψ(g) if g ∈ H,

0 if g 6∈ H.

Using this formula, one can prove the following fact:

Theorem 3.4. Let G be a group with subgroup H, and let g ∈ G. Then IndGH1 ∼=

IndGgHg−11.


Proof. Consider the value of the induced characters on an arbitrary element x ∈ G.By the above formula,

(IndGgHg−11)(x) =

1

|gHg−1|∑y∈G

1̇gHg−1(y−1xy).

Since conjugation of a subgroup does not change its order, and since y−1xy ∈gHg−1 implies y−1xy ∈ H,

(IndGgHg−11)(x) =

1

|H|∑y∈G

1̇H(y−1xy).

As y runs over G, yg also runs over G, and so

(IndGgHg−11)(x) =

1

|H|∑y∈G

1̇H((yg)−1x(yg)) =1

|H|∑y∈G

1̇H(g−1y−1xyg).

But y−1xy ∈ gHg−1 ⇔ g−1y−1xyg ∈ H, which implies that 1̇H((yg)−1x(yg)) willbe nonzero exactly when 1̇gHg−1(y−1xy) is nonzero. Hence,

(IndGgHg−11)(x) = (IndG

H1)(x),

and so the characters of the induced representations agree. Therefore, by Theorem3.3, the induced representations are equivalent.

3.7 Inner Products

Another important concept in representation theory is that of an inner product.In general, an inner product associates to any two vectors φ, ψ from a givenvector space V a number 〈φ, ψ〉 from a given field F such that the following threeproperties hold:(1) 〈φ, λ〉 = 〈λ, φ〉 for all φ, λ ∈ V(2) 〈c1φ1 + c2φ2, λ〉 = c1〈φ1, λ〉+ c2〈φ2, λ〉 for all c1, c2 ∈ F and φ1, φ2, λ ∈ V(3) 〈φ, φ〉 > 0 for all φ 6= ~0

Looking in particular at the vector space of functions from a group G to thefield C, an inner product of two characters χ, ψ of G may be defined as follows:

〈χ, ψ〉 =1

|G|∑g∈G

χ(g)ψ(g−1).


It is straightforward to check that this definition satisfies the three properties thatdefine an inner product.

Even more can be said about taking the inner product of two irreduciblecharacters, that is, characters which correspond to irreducible representations.As shown in the proof of Theorem 14.12 in [2], if χ, ψ are irreducible charactersof a finite group G, then

〈χ, ψ〉 =

{1 if χ = ψ,

0 if χ 6= ψ.

This important equation is called the Schur orthogonality relation.Finally, since any representation may be decomposed into a direct sum of ir-

reducible representations by Theorem 3.2, any character of G can be written asa sum irreducible characters. That is, if {χ1, χ2, . . . , χk} is the set of all irre-ducible characters of G (see Theorem: 3.6), then an arbitrary character ψ of Gcan be written as ψ =

∑ki=1mχi

χi for some nonnegative integers mχifor which

mχi= 〈ψ, χi〉. By extension, this also shows that to the inner product of any two

characters of G must be an integer (see [2], Theorem 14.24). This integer givesinformation about the multiplicities of the irreducible representations common tothe decompositions of the two representations corresponding to the characters inthe inner product.

3.8 Frobenius Reciprocity

One very elegant theorem of representation theory is the Frobenius ReciprocityTheorem, which describes an important relationship between restricted and in-duced representations. A proof of this theorem may be found in [6, Theorem1.12.6]. If H is a subgroup of G, χ is a character of G, and ψ is a character of H,then the Frobenius Reciprocity Theorem states:

Theorem 3.5. (Frobenius Reciprocity Theorem)

〈IndGHψ, χ〉G = 〈ψ,ResHχ〉H .

3.9 The Number of Irreducible Representations

For a given group G with an element x, recall that the conjugacy class of x inG is the set

xG = {g−1xg | g ∈ G}.


For any finite group G, it is possible to determine the exact number of (non-isomorphic) irreducible representations (or, equivalently, characters) of G via thefollowing fact, the proof of which may be found in Chapter 15 of [2]:

Theorem 3.6. The number of distinct irreducible characters of G is equal to thenumber of distinct conjugacy classes of G.

3.10 Double Cosets

Before stating the last theorem of representation theory needed for the remainderof this paper, it is necessary to understand double cosets. Given an element s ina group G, which has subgroups H,K, the (H,K)-double coset of s in G is theset HsK = {hsk | h ∈ H, k ∈ K}.

Theorem 3.7. Define a relation ∼ by x ∼ y if and only if x and y are in thesame (H,K)-double coset, that is, y = hxk for some h ∈ H, k ∈ K. Then ∼ is anequivalence relation.

Proof. (Reflexive:) Let x ∈ G. Then x = 1x1 and, since H and K are subgroups,1 ∈ H and 1 ∈ K; thus x ∼ x.

(Symmetric:) Let x, y ∈ G such that x ∼ y. Then y = hxk for some h ∈ Hand k ∈ K, so x = h−1yk−1. But h−1 ∈ H and k−1 ∈ K, and hence y ∼ x.

(Transitive:) Let x, y, z ∈ G such that x ∼ y and y ∼ z. Then y = h1xk1

and z = h2yk2 for some h1, h2 ∈ H and k1, k2 ∈ K. Then z = h2(h1xk1)k2 =(h2h1)x(k1k2), and since h2h1 ∈ H and k1k2 ∈ K, x ∼ z.

Hence, just like left or right cosets, the (H,K)-double cosets of G form apartition of G. Furthermore, the left and right cosets are a sub-partition of the(H,K)-double cosets. To see this, recall that y = xk for some k ∈ K if and onlyif x and y are in the same left coset of K in G, or y ∈ xK. But then it is alsotrue that y = 1xk, and since 1 ∈ H, this implies that y ∈ HxK, and thus x andy are in the same (H,K)-double coset. A similar argument shows the same resultfor right cosets.

3.11 Mackey’s Theorem

One question one might ask is what happens when a representation is inducedfrom one subgroup up to a larger group and then the resulting representation isrestricted back down to another subgroup. The answer to this question is givenby an important theorem known as Mackey’s theorem. Let G be a finite group,

4 REPRESENTATIONS OF THE SYMMETRIC GROUPS 18

H and K be subgroups of G, and ρ be a representation of H. Given s in theset of (H,K)-double coset representatives, define Hs = sHs−1 ∩K and define therepresentation ρs of Hs by ρs(x) = ρ(sxs−1). Mackey’s theorem, a proof of whichmay be found in Section 7.3 of [7], then states the following:

Theorem 3.8. (Mackey’s Theorem)

ResKIndGHρ

∼=⊕

s∈K\G/H

IndKHsρs.

4 Representations of the Symmetric Groups

With an understanding of the representation theory presented above, it is nowpossible to begin looking at a recent paper by A. M. Vershik. His 2006 articleentitled “A New Approach to the Representation Theory of the Symmetric Groups,III: Induced Representations and the Frobenius-Young Correspondence” attemptsto use a purely combinatorial argument to present a new proof of a previously-known theorem.

The aforementioned article draws on earlier articles written by Vershik andOkounkov (see [5], [9]). In the introduction to [9], Vershik and Okounkov state thatthey seek a “more direct and natural” method of realizing the main combinatorialideas of the representation theory of Sn. Specifically, the paper developed a newway of looking at the representations of Sn which considers the whole chain S1 ⊂S2 ⊂ · · · ⊂ Sn and builds inductively to Sn+1. Perhaps the most importantdifference between Vershik and Okounkov’s method in contrast to the older methodis that the new perspective gives the branching rules (see Section 4.2, below)immediately as a natural result rather than as a final corollary of a quite technicalargument.

Relying on these same ideas (the chain of embedded symmetric groups and thebranching rules), Vershik strives in [10] to develop a purely combinatorial proofof the famous Frobenius-Young correspondence (see Section 4.4). This correspon-dence is known to be true from earlier proofs, such as the one given in Lecture 4of [1], but Vershik desires to reach the same conclusion via a different (inductive)method. However, as will soon be demonstrated, Vershik’s argument is flawed; inhis sequence of lemmas, Vershik asserts that two statements are equivalent whenin fact the implication only is true in one direction. This mistake invalidatesVershik’s attempted proof, as shown through various counterexamples.


4.1 The Irreducible Representations of Sn

It is known (see, for instance, Chapter 12 of [2]) that for the group G = Sn,conjugacy classes correspond to cycle-type, that is to say:

Theorem 4.1. For x ∈ Sn, the conjugacy class xSn consists of all permutationsin Sn which have the same cycle-type as x.

Cycle-types in Sn correspond to partitions of the integer n. Hence, applyingTheorem 3.6, the number of irreducible characters of Sn is equal to the number ofpartitions of n.

Thus, for any partition λ ` n, one can define an irreducible representation ofSn, denoted by πλ, which corresponds to the given partition. There are severalmethods to construct these πλ, one of which was the method developed by Vershikand Okounkov in [5] and [9], and it is this point of view which will be assumed forthe remainder of this paper.

By way of example, consider the partition λ = (3, 2, 1) of n = 6, which corre-sponds to the cycle-type (3,2,1), which in turn corresponds to the conjugacy classof S6 of all elements with cycle-type (3,2,1), which in turn corresponds to someirreducible representation πλ of S6.

4.2 Branching Rule

A central idea of Vershik’s two earlier papers is the development of a knowntheorem (for example, see Theorem 2.8.3 of [6]) which gives information about therestriction or induction of an irreducible representation ρ of Sn to Sn−1 or Sn+1,respectively. This theorem is key to Vershik’s overall proof strategy because heargues using induction on n, and so the branching rule gives him a way to movefrom Sn up to Sn+1.

Theorem 4.2. (Branching Rule) If λ ` n, then

ResSn−1πλ∼=⊕γ≺λ

πγ, and

IndSn+1

Snπλ∼=⊕λ≺µ

πµ.

4.3 Young Subgroups

For each symmetric group Sn, a subgroup is called a Young subgroup if it isisomorphic to Sλ1 × Sλ2 × · · · × Sλk

for some partition λ = (λ1, λ2, . . . , λk) of the


integer n. One specific Young subgroup, denoted by Sλ, is the Young subgroup ofSn for which

Sλ1 acts on the set {1, 2, . . . , λ1}...

Sλiacts on the set

{(i−1∑j=1

λj

)+ 1,

(i−1∑j=1

λj

)+ 2, . . . ,

(i−1∑j=1

λj

)+ λi

}...

Sλkacts on the set

{(k−1∑j=1

λj

)+ 1,

(k−1∑j=1

λj

)+ 2, . . . , n

}.

Now, for any m ≤ n, there are

(n

m

)ways to embed the group Sm within Sn

such that both symmetric groups act on individual elements (as opposed to pairsof elements or subgroups, for instance), but for the remainder of this paper, theassumption will be that an element of Sm will act on the numbers 1 through mand leave the remaining (n−m) elements fixed. Based on facts known from grouptheory about the conjugation of subgroups, given an element σ ∈ Sn, the Youngsubgroup σSλσ

−1 is isomorphic to the Young subgroup Sλ. For this conjugatesubgroup, the Sλ1 term in the direct product expansion acts on the elements{σ(1), σ(2), . . . , σ(λ1)}, and similarly for any i, the Sλi

term in the direct productexpansion acts on the elements{

σ

([i−1∑j=1

λj

]+ 1

), σ

([i−1∑j=1

λj

]+ 2

), . . . , σ

([i−1∑j=1

λj

]+ λi

)}.

Also note that given any partition λ = (λ1, λ2, . . . , λk) of n, each Young subgroupisomorphic to Sλ1 ×Sλ2 × · · · ×Sλk

corresponds to a Young tableau of shape λ byfilling in the row of [λ] corresponding to the part λi with the subset of {1, 2, . . . , n}on which the Sλi

factor acts in the Young subgroup. The tableau correspondingto a subgroup will be denoted with brackets; for instance, the tableau for Sλ is[Sλ].

In any of these Young tableaux, the convention will be to write the numbers ina given row in strictly increasing order from left to right and, if multiple rows haveequal length, to order them so that the rightmost boxes of these rows are strictlyincreasing from top to bottom. Finally, note that if any two Young tableaux have


the same shape, then they must arise from isomorphic Young subgroups, bothisomorphic to Sλ, where λ is the shape of the Young tableau.

To help illustrate these concepts, consider the partition λ = (6, 3, 2, 1) of n =12. A corresponding Young subgroup will then be isomorphic to S6×S3×S2×S1.The Young subgroup can then be translated into a Young tableau; given sometransposition σ = (3, 12) ∈ S12, for example,

[Sλ] =

1 2 3 4 5 67 8 9101112 and [σSλσ

−1] =

1 2 4 5 6 127 8 910113

As is the case in this particular example, the Young tableau corresponding to anySλ will have the integers 1 through n in successive order moving from left to rightand top to bottom.

4.4 Frobenius-Young Correspondence

Using the definitions of the previous subsection, consider the sets of irreduciblecomponents of the decompositions of IndSn

Sλ1 and IndSn

Sλ′sgn with multiplicity. The

Frobenius-Young correspondence, given as Corollary 4.39 in [1], asserts that thesetwo sets have exactly one representation in common, which is πλ, and that thisrepresentation occurs in each set with multiplicity one.

Theorem 4.3. (Frobenius-Young correspondence) If the decomposition of IndSnSλ

1

into irreducible representations is⊕p

i=1 ρi and the decomposition of IndSnSλ′

sgn into

irreducible representations is⊕q

j=1 γj, then {ρi | 1 ≤ i ≤ p} ∩ {γj | 1 ≤ j ≤ q} ={πλ}.

4.5 Mackey’s Theorem Applied to Sn

In his attempt to prove the Frobenius-Young correspondence using the branchingrule, Vershik uses a sequence of three lemmas from which the desired correspon-dence is supposed to follow. The first of these is an application of Mackey’sTheorem (Theorem 3.8) to the symmetric group.

Theorem 4.4. (Lemma 1 in [10])

ResSn−1IndSnSλ

1 ∼=⊕

γ:γ≺λ

c(λ, γ)IndSn−1

Sγ1


where c(λ, γ) is the number of rows of the diagram [λ] which have the same lengthas the row of [λ] which is being modified to yield the diagram [γ].

Proof. Mackey’s Theorem yields the following (taking G = Sn, H = Sλ, K = Sn−1,and ρ equal to the trivial representation 1):

ResSn−1IndSnSλ

1 ∼=⊕

σ∈Sn−1\Sn/Sλ

IndSn−1

σSλσ−1∩Sn−11.

Note that ρs = 1 since the value of the trivial character is 1 for all g ∈ G, andhence for all g ∈ Hs.

Next, it is necessary to find the distinct representatives of the double cosetsSn−1\Sn/Sλ. The set C = {(1), (1, n), (2, n), . . . , (n − 1, n)} is a complete setof coset representatives for the right cosets Sn−1\Sn. Now, for a given Sλ cor-responding to the partition λ = (λ1, λ2, . . . , λk) of n, consider two representa-tives σ1 and σ2 from the same (Sn−1, Sλ)-double coset. Then it must be thecase that τ2σ1τ1 = σ2 for some τ2 ∈ Sn−1 and τ1 ∈ Sλ. Suppose that σ1(i) =n, and τ1(j) = i, which occurs if and only if i and j are in the same subset{(∑i−1

j=1 λj

)+ 1,

(∑i−1j=1 λj

)+ 2, . . . ,

(∑i−1j=1 λj

)+ λi

}on which Sλi

acts. Since

τ2 ∈ Sn−1, τ2 fixes n, and hence τ2σ1τ1(j) = n, and so also σ2(j) = n. Hence, if iand j are in the same row of the Young tableau [Sλ], then (i, n) and (j, n) representthe same (Sn−1, Sλ)-double coset. Also, if i and j are in different rows of [Sλ], then(i, n) and (j, n) represent distinct (Sn−1, Sλ)-double cosets. Thus, a complete set ofall (Sn−1, Sλ)-double coset representatives is any subset {(i1, n), (i2, n), . . . , (ik, n)}of the set C such that each i` is in a different row of the Young tableau [Sλ].

Taking any σ = (i`, n) from the set of (Sn−1, Sλ)-double coset representatives,the Young subgroup σSλσ

−1 gives a Young tableau which has the same shape as[Sλ] but which reverses the positions of i` and n. Intersecting this with Sn−1 hasthe effect of removing from the Young tableau the last box of the row j containingn, since by the construction of the Young tableau, n is written in the last boxof a collection of rows of the same size. This gives a Young tableau for Sn−1 ofshape γ = (λ1, λ2, . . . , λj − 1, . . . , λk). For each row of [Sλ] which has the samelength as row j, conjugating Sλ by a transposition corresponding to an elementin that row and intersecting with Sn−1 gives a Young subgroup isomorphic toσSλσ

−1 intersected with Sn−1, since the resulting Young tableau has the sameshape γ. It then follows from Theorem 3.4 that each of the corresponding inducedrepresentations are isomorphic. So, for each partition γ ≺ λ, there are exactlyc(λ, γ) copies of this induced representation. Thus, the direct sum is taken over allpossible γ ≺ λ, where each of the respective induced representations is multiplied


by the factor c(λ, γ).

4.6 The Snapper Conjecture

The mistake that Vershik makes occurs when he sets out to prove his third lemma,which can be stated as follows:

Theorem 4.5. (Lemma 3 in [10]) The irreducible representations πµ appearing inthe decomposition of the induced representation IndSn

Sλ1 correspond to partitions µ

which dominate λ.

This lemma is previously known to be true. The original formulation of thisproblem was put forth by Snapper, who proved that

[〈IndSnSλ

1, πµ〉 > 0] ⇒ [µ D λ]

and conjectured, but did not prove, that the converse was also true (see [8]).Two years later, the converse was in fact shown to be true by Liebler and Vitale(see [3]). Because of its history, Theorem 4.5 is sometimes referred to as theSnapper conjecture. Despite this background knowledge, Vershik seeks a newinductive proof relying on combinatorics and the branching rules. However, inthe translation of the representation theory into combinatorics, Vershik asserts anequivalence between two statements where there is actually only an implication inone direction.

4.6.1 Combinatorial Statement

In Vershik’s attempted proof of Theorem 4.5, he makes one central claim, namely,he gives the following two statements about the partitions λ ` n, µ ` n and assertsthat they are equivalent:

Statement 4.6. µ D λ

Statement 4.7. For all ρ ` (n− 1) for which ρ ≺ µ, there exists γ ` (n− 1) forwhich γ ≺ λ such that ρ D γ.

One of these two implications is true, that is:

Theorem 4.8. Statement 4.6 implies Statement 4.7.


Proof. Define µ = (µ1, µ2, . . . , µa) and define λ = (λ1, λ2, . . . , λb). If µ D λ, thenby definition, for all 1 ≤ i ≤ max {a, b}

i∑j=1

µj ≥i∑

j=1

λj.

Now define the diagram [ρ] to be the diagram [µ] with one block removed fromthe end of row k. Using this definition, one may see that [λ] must also have atleast k rows, otherwise

k−1∑j=1

µj < n =k−1∑j=1

λj,

which would violate the dominance ordering.Define a removable block to be a block of a diagram [λ] which may be

removed to yield another valid diagram [γ] (such that γ ≺ λ). Every diagramhas at least one removable block, since the rightmost cell of the bottom row ofany diagram must be removable. Also, except for the removable block in the finalrow, there is a removable block in row i of [λ] if and only if row (i+ 1) has lengthstrictly less than row i. With this definition in hand, consider two cases.

Case 1: Suppose that [λ] has a removable block in row ` ≤ k and form thediagram [γ] by removing the removable block in row ` of [λ]. Then:

for m < `, µm = ρm and λm = γm,

for ` ≤ m < k, µm = ρm and (m∑

j=1

λj)− 1 =m∑

j=1

γj,

for m ≥ k, (m∑

j=1

µj)− 1 =m∑

j=1

ρj and (m∑

j=1

λj)− 1 =m∑

j=1

γj.

Since for all i,∑i

j=1 µj ≥∑i

j=1 λj, one can use these three equations to show that,

for all i,∑i

j=1 ρj ≥∑i

j=1 γj, which in turn implies that ρ D γ as desired.Case 2: Now assume that [λ] has no removable block in any row r ≤ k. Define

` to be the uppermost row of [λ] which has a removable block. Then it must betrue that ` > k. Also, for all 1 ≤ m ≤ `:

m∑j=1

λj = mλ1 ≤m∑

j=1

µj,


where the equality on the left is due to the fact that having no removable blockin the first (`− 1) ≥ k rows implies that the first ` rows all have the same length,namely that of λ1, and where the inequality on the right is due to the dominanceordering.

Now, consider∑k

j=1 µj. Since [µ] has a removable block in row k, it followsthat µk > µk+1. Also, since [λ] does not have a removable block in row k, it mustbe true that λk = λk+1. Combining these two facts with the dominance orderingon µ and λ, it must be true that

k∑j=1

λj <k∑

j=1

µj

since equality would imply that∑k+1

j=1 λj >∑k+1

j=1 µj, in violation of the dominanceordering. Following the same line of reasoning used in Case 1, it may now beenseen that

for m < k, µm = ρm and λm = γm,

for k ≤ m < `, (m∑

j=1

µj)− 1 =m∑

j=1

ρj and λm = γm,

for m ≥ `, (m∑

j=1

µj)− 1 =m∑

j=1

ρj and (m∑

j=1

λj)− 1 =m∑

j=1

γj.

Hence, since [∑k

j=1 λj <∑k

j=1 µj] ⇒ [∑k

j=1 λj ≤ (∑k

j=1 µj) − 1], these threestatements together imply that ρ D λ.

4.6.2 Combinatorial Counterexamples

Despite Theorem 4.8, a relatively simple counterexample shows that the equiva-lence which Vershik asserts is in fact false. Consider the partitions µ = (2, 2, 2)and λ = (3, 1, 1, 1) of n = 6. Recall that, as noted in Section 2.3, these twopartitions cannot be compared using the dominance ordering.

[µ] = , [λ] =


For the given µ, the only possible ρ ≺ µ corresponds to the partition ρ = (2, 2, 1).There are two choices for γ, but choosing the partition γ = (2, 1, 1, 1) it is clearthat ρ D γ.

D

Hence, Statement 4.6 is false while Statement 4.7 is true, and so the two statementsare not equivalent.

While this particular counterexample occurs when n = 6, there exist counterex-amples for greater values of n as well. The value n = 6 yields the smallest possiblecounterexample, since, as mentioned in Section 2.3, the dominance ordering is atotal ordering for n ≤ 5, so in these instances it would not be possible to find twopartitions µ, λ such that µ 4 λ and λ 4 µ. Surprisingly enough, however, onecan find an infinite set of counterexamples when n ≥ 6. For instance, the aboveargument can be extended to any even n ≥ 6 (specifically, define n = 2k for somek ∈ N) by defining the diagram [µ] to have k rows of length 2 and defining [λ] tobe the diagram with one row of length 3 and (n−3) rows of length 1. Then µ 4 λ,since 2 < 3. But now choose [γ] to be the diagram [λ] with one block removedfrom the uppermost row; [ρ] must be be the diagram [µ] with one block removedfrom the bottommost row. Hence, ρ D γ, leading to the previously-establishedcontradiction, and so as n ranges over the positive integers greater than or equalto 6, this construction can be used to create an infinite class of counterexamples.

4.6.3 Representation Theory Counterexamples

With this background, one now can translate the combinatorial counterexamplesgiven in Section 4.6.2 into representation theoretic counterexamples. In whatfollows, define the value of the inner product of two representations to be the valueof the inner product of the corresponding characters, as defined in Section 3.7.Recall that, as noted in the discussion following Theorem 4.5, Vershik attemptsto use the branching rules to prove the known fact (see [8], [3]) that

〈πµ, IndSmSλ

1〉 6= 0 ⇔ µ D λ,

which is nothing more than a restatement of Theorem 4.5.


To begin, note that the implication

〈πµ, IndSmSλ

1〉 6= 0 ⇒ for all ρ ≺ µ, 〈⊕ρ≺µ̃

πµ̃, IndSmSλ

1〉 6= 0 (4)

is valid, since one may break the inner product of the direct sum into a sum ofinner products, each of which are greater than or equal to zero. For at least oneµ̃ in this sum (specifically, µ̃ = µ) the inner product is not equal to zero, hence itmust be greater than zero, and thus the sum of the inner products must also begreater than zero.

Then, using the branching rule for induced representations (Theorem 4.2),one can see that the right side of this implication is equivalent to the followingstatement:

for all ρ ≺ µ, 〈IndSmSm−1

πρ, IndSmSλ

1〉 6= 0.

By Frobenius Reciprocity (see Theorem 3.5), this statement is equivalent tothe assertion below:

for all ρ ≺ µ, 〈πρ,ResSm−1IndSmSλ

1〉 6= 0.

Theorem 4.4 can now be applied to change the restriction of the induction ofthe character 1 into a direct sum of representations:

for all ρ ≺ µ, 〈πρ,⊕γ≺λ

c(λ, γ)IndSm−1

Sγ1〉 6= 0.

Since an inner product is sesquilinear, the direct sum within the inner productcan be rewritten as a sum of inner products. Additionally, since each c(λ, γ) isa positive integer, these coefficients may be pulled out of each inner product justas they are, without needing to be conjugated. To be sure that the sum of theresulting inner products is not equal to zero, there must be at least one term ofthe sum which is not equal to zero, hence:

for all ρ ≺ µ, there exists a γ ≺ λ such that 〈πρ, IndSm−1

Sγ1〉 6= 0.

Finally, applying the equivalence in Lemma 4.5 shown by Snapper, Liebler,and Vitale, the statement about inner products may be converted into a statementabout partitions of (n− 1):

for all ρ ≺ µ, there exists a γ ≺ λ such that ρ D γ.

5 FURTHER QUESTIONS 28

This is none other than Statement 4.7 from the section above. Now, Vershik’serroneous claim is that this statement is equivalent to Statement 4.6, but as shownin Section 4.6.2, this is false. Still, by Theorem 4.5,

[〈IndSnSλ

1, πµ〉 > 0] ⇔ [µ D λ]

is true, as noted in the introduction to Section 4.6.All of the above equivalences and implications can be used to show that the

converse of statement (4) is false. First, recall that Statement 4.7 does not implyStatement 4.6 due to the combinatorial counterexamples given in Section 4.6.2.Now, following the above chain of equivalences from the bottom upwards, it mustbe the case that from Statement 4.7 implies the right side of statement (4). Ifthe converse of statement (4) were true, then this chain of implications wouldcontinue: the right side of statement (4) would imply the left side, which in turnwould imply Statement 4.6 by Snapper’s proof. Therefore, were the converse ofstatement (4) true, then Statement 4.7 would imply Statement 4.6, which hasalready been shown to be false. So the converse of statement (4) must be false.

In summary, for every combinatorial counterexample in which for all ρ ≺ µthere exists γ ≺ λ such that ρ D γ, and yet µ 4 λ, one can find that for allρ ≺ µ, 〈

⊕ρ≺µ̃ πµ̃, IndSm

Sλ1〉 6= 0, and yet the representation πµ does not appear in

the decomposition of IndSmSλ

1 into irreducible representations.

4.7 Frobenius-Young Correspondence, Revisited

As noted in the introduction to this section, despite the error in Vershik’s proof,the Frobenius-Young correspondence is still known to be true (see [1]). It mayeven be that there is a combinatorial proof of the correspondence which uses thebranching rules, as Vershik attempted to do, but such a proof would require adifferent method of argument. Any valid combinatorial method would need toarrive at the intended result by a proof which can altogether avoid the equivalenceVershik tried, and failed, to show.

5 Further Questions

The most obvious and important question which arises from the given considera-tions of Vershik’s paper is whether there is indeed a a combinatorial proof of theFrobenius-Young correspondence using the branching rules. Another question toinvestigate would be whether other known theorems on the representation the-

REFERENCES 29

ory of Sn, such as the Littlewood-Richardson Rule or the Murnaghan-NakayamaRule (see Sections 4.9 and 4.10 in [6]), can be proven using the point of viewestablished in Vershik and Okounkov’s earlier papers ([5] and [9]). One wouldhope that answers to such questions could help to complete another proof of theFrobenius-Young correspondence and to continue to broaden the perspective ofthe symmetric groups as pioneered by Vershik and Okounkov.

References

[1] W. Fulton and J. Harris, Representation Theory: A First Course. GraduateTexts in Mathematics, 129. Readings in Mathematics. Springer-Verlag, NewYork, 1991.

[2] G. James and M. Liebeck, Representations and Characters of Groups. Secondedition. Cambridge University Press, New York, 2001.

[3] R.A. Liebler and M.R. Vitale, Ordering the Partition Characters of the Sym-metric Group, J. of Algebra 25 (1973), 487–489.

[4] I.G. Macdonald, Symmetric functions and Hall polynomials. Oxford Math-ematical Monographs. The Clarendon Press, Oxford University Press, NewYork, 1979.

[5] A. Okounkov and A.M. Vershik, A New Approach to Representation Theoryof Symmetric Groups, Selecta Math. (N.S.) 2 (1996), no. 4, 581–605.

[6] B. Sagan, The Symmetric Group: Representations, Combinatorial Algo-rithms, and Symmetric Functions. Second edition. Graduate Texts in Math-ematics, 203. Springer-Verlag, New York, 2001.

[7] J.-P. Serre, Linear Representations of Finite Groups. Translated from thesecond French edition by Leonard L. Scott. Graduate Texts in Mathematics,42. Springer-Verlag, New York-Heidelberg, 1977.

[8] E. Snapper, Group characters and nonnegative integral matrices, J. of Algebra19 (1971), 520–535.

[9] A.M. Vershik and A. Okounkov, A New Approach to Representation Theoryof Symmetric Groups, II. (Russian) Zap. Nauchn. Sem. S.-Peterburg. Otdel.Mat. Inst. Steklov. (POMI) 307 (2004), 57–98.

REFERENCES 30

[10] A.M. Vershik, A New Approach to the Representation Theory of the Sym-metric Groups, III: Induced Representations and the Frobenius-Young Cor-respondence, Mosc. Math. J. 6 (2006), no. 3, 567–585.

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