On verti es of ompletely splittable modules for symmetri groups and simple modules labelled by two part partitionsSusanne DanzAbstra tWe prove that the verti es of a ompletely splittable module D for a �nite symmetri group over an algebrai ally losed �eld of hara teristi p > 0 are pre isely the defe tgroups of the blo k ontaining D. Moreover, we also determine the verti es of simplemodules for the symmetri groups in hara teristi p > 2 whi h are labelled by two partpartitions (n−m, m) su h that m < p(p + 1)/2.Subje t lassi� ation: 20C20, 20C30Keywords: simple module, symmetri group, two part partition, vertex1 Introdu tionIn 1959 A. J. Green [9℄ introdu ed the notion of verti es of inde omposable modules overgroup algebras. Given a �nite group G and an algebrai ally losed �eld F of positive har-a teristi p, by a vertex of an inde omposable FG-module M we understand a subgroup Pof G whi h is minimal subje t to the ondition that M is isomorphi to a dire t summandof IndGP (ResG

P (M)). It is well-known that the verti es of M are p-subgroups of G and thatthey are determined up to G- onjuga y. The on ept of verti es thus enables us to relateinde omposable and, in parti ular, simple FG-modules to modules for p-subgroups of G andnormalizers of these subgroups whi h is one of the major on erns in modular representationtheory. However, there are only very few general results on verti es of simple modules forarbitrary �nite groups available in the literature. For this reason, it is worthwhile to fo us on ertain lasses of �nite groups and their simple modules �rst.In this arti le we aim to determine the verti es of two lasses of simple modules for the sym-metri group Sn of degree n ∈ N over an algebrai ally losed �eld F of hara teristi p > 0.As usual, we denote the simple FSn-module parametrized by the p-regular partition λ ofn by Dλ. The �rst lass of modules under onsideration onsists of the so- alled ompletelysplittable FSn-modules, introdu ed by Klesh hev in [14℄. Following Klesh hev, a simple FSn-module D is ompletely splittable, provided the restri tion of D to Sl ≤ Sn is semisimple, forall l < n. The lass of ompletely splittable FSn-modules in parti ular ontains the Fibona imodules, studied by Ryba [22℄. In Se tion 3 we will prove that the verti es of a ompletelysplittable FSn-module are the defe t groups of its blo k.Besides the ompletely splittable modules, we will fo us on simple FSn-modules in hara -teristi p > 2, labelled by two part partitions. These had been studied before by B. Fotsing[7℄ and R. Zimmermann [23℄. Zimmermann determined the verti es of simple FSn-modulesin odd hara teristi whose labelling partitions are of the form (n−m,m), for m ≤ 3. Later,Fotsing extended those results to the ases where m < 2p, or m = 2p and p > 3. In Se tion 4,we will generalize to the ase where m < p(p+1)/2, and will show that the verti es of the sim-ple FSn-module D(n−m,m) are pre isely the defe t groups of the blo k ontaining D(n−m,m),1

unless D(n−m,m) ∼= S(n−m,m) where S(n−m,m) is the FSn-Spe ht module orresponding to(n−m,m). In the latter ase, the verti es of D(n−m,m) are known, by Grabmeier's Theorem[8℄. Our method of proof basi ally builds on extensive use of modular bran hing rules and aresult due to K. Erdmann [5℄. Erdmann's Theorem enables us to determine the dimensionsof the simple modules in question, without omputing any de omposition numbers. In ombi-nation with our results on ompletely splittable FSn-modules, we are able to determine theverti es of all simple FSn-modules labelled by two part partitions, provided n ≤ 2p2 − 1 andp > 3, or p = 3 and n ≤ 26.It shall be said at this point that the situation in hara teristi 2 is more ompli ated. Forp = 2, in [18℄ J. Müller and R. Zimmermann have shown that the simple FSn-module D(n−1,1)has the defe t groups of its blo k as its verti es, unless n = 4. Namely, D(3,1) has the unique Sy-low 2-subgroup of the alternating group A4 as its vertex. But, already for m = ⌊p+1

2 ⌋p−1 = 2,the information on the orresponding module D(n−m,m) = D(n−2,2) is far too s ar e to deter-mine its verti es in a similar manner. Currently, Müller/Zimmermann's result seems to be theonly general result on verti es of simple FSn-modules orresponding to two part partitions,for p = 2. Nevertheless, we an provide some omputational data in hara teristi 2. In [2℄ theverti es of the simple FSn-modules D(n−2,2) have been omputed, for n ≤ 47. The results willalso appear in [3℄, and they suggest that, in analogy to the ase of odd hara teristi , D(n−2,2)has the Sylow 2-subgroups of Sn as verti es, unless S(n−2,2) ∼= D(n−2,2) or n = 5. In the latter ase the Sylow 2-subgroup of A4 is a vertex of D(n−2,2) = D(3,2). If S(n−2,2) ∼= D(n−2,2) thenthe Sylow 2-subgroups of Sn−4×S2×S2 are known to be verti es of D(n−2,2), by Grabmeier'sTheorem.The present paper is organized as follows: we begin by olle ting together some general nota-tion and known results on verti es of simple modules over group algebras. Furthermore, we re- all the fa ts about FSn-modules whi h will be ne essary throughout the subsequent se tions,in luding Erdmann's Theorem mentioned above. In Se tions 3, 4 and 5 we present our resultson verti es of ompletely splittable FSn-modules and simple FSn-modules parametrized bytwo part partitions. In what follows, F will always denote an algebrai ally losed �eld of hara teristi p > 0. Moreover, given a �nite group G, any FG-module is understood to be a�nitely generated left FG-module. Given any integer z su h that z is divisible by pl but notby pl+1, we write νp(z) = l.Detailed information on erning the representation theory of the symmetri groups is givenin [12℄ and [13℄ where we also take most of our notation from. Our omputations in Se tion5 have been arried out using the omputer algebra system MAGMA [1℄; des riptions of thealgorithms a tually used to perform these omputations an be found in [2℄ and [23℄, and willalso appear in [4℄.A knowledgements. Parts of the results presented here are taken from the author's PhDthesis whi h was supported by the �Deuts he Fors hungsgemeins haft� (DFG). The authorwishes to thank her supervisor Prof. Burkhard Külshammer for numerous valuable dis ussionsand his support throughout her PhD studies. Moreover, the author is grateful to the refereefor his or her omments on an earlier version of the manus ript.2

2 Preliminaries and notation2.1 Verti es of simple modulesWe begin by �xing some notation. Given a �nite group G and FG-modules M and N su hthat N is isomorphi to a dire t summand of M , we write N |M . Suppose that H is a subgroupof G su h that M | IndGH(ResG

H(M)). Then M is said to be relatively H-proje tive. ProvidedM is inde omposable, a vertex of M is a subgroup P of G whi h is minimal with respe t tothe ondition that M is relatively P -proje tive. A vertex of M is a p-subgroup of G, and isdetermined up to G- onjuga y. Moreover, for any vertex P of M , there are a Sylow p-subgroupR of G and a defe t group ∆ of the blo k ontaining M su h that P ≤ ∆ ≤ R. The followingfa ts about verti es of inde omposable modules are well-known, and will be used repeatedlythroughout the next se tions:Lemma 2.1 ([20℄, L. 4.3.4, Thm. 4.7.5). Let M be an inde omposable FG-module with vertexP ≤ R, for a Sylow p-subgroup R of G. Furthermore, let H be a subgroup of G, and let N bean inde omposable FH-module with vertex Q. Then(i) |R : P |

∣∣ dim(M).(ii) If N |ResG

H(M) then Q ≤G P .(iii) If M | IndGH(N) then P ≤G Q.Theorem 2.2 (Knörr [16℄, Thm. 3.3). Let D be a simple FG-module belonging to the blo k

B of FG. Let further P be a vertex of D. Then there exists a blo k b of F [PCG(P )] su hthat bG = B, and P is a defe t group of b. Hen e there is a defe t group ∆ of B su h thatC∆(P ) ≤ P ≤ ∆.In the ase of the symmetri groups even more is true. Therefore we now suppose that G = Sn,for some n ∈ N, and that the blo k B has p-weight w. By [13℄, Thm. 6.2.45, the defe t groupsof B are exa tly the Sn- onjugates of the Sylow p-subgroups of Spw. Then the followinglemma is a dire t onsequen e of Knörr's Theorem and a result by Olsson ( f. [21℄, Prop.1.4).Lemma 2.3. Let D be a simple FSn-module belonging to a blo k B of p-weight w. Moreover,let P ≤ Spw be a vertex of D. Then CSpw

(P ) = Z(P ). In parti ular, D is not relativelySpw−1-proje tive.We �nally mention what the Sylow p-subgroups of the symmetri group Sn look like. For this,let n =

∑ri=0 αip

i be the p-adi expansion of n. Furthermore, set P1 := {1}, Pp := 〈(1, . . . , p)〉,and Ppi := Ppi−1 ≀ Pp, for i ≥ 2. When onsidered as a subgroup of Spi in the obvious way,Ppi is a Sylow p-subgroup of Spi , for all i ∈ N0. In general the Sylow p-subgroups of Sn arepre isely the Sn- onjugates of

Pn := Pαrpr × · · · × Pα1

p ≤ Sαrpr × · · · ×S

α1p ≤ Sn,with di�erent dire t fa tors a ting on disjoint subsets of {1, . . . , n}. A proof for this is givenin [13℄, 4.1.22, 4.1.24. We will keep the above notation, for the remainder of this arti le.3

2.2 Modules for the symmetri groupsIn this subse tion, �x n ∈ N. The set of partitions of n will be denoted by Pn, and the subsetof p-regular partitions of n by Pn,p. Given λ = (λ1, . . . , λs) ∈ Pn, we denote the orrespondingFSn-Spe ht module by Sλ. Moreover, we onsider the Young subgroup

Sλ1 ×Sλ2 × · · · ×Sλsof Sn. Here Sλiis understood to be a ting on the set {λ1 + · · ·+ λi−1 + 1, . . . , λ1 + · · ·+ λi},for all i = 1, . . . , s. Then there is an inde omposable FSn-module Y λ whi h is unique up toisomorphism satisfying the following two onditions:(1) Y λ |Mλ and (2) Sλ ⊆ Y λ,where Mλ := IndSn

Sλ(F ) is the FSn-permutation module orresponding to λ. This module

Y λ is alled the Young module orresponding to the partition λ. Further details on erningYoung modules an, for example, be found in [8℄ and in [17℄, Se . 4.6. The following theorem,due to J. Grabmeier, determines the verti es of Y λ.Theorem 2.4 (Grabmeier [8℄, S. 7.8). Let n ∈ N, and let λ = (λ1, . . . , λs) be a partition ofn su h that λs > 0. Additionally, set λs+1 := 0. Then the Sylow p-subgroups of the Youngsubgroup

s∏

i=1

(Sλi−λi+1)iof Sn are verti es of the FSn-Young module Y λ.Remark 2.5. Note that in ase λ ∈ Pn,p and Sλ ∼= Dλ, we also have Sλ ∼= Y λ. This followsimmediately from the fa t that the module Y λ has a Spe ht �ltration

0 ⊆ Sλ = Yk ⊆ Yk−1 ⊆ . . . ⊆ Y0 = Y λ,where Yj/Yj+1∼= Sµj , for some µj ⊲ λ and j = 0, . . . , k − 1. Sin e Y λ is, by onstru tion,inde omposable and selfdual, this implies Y λ ∼= Dλ ∼= Sλ. With this in mind, we will applyGrabmeier's Theorem several times in Se tions 4 and 5. In order to determine when Sλ ∼= Dλholds, we will make use of Carter's Criterion a proof of whi h an, for instan e, be found in[13℄, Thm. 7.3.23:Theorem 2.6 (Carter's Criterion). Let λ be a p-regular partition of n. Then the orresponding

FSn-Spe ht module Sλ is simple if and only if the highest p-power dividing the hook lengthhxz equals the highest p-power dividing the hook length hyz, for all nodes (x, z), (y, z) of theYoung diagram [λ].For determining the verti es of ompletely splittable modules and simple FSn-modules la-belled by ertain two part partitions, we will extensively use modular bran hing rules, due toA. Klesh hev. A detailed introdu tion to this subje t is given in [15℄. Here, we just re all thefa ts needed in the ourse of the next se tions.Remark 2.7. Consider a partition λ of n with Young diagram [λ]. Moreover, let i ∈{0, . . . , p − 1}. Ea h i-removable node (x, y) ∈ [λ], i.e. y − x ≡ i (mod p), is marked by4

a �−�. Analogously, any i-addable node (u, v) 6∈ [λ], i.e. v − u ≡ i (mod p), is marked by a�+�. Going along the rim of [λ] from bottom left to top right, we then obtain a sequen e ofpluses and minuses whi h is alled the i-signature of λ. Su essively an elling all terms ofthe form �−+�, we dedu e the redu ed i-signature of λ whi h has, by onstru tion, the form�+ · · ·+− · · · −�.An i-removable node of [λ] orresponding to a �−� in the redu ed i-signature of [λ] is alledi-normal, and the total number of these is denoted by εi(λ). An i-addable node of [λ] or-responding to a �+� in the redu ed i-signature is alled i- onormal, and the total number ofthese is denoted by ϕi(λ).Keeping this notation, the following result will be substantial:Theorem 2.8 (Klesh hev [15℄, Thms. 11.2.10, 11.2.11). Let λ ∈ Pn,p, and i ∈ {0, . . . , p− 1}.Let µ ∈ Pn−εi(λ),p be the p-regular partition obtained after removing all i-normal nodes from[λ]. Furthermore, let ν ∈ Pn+ϕi(λ),p be the p-regular partition obtained after adding all i- onormal nodes to [λ]. Then

(⊕

εi(λ)!

Dµ)∣∣ ResSn

Sn−εi(λ)(Dλ) and (

⊕

ϕi(λ)!

Dν)∣∣ Ind

Sn+ϕi(λ)

Sn(Dλ).Remark 2.9. In the situation of the previous theorem, any vertex of Dλ ontains ver-ti es of both Dµ and Dν , by Lemma 2.1. If also Dλ

∣∣ IndSn

Sn−εi(λ)(Dµ) then Dλ and Dµhave ommon verti es, by Lemma 2.1. We indi ate this by writing Dλ ←→i Dµ. Analo-gously, if Dλ

∣∣Res

Sn+ϕi(λ)

Sn(Dν) then Dλ and Dν have ommon verti es. In this ase we write

Dν ←→i Dλ.Remark 2.10. As mentioned in the introdu tion, the basi ingredients for the proof of ourmain theorem in Se tion 4 will be the modular bran hing rules and a result by K. Erdmann[5℄. For this reason, we re all the ne essary fa ts from [5℄.Let (n−m,m) be a p-regular partition of n, and let s := n−2m. Consider the p-adi expansions + 1 =

∑li=1 dip

ti where 0 ≤ t1 < . . . < tl =: t and 1 ≤ di ≤ p− 1, for i = 1, . . . , l. Moreover,set δ := (p − d1)pt1 , and

Γs(q) :=

l∏

i=2

(1− q(p−di)pti )

(1− qpti−1+1)

=

l∏

i=2

(1 + qpti−1+1

+ q2pti−1+1

+ · · · + qeipti−1+1

),where ei := (p− di)pti−ti−1−1 − 1. Writing Γs(q) =

∑∞i=0 ciq

i, the following holds:Theorem 2.11 (Erdmann [5℄, L. 5.4). The dimension of D(n−m,m) is equal to∞∑

i=0

ci(

⌊m/pt+1⌋∑

j=0

Li,j −Mi,j),

5

whereLi,j :=

(n

m− i− jpt+1

)

−

(n

m− i− jpt+1 − 1

) andMi,j :=

(n

m− i− jpt+1 − δ

)

−

(n

m− i− jpt+1 − δ − 1

)

.Note that, in the previous theorem, it su� es to determine the oe� ients ci, for i ≤ m. Fori > m, the respe tive summands Li,j −Mi,j in the dimension formula learly vanish.3 Completely splittable modulesIn this se tion we will determine the verti es of the ompletely splittable modules for symmetri groups as introdu ed by Klesh hev in [14℄. We brie�y re all the notation needed throughout.Following [14℄, a simple FSn-module Dλ is alled ompletely splittable if its restri tion to anyYoung subgroup Sµ of Sn remains semisimple.Given λ ∈ Pn with λ = (λ1, . . . , λs) and λs > 0, we set h(λ) := s and χ(λ) := λ1 − λs + s.With this de�nition, the following holds:Theorem 3.1 (Klesh hev [14℄, Thm. 2.1, L. 1.7). Let Dλ be a simple FSn-module. Thenthe following assertions are equivalent:(i) Dλ is ompletely splittable.(ii) ResSn

Sl(Dλ) is semisimple, for all l < n.(iii) χ(λ) ≤ p.De�nition 3.2. For ea h n ∈ N, let Φn := {Dn

1 , . . . ,Dnz(n)} be a set of simple FSn-modulessatisfying the following onditions:(1) Provided n ≥ 2 and j ∈ {1, . . . , z(n)}, we have ResSn

Sn−1(Dn

j ) ∼= a1Dn−11 ⊕ · · · ⊕

az(n−1)Dn−1z(n−1), for appropriate a1, . . . , az(n−1) ∈ N0.(2) For ea h n ≥ 2 and i ∈ {1, . . . , z(n − 1)}, there is some j ∈ {1, . . . , z(n)} su h that

Dn−1i | ResSn

Sn−1(Dn

j ).We then de�ne Φ :=⋃

n∈NΦn, and all Φ a semisimple indu tive system.Remark 3.3. (a) By Theorem 3.1, all simple modules belonging to su h a semisimple in-du tive system are ompletely splittable. In [14℄, Klesh hev investigates spe ial lasses ofsemisimple indu tive systems whi h are onstru ted as follows: let n ∈ N and s ∈ {1, . . . , p−1}.Furthermore, let

ξsn := {Dµ|µ ∈ Pn,p, h(µ) = s, χ(µ) ≤ p} and ωs

n := {Dµ|µ ∈ Pn,p, h(µ) < s, µ1 ≤ p− s}.Now set Φn(s) := ξsn∪ωs

n and Φ(s) :=⋃

n∈NΦn(s). By [14℄, Thm. 2.8 (i), Φ(s) is a semisimpleindu tive system, for ea h s ∈ {1, . . . , p− 1}. 6

(b) For s ∈ {1, . . . , p−1}, the semisimple indu tive system Φ(s) in parti ular ontains all om-pletely splittable modules parametrized by p-regular partitions with exa tly s non-zero parts.Conversely, ea h ompletely splittable module is parametrized by some p-regular partition withexa tly s non-zero parts, where s ∈ {1, . . . , p− 1}. This follows from Theorem 3.1. Thus ea h ompletely splittable module has to be ontained in Φ(s), for an appropriate s ∈ {1, . . . , p−1}.( ) Given a p-regular partition λ := (n − m,m) of n su h that m > 0, the orrespondingsimple FSn-module Dλ is ompletely splittable if and only if n − 2m + 2 = χ(λ) ≤ p, i.e. ifand only if n− 2m ≤ p− 2, by Theorem 3.1 (iii). In onsequen e, in hara teristi 2 there areno ompletely splittable modules parametrized by partitions with exa tly two non-zero parts.Assuming p > 2 and n > p − 2, there are exa tly p−12 ompletely splittable FSn-moduleslabelled by partitions with exa tly two non-zero parts. If n = 2r then the orrespondingpartitions have the form (r+ i, r− i), for 0 ≤ i ≤ p−1

2 −1. If n = 2r+1 then the orrespondingpartitions have the form (r + 1 + i, r − i), for 0 ≤ i ≤ p−12 − 1. In parti ular, for p = 3and n ≥ 2, there is only one ompletely splittable FSn-module labelled by a partition withexa tly two non-zero parts, namely the alternating module.For p = 5 and n ≥ 4, there are pre isely two ompletely splittable FSn-modules labelledby partitions with exa tly two non-zero parts. These are the so- alled Fibona i modules,introdu ed by Ryba in [22℄, whose dimensions equal the nth and (n− 1)st Fibona i number,respe tively.We are now prepared to determine the verti es of the ompletely splittable modules:Theorem 3.4. Let n ∈ N, and let Dλ be a ompletely splittable FSn-module. Then theverti es of Dλ are pre isely the defe t groups of the blo k of FSn ontaining Dλ.Proof. We will show that, given s ∈ {1, . . . , p − 1}, ea h FSn-module belonging to Φn(s)has the defe t groups of its blo k as its verti es. By Knörr's Theorem we may assume that

n ≥ p2 > (p− s)(s− 1). Then, in parti ular, Φn(s) = ξsn, i.e. ea h module in Φn(s) is labelledby a partition with exa tly s non-zero parts. In ase that n ≡ 0 (mod p) exa tly one of the

FSn-modules belonging to Φn(s) is ontained in the prin ipal blo k of FSn, by [10℄, Thm.3.3, L. 3.5. Moreover, the dimension of this module is not divisible by p, by [11℄, Thm. 5.7.1,so that its verti es are the Sylow p-subgroups of FSn.Let now D be any of the remaining FSn-modules belonging to Φn(s). Then D is relativelySn−1-proje tive, by [10℄, Thm. 3.3., and is thus ontained in a blo k of p-weight w < n/p.If w < p then the verti es of D are learly the defe t groups of its blo k, by Knörr's The-orem. Hen e we may assume that w ≥ p. Then we have pw ≥ p2 > (p − s)(s − 1) andthus Φpw(s) = ξs

pw. Sin e D is ompletely splittable, ResSn

Spw(D) is semisimple, and its simpledire t summands belong to Φpw(s). One of those has a vertex in ommon with D. By [10℄,Thm. 3.3, L. 3.5. and [11℄, Thm. 5.7.1., the only module in Φpw(s) whi h is not relatively

Spw−1-proje tive belongs to the prin ipal blo k of FSpw, has a dimension oprime to p andtherefore vertex Ppw. Consequently, D has vertex Ppw as well.Let now n 6≡ 0 (mod p), and let D be a simple FSn-module in Φn(s) belonging to a blo kof weight w. In view of Knörr's Theorem, we may again assume that w ≥ p. As above, wethen dedu e that there is up to isomorphism only one not relatively Spw−1-proje tive simpledire t summand of ResSn

Spw(D). That summand belongs to the prin ipal blo k of FSpw andhas vertex Ppw. Hen e also D has vertex Ppw, and the assertion follows.7

4 Simple FSn-modules labelled by two part partitionsIn this se tion, let p be odd. For the sake of simpli ity, a two part partition of n is understoodto be a partition having either one or two non-zero parts. We will determine the verti es ofsimple FSn-modules labelled by ertain two part partitions of n. This work was already begunin [23℄ where R. Zimmermann determined the verti es of simple FSn-modules orrespondingto the partitions (n −m,m) with m ≤ 3. Later, in [7℄ B. Fotsing determined the verti es ofsimple FSn-modules orresponding to the partitions (n−m,m) where m ≤ 2p−1, or m = 2pand p > 3. We will extend these results and show that the following holds:Theorem 4.1. Let p > 2, m ∈ {0, . . . , p+12 p − 1} and n ≥ 2m. Furthermore, let λ :=

(n−m,m). Then exa tly one of the following ases o urs:(1) The verti es of Dλ are pre isely the defe t groups of its blo k.(2) m = kp + r, for some k ≥ 1 and some r ∈ {0, . . . , p − 1}, su h that n ≡ bp + 2r − 1(mod p2), for some b ∈ {1, . . . , k − 1, 2k}. In this ase we have Dλ ∼= Sλ ∼= Y λ withvertex P := Pn−2m × (Pm)2. Moreover, P is not a defe t group of the blo k of FSn ontaining Dλ.Remark 4.2. In order to prove the theorem we will pro eed as follows:

• We determine when Dλ ∼= Sλ holds true, and when Dλ ∼= Sλ has vertex P whi h is nota defe t group of its blo k.• We prove the assertion of the theorem, by indu tion on m. For this, we will basi allydistinguish between two ases, depending on whether or not m is divisible by p.For the remainder of this se tion, we write m = kp + r, for appropriate k ∈ {0, . . . , p−1

2 } andr ∈ {0, . . . , p − 1}. Moreover, for any integer z, let z be its residue lass modulo p. We willoften make use of both the p-residue diagram and the hook diagram of λ. These are of thefollowing shapes:[λ]p= 0 · · · r − 2 r − 1 · · · n− r − 1 n− r

p− 1 · · · r − 3 r − 2 r − 1

p− 2

[λ]hook =n− kp− r + 1 · · · n− kp− 2r + 1 · · · n− (2k − 1)p − 2r + 1 · · · · · ·

kp + r · · · kp · · · p · · · 1We will now break up the entire proof of Theorem 4.1 into a series of propositions. As a �rststep in the proof of Theorem 4.1, we will determine when D(n−m,m) ∼= S(n−m,m) holds, andwill show that the assertion of Theorem 4.1 holds, for m < p. The �rst three propositions are onsequen es of Theorem 2.8 and Theorem 2.6, and do in fa t also hold for k > p−12 .Proposition 4.3. Let k > 0 and r = 0. Furthermore, let −1 6≡ n 6≡ 0 (mod p). Then Dλ and

D(n−kp−1,kp) have ommon verti es. Moreover, Dλ 6∼= Sλ, and D(n−kp−1,kp) 6∼= S(n−kp−1,kp).8

Proof. Again we denote the residue lass of n−1modulo p by n− 1 =: i. In dire t onsequen eof Theorem 2.6, Dλ 6∼= Sλ and D(n−kp−1,kp) 6∼= S(n−kp−1,kp). Furthermore, λ has redu ed i-signature �−�, and (n− kp− 1, kp) has redu ed i-signature �+�. Thus Dλ ←→i D(n−kp−1,kp),and the assertion follows.Proposition 4.4. Let k ≥ 0, and let r ≥ 1.(i) If 2r − 1 6≡ n 6≡ 2r − 2 (mod p) then Dλ and D(n−kp−r,kp+r−1) have ommon verti es.(ii) If n ≡ 2r − 1 (mod p) then Dλ and D(n−kp−r−1,kp+r−1) have ommon verti es.(iii) If r > 1 and n ≡ 2r−2 (mod p) then Dλ and D(n−kp−r−1,kp+r−1) have ommon verti es.Proof. (i) In ase that 2r−1 6≡ n 6≡ 2r−2 (mod p) we have r−2 6≡ n−r−1 6≡ r−3 (mod p).Thus, in this ase, λ has redu ed (r − 2)-signature �−�, and (n − kp − r, kp + r − 1) has re-du ed (r − 2)-signature �+�. Therefore, we have Dλ ←→r−2 D(n−kp−r,kp+r−1), and (i) follows.(ii) In ase that n ≡ 2r− 1 (mod p), we have n− r− 1 ≡ r− 2 (mod p) so that λ has redu ed(r − 2)-signature �−−�, and (n− kp− r − 1, kp + r − 1) has redu ed (r − 2)-signature �++�.Consequently, Dλ ←→r−2 D(n−kp−r−1,kp+r−1), and we thus obtain (ii).(iii) In ase that n ≡ 2r − 2 (mod p) we have n − r − 1 ≡ r − 3 (mod p) so that λhas redu ed (r − 3)-signature �−�, and redu ed (r − 2)-signature ∅. Thus ResSn

Sn−1(Dλ) ∼=

D(n−kp−r−1,kp+r), by Theorem 2.8. Furthermore, r − 1 6≡ r − 3 6≡ p − 2 (mod p), sin er > 1. Hen e (n − kp − r − 1, kp + r) has redu ed (r − 3)-signature �+� so that Dλ ←→r−3

D(n−kp−r−1,kp+r). Sin e n− 1 ≡ 2r− 3 (mod p) and 2r− 1 6≡ 2r − 3 6≡ 2r− 2 (mod p), from(i) we dedu e that Dλ and D(n−kp−r−1,kp+r−1) have ommon verti es.Proposition 4.5. Let k ≥ 0 and r ≥ 1.(i) If 2r − 1 6≡ n 6≡ 2r − 2 (mod p) then Dλ ∼= Sλ if and only if D(n−kp−r,kp+r−1) ∼=S(n−kp−r,kp+r−1). In parti ular, D(n−kp−r,kp+r−1) 6∼= S(n−kp−r,kp+r−1), for k > 0.(ii) If n ≡ 2r − 1 (mod p) then Dλ ∼= Sλ if and only if

D(n−kp−r−1,kp+r−1) ∼= S(n−kp−r−1,kp+r−1).(iii) If r > 1 and n ≡ 2r − 2 (mod p) then Dλ 6∼= Sλ, andD(n−kp−r−1,kp+r−1) 6∼= S(n−kp−r−1,kp+r−1).Proof. The assertion follows immediately from Theorem 2.6.Proposition 4.6. With the notation of Remark 4.2, the following holds:(i) If k ≥ 1 then Sλ ∼= Dλ if and only if n ≡ bp+2r− 1 (mod p2), for some b ∈ {0, . . . , k−

1, 2k, . . . , p− 1}.(ii) If k = 0 and r ≥ 1 then Sλ ∼= Dλ if and only if n 6≡ r + l − 1 (mod p), for alll ∈ {0, . . . , r − 1}. 9

Proof. Let k ≥ 1. By Theorem 2.6, we then have that Sλ ∼= Dλ if and only if n− (2r− 1) ≡ 0(mod p) and νp(n − kp − lp − 2r + 1) = νp((k − l)p) = 1, for all l = 0, . . . , k − 1. The latter ondition is equivalent to n = ap2 + bp + 2r − 1 with a ∈ N0 and b ∈ {0, . . . , p− 1} su h thatνp(ap2 + (b − k − l)p) = νp((k − l)p) = 1, for all l = 0, . . . , k − 1. Consequently, Sλ ∼= Dλ ifand only if n = ap2 + bp + 2r − 1, for some a ∈ N0 and some b ∈ {0, . . . , k − 1, 2k, . . . , p− 1}.This proves (i).Now let k = 0 and r ≥ 1. Then we immediately get that Sλ ∼= Dλ if and only if n−r−l+1 6≡ 0(mod p), for all l = 0, . . . , r − 1, by Theorem 2.6. This proves (ii).Remark 4.7. Note that in the ase where k = 0 = r we have λ = (n) and thus Dλ ∼= Sλ, forall n ∈ N.Proposition 4.8. With the notation of Remark 4.2, assume that Dλ ∼= Sλ. Moreover, let Pbe a vertex of Dλ.(i) If k = 0 then P is a defe t group of the blo k of FSn ontaining Dλ.(ii) If k ≥ 1 then P is a defe t group of the blo k of FSn ontaining Dλ if and only if

n ≡ bp + 2r − 1 (mod p2), for some b ∈ {0, 2k + 1, . . . , p− 1}.Proof. (i) By Proposition 4.6, we have n 6≡ r + l − 1 (mod p), for all l ∈ {0, . . . , r − 1}. Fur-thermore, Dλ has vertex Pn−2r, by Theorem 2.4. In order to show that Dλ belongs to a blo kwith defe t group Pn−2r, we assume that λ has p- ore κ su h that |κ| < 2r. Then κ = (r−1, l),for some l ∈ {0, . . . , r− 1}. Consequently, n− r + 1− l = n− |κ| ≡ 0 (mod p) whi h leads tothe ontradi tion n ≡ r+ l−1 (mod p). Hen e Dλ belongs to a blo k B of FSn with p- ore κsu h that |κ| ≥ 2r, and the defe t groups of B are onjugate to Pn−|κ| = Pn−2r. This proves (i).(ii) Let now k ≥ 1. By Proposition 4.6, we have n = ap2 + bp + 2r − 1, for some a ∈ N0 andsome b ∈ {0, . . . , k − 1, 2k, . . . , p− 1}. We distinguish between three ases.Case 1: 1 ≤ b ≤ 2k. Then, in parti ular, a ≥ 1. Moreover, λ has p- ore (p + r − 1, r)and p-weight w := ap + b − 1 so that Ppw = Pap2 × (Pp)b−1. Furthermore, Dλ has vertex

P := Pn−2kp−2r × (Pkp+r)2 =Sn

Pn−2kp−2r × (Pp)2k, by Theorem 2.4. Sin e

n− 2kp− 2r = ap2 + (b− 2k)p − 1 = (a− 1)p2 + (p + b− 2k − 1)p + p− 1,we obtain P =SnP(a−1)p2 × (Pp)

p+b−1 <SnPpw.Case 2: b = 0. Then a ≥ 1, and λ has p- ore (p + r− 1, r) and p-weight w := (a− 1)p + p− 1.Hen e Ppw = P(a−1)p2 × (Pp)

p−1. Sin e, in this ase,n− 2kp − 2r = ap2 − 2kp − 1 = (a− 1)p2 + (p− 2k − 1)p + p− 1,we obtain that Dλ has vertex P = Pn−2kp−2r × (Pp)

2k =SnP(a−1)p2 × (Pp)

p−1 = Ppw.Case 3: b ≥ 2k + 1. Then λ has p- ore (p + r − 1, r) and p-weight w := ap + b − 1 so thatPpw = Pap2 × (Pp)

b−1. Sin en− 2kp − 2r = ap2 + (b− 2k)p − 1 = ap2 + (b− 2k − 1)p + p− 1,10

also in this ase, Dλ has vertex P := Pn−2kp−2r × (Pp)2k =Sn

Pap2 × (Pp)b−1 = Ppw. Thisproves the proposition.Corollary 4.9. The assertion of Theorem 4.1 holds, for m ∈ {0, . . . , p− 1}.Proof. The verti es of D(n) and D(n−1,1) are pre isely the defe t groups of the orrespondingblo ks of FSn, by [18℄. By Proposition 4.4, the verti es of D(n−m,m) are then also the defe tgroups of its blo k, for all m = 2, . . . , p− 1.Remark 4.10. We re all that we aim to prove Theorem 4.1 by indu tion on m. For this,Corollary 4.9 is the start of the indu tion argument. In order to do the indu tion step, we �x

m ∈ {1, . . . , p+12 p − 1}, and assume that the statement of Theorem 4.1 holds for all two partpartitions whose se ond part is stri ty less than m. We need to show that the assertion ofTheorem 4.1 then also holds for λ = (n−m,m). It will be helpful to distinguish between the ases m ≡ 0 (mod p) and m 6≡ 0 (mod p).4.1 Indu tion step - the ase m ≡ 0 (mod p)Remark 4.11. In this subse tion, we do the indu tion step, for m ≡ 0 (mod p). In this ase,

λ = (n − kp, kp), for some k ∈ {1, . . . , p−12 }, so that Dλ is ontained in the prin ipal blo kof FSn. By Proposition 4.8 we already know that the verti es of Dλ di�er from the Sylow

p-subgroups of Sn, provided n ≡ bp − 1 (mod p2), for some b ∈ {1, . . . , k − 1, 2k}. In thefollowing, we will show that in all other ases Dλ has pre isely the Sylow p-subgroups of Snas its verti es. By Proposition 4.3, we also know that it su� es to onsider the ases n ≡ −1(mod p), and n ≡ 0 (mod p) expli itly.Proposition 4.12. Let n ≡ −1 (mod p). Then either the verti es of Dλ are the Sylow p-subgroups of Sn, or n ≡ bp− 1 (mod p2), for some b ∈ {1, . . . , k − 1, 2k}. In the latter ase,Dλ has vertex Pn−2kp × (Pp)

2k /∈ Sylp(Sn).Proof. By Proposition 4.6, Dλ ∼= Sλ if and only if n = ap2 + bp − 1, for appropriate a ∈ N0and b ∈ {0, . . . , k − 1, 2k, . . . , p− 1}. By Proposition 4.8, Dλ ∼= Sλ has vertex Pn if and onlyif b ∈ {0, 2k + 1, . . . , p− 1}.It thus remains to show that Dλ has vertex Pn, for n = ap2 + bp − 1 where a ∈ N0and b ∈ {k, . . . , 2k − 1}. Therefore we write n = ap2 + bp + p − 1, for a ∈ N0 andb ∈ {k − 1, . . . , 2k − 2}. In addition, we may assume that a > 0, sin e otherwise n < p2,and the assertion then holds, by Knörr's Theorem. We distinguish between two ases.Case 1: k = 1. Then b = 0, i.e. n = ap2 + (p − 1). With the notation of Remark 2.10, weobtain the following:

s + 1 = n− 2kp + 1 = ap2 + p− 2p = (a− 1)p2 + (p− 1)p,

t1 = 1, d1 = p− 1, δ = p, e2 = (p − d2)pt2−2 − 1, t ≥ 1,

⌊kp

pt+1

⌋

=

⌊p

pt+1

⌋

= 0,

∞∑

i=0

ciqi = Γs(q) = (1 + qp2

+ q2p2+ · · ·+ qe2p2

)(1 + qpt2+1+ q2pt2+1

+ · · · + qe3pt2+1) · · · .11

Obviously, c0 = 1 and c1 = . . . = cp = 0. Hen e we havedim(Dλ) =

(n

p

)

−

(n

p− 1

)

−

(n

p− δ

)

+

(n

p− δ − 1

)

=

(n

p

)

−

(n

p− 1

)

− 1,by Theorem 2.11. This shows that dim(Dλ) ≡ −2 (mod p).Case 2: k > 1. Then b ≥ 1. Using the notation of Remark 2.10 and Theorem 2.11, we nowhaves + 1 = n− 2kp + 1 = (a− 1)p2 + (p + b− 2k + 1)p,

t1 = 1, d1 = p + b− 2k + 1, δ = (2k − b− 1)p, t ≥ 1,⌊

kp

pt+1

⌋

= 0, e2 = (p− d2)pt2−2 − 1,

∞∑

i=0

ciqi = Γs(q) = (1 + qp2

+ q2p2+ · · ·+ qe2p2

)(1 + qpt2+1+ q2pt2+1

+ · · ·+ qe3pt2+1) · · · .Consequently, also here we have c0 = 1 and c1 = . . . = ckp = 0. Hen e

dim(Dλ) =

(n

kp

)

−

(n

kp− 1

)

−

(n

p(b− k + 1)

)

+

(n

p(b− k + 1)− 1

)

.We subsequently show that dim(Dλ) 6≡ 0 (mod p). Firstly, let b = k − 1 so thatdim(Dλ) =

(n

kp

)

−

(n

kp− 1

)

−

(n

0

)

.An easy al ulation shows that ( nkp

)≡ 0 (mod p) and ( n

kp−1

)≡ 1 (mod p) so that dim(Dλ) ≡

−2 (mod p).We now assume that b ≥ k. Then we havedim(Dλ) =

n− 2kp + 1

(kp)!

kp−2∏

i=0

(n− i)−n− 2(b− k + 1)p + 1

((b− k + 1)p)!

(b−k+1)p−2∏

i=0

(n− i) =z

(kp)!,where

z := (n− 2kp+1)

kp−2∏

i=0

(n− i)− (n− 2(b− k +1)p+1)

(b−k+1)p−2∏

i=0

(n− i)((b− k +1)p+1) · · · kp.Sin e νp((kp)!) = k, we need to show that z 6≡ 0 (mod pk+1). Firstly, we have:(n− 2kp + 1)n(n− 1) · · ·(n− kp + 2) = (ap2 + (b− 2k + 1)p)n(n − 1) · · · (n− kp + 2)

︸ ︷︷ ︸

≡0 (mod pk−1)

≡ (b− 2k + 1)pn(n− 1) · · · (n− kp + 2) (mod pk+1)

≡ pk[(b− 2k + 1)((p − 1)!)k b · · · (b− (k − 2))] (mod pk+1).12

Se ondly, we have:(n− 2(b− k + 1)p + 1)n(n− 1) · · · (n − (b− k + 1)p + 2)((b − k + 1)p + 1) · · · kp

= (ap2 + (2k − b− 1)p)n(n − 1) · · · (n− (b− k + 1)p + 2)︸ ︷︷ ︸

≡0 (mod pb−k)

((b− k + 1)p + 1) · · · kp︸ ︷︷ ︸

≡0 (mod p2k−b−1)

≡ (2k − b− 1)pn(n − 1) · · · (n− (b− k + 1)p + 2)((b − k + 1)p + 1) · · · kp (mod pk+1)

≡ pk[(2k − b− 1)((p − 1)!)k b(b− 1) · · · (k + 1)(b− (k − 2))(b− (k − 3)) · · · k] (mod pk+1)

≡ −pk[(b− 2k + 1)((p − 1)!)k b(b− 1) · · · (b− (k − 2))] (mod pk+1).Thus we �nally dedu e thatz ≡ 2pk((p − 1)!)k b(b− 1) · · · (b− (k − 2))(b − 2k + 1) (mod pk+1)

6≡ 0 (mod pk+1).We have now shown that also in this ase dim(Dλ) 6≡ 0 (mod p). Consequently, Dλ has vertexPn, for n ≡ bp− 1 (mod p2) with b ∈ {k, . . . , 2k − 1}, and the proposition is proved.Proposition 4.13. Let n ≡ 0 (mod p). Then Dλ = D(n−kp,kp) has vertex Pn.Proof. By Theorem 2.8, we get ResSn

Sn−1(Dλ) ∼= D(n−kp−1,kp) ⊕D(n−kp,kp−1). Here

D(n−kp,kp−1) = D(n−1−(k−1)p−(p−1),(k−1)p+p−1)belongs to the blo k of FSn−1 with p- ore (p − 2, 1) and p-weight np − 1. Furthermore,

n− 1 ≡ −1 6≡ 2(p− 1)− 1 (mod p). ThusD(n−1−(k−1)p−(p−1),(k−1)p+p−1) 6∼= S(n−1−(k−1)p−(p−1),(k−1)p+p−1),by Proposition 4.6, and has vertex Pn−p, by our indu tive hypothesis. In the ase where n 6≡ 0

(mod p2) we have |Pn : Pn−p| = p, and the assertion then immediately follows from Lemma 2.3.For the remainder of the proof, we may thus assume that n ≡ 0 (mod p2). Therefore we writen = ap2, for some a ∈ N. With the notation of Remark 2.10 and Theorem 2.11, we then get:

s + 1 = n− 2kp + 1 = (a− 1)p2 + (p− 2k)︸ ︷︷ ︸

>0

p + 1,

t1 = 0, d1 = 1, δ = p− 1, t2 = 1, d2 = p− 2k, t ≥ 1,⌊

kp

pt+1

⌋

= 0, e2 = 2k − 1,

∞∑

i=0

ciqi = Γs(q) = (1 + qp + q2p + · · ·+ qe2p)(1 + qp2

+ q2p2+ · · ·+ qe3p2

) · · · .Consequently, ci = 1 for i ∈ {0, p, 2p, . . . , kp}, and ci = 0 for i ∈ {0, . . . , kp}\{0, p, 2p, . . . , kp},and hen edim(Dλ) =

k∑

l=0

(n

lp

)

−

(n

lp− 1

)

−

(n

lp− p + 1

)

+

(n

lp− p

)

.13

We setσj :=

j∑

l=0

(n

lp

)

−

(n

lp− 1

)

−

(n

lp− p + 1

)

+

(n

lp− p

)

,and show that σj 6≡ 0 (mod p), for all 0 ≤ j ≤ p−12 . We learly have

σ0 =

(n

0

)

= 1 and σ1 = 2 +

(n

p

)

−

(n

p− 1

)

− n.It is easily veri�ed that (np

)≡ 0 ≡

(n

p−1

)(mod p) so that σ1 ≡ 2 (mod p). Thus we may nowassume that j > 1, and argue by indu tion on j. Then

σj = σj−1 +

(n

jp

)

−

(n

jp− 1

)

−

(n

(j − 1)p + 1

)

+

(n

(j − 1)p

)

.A small al ulation shows that(

n

jp

)

≡

(n

jp− 1

)

≡

(n

(j − 1)p + 1

)

≡

(n

(j − 1)p

)

≡ 0 (mod p).Consequently, σj ≡ σj−1 ≡ 2 (mod p).Hen e we have proved that dim(Dλ) 6≡ 0 (mod p), provided n ≡ 0 (mod p2). Thus Dλ hasvertex Pn in this ase, and the proposition is proved.Proposition 4.14. If −1 6≡ n 6≡ 0 (mod p) then Dλ = D(n−kp,kp) has vertex Pn.Proof. This follows immediately from Propositions 4.3, 4.12 and 4.13.Remark 4.15. In onsequen e of Propositions 4.12, 4.13 and 4.14, we have now ompletedthe indu tion, in the ase where m ≡ 0 (mod p). Therefore we may now assume that m 6≡ 0(mod p), and prove the assertion of Theorem 4.1 for this ase. This will be done in the nextsubse tion.4.2 Indu tion step - the ase m 6≡ 0 (mod p)Remark 4.16. Throughout this subse tion, we assume that m 6≡ 0 (mod p). Again we writem = kp + r, for appropriate k ∈ {0, . . . , p−1

2 } and r ∈ {1, . . . , p − 1}. By Proposition 4.4, weknow that it su� es to determine the verti es of D(n−kp−r,kp+r), for r = 1 and n ≡ 0 (mod p).In parti ular, we then have S(n−kp−r,kp+r) 6∼= D(n−kp−r,kp+r), and D(n−kp−r,kp+r) is ontainedin the prin ipal blo k of FSn.Proposition 4.17. With the assumptions of the previous remark, let r = 1 and n ≡ 0(mod p). Then Dλ = D(n−kp−r,kp+r) has vertex Pn.Proof. By Theorem 2.8, we have ResSn

Sn−1(Dλ) ∼= D(n−kp−2,kp+1). Furthermore, D(n−kp−2,kp+1)is ontained in the blo k of FSn−1 with p- ore (p− 2, 1) and p-weight n

p − 1. Sin e r = 1 and2r − 1 6≡ n − 1 6≡ 2r − 2 (mod p), the simple modules D(n−kp−2,kp+1) and D(n−kp−2,kp) have ommon verti es, by Proposition 4.4. By our indu tive hypothesis and Proposition 4.6, theFSn−2-module D(n−kp−2,kp) has vertex Pn−p. Hen e D(n−kp−2,kp+1) has vertex Pn−p as well.14

In the ase where n 6≡ 0 (mod p2), we have |Pn : Pn−p| = p so that then Dλ has vertex Pn,by Lemma 2.3.For this reason, we now assume that n ≡ 0 (mod p2) and write n = ap2, for some a ∈ N.In the following, we will show that dim(Dλ) ≡ −2 (mod p). For this we use the notation ofRemark 2.10 and Theorem 2.11 again, and obtains + 1 = ap2 − 2kp− 1 = (a− 1)p2 + (p− 2k − 1)

︸ ︷︷ ︸

≥0

p + p− 1.We distinguish between three ases.Case 1: a = 1, k = p−12 . Then we get

s + 1 = p− 1, t1 = 0, d1 = p− 1, δ = 1, t = 0,∞∑

i=0

ciqi = Γs(q) = 1,

⌊kp + 1

pt+1

⌋

=

⌊kp + 1

p

⌋

= k.Consequently, c0 = 1 and ci = 0, for i ≥ 1 so thatdim(Dλ) =

k∑

l=0

[(n

lp + 1

)

− 2

(n

lp

)

+

(n

lp− 1

)]

.Case 2: k < p−12 , i.e. p− 2k − 1 > 0. Thus we then gett1 = 0, d1 = p− 1, δ = 1, t2 = 1, d2 = p− 2k − 1, t ≥ 1,

⌊kp + 1

pt+1

⌋

= 0, e2 = 2k,

∞∑

i=0

ciqi = Γs(q) = (1 + qp + q2p + · · ·+ qe2p)(1 + qp2

+ q2p2+ · · ·+ qe3p2

) · · ·This shows that c0 = cp = c2p = . . . = ckp = 1 and ci = 0, for i ∈ {0, . . . , kp + 1} \{0, p, 2p, . . . , kp}, and also here we get

dim(Dλ) =

k∑

l=0

[(n

lp + 1

)

− 2

(n

lp

)

+

(n

lp− 1

)]

.Case 3: k = p−12 and a ≥ 2. Then we have

s + 1 = (a− 1)p2 + p− 1, t1 = 0, d1 = p− 1, δ = 1, t2 ≥ 2, t ≥ 2,⌊

kp + 1

pt+1

⌋

= 0, e2 = (p− d2)pt2−1 − 1 > k,

∞∑

i=0

ciqi = Γs(q) = (1 + qp + q2p + · · ·+ qe2p)(1 + qpt2+1

+ · · ·+ qe3pt2+1) · · · ,15

Again we obtain c0 = cp = . . . = ckp = 1 and ci = 0, for i ∈ {0, . . . , kp+1}\{0, p, 2p, . . . , kp},anddim(Dλ) =

k∑

l=0

[(n

lp + 1

)

− 2

(n

lp

)

+

(n

lp− 1

)]

.We now setσj :=

j∑

l=0

[(n

lp + 1

)

− 2

(n

lp

)

+

(n

lp− 1

)]

,and show that σj ≡ −2 (mod p), for all 0 ≤ j ≤ p−12 . We argue by indu tion on j. If j = 0,then σj = σ0 = n− 2 ≡ −2 (mod p). For j ≥ 1, we obviously have

σj = σj−1 +

(n

jp + 1

)

− 2

(n

jp

)

+

(n

jp − 1

)

.An easy al ulation shows that(

n

jp + 1

)

≡

(n

jp

)

≡

(n

jp− 1

)

≡ 0 (mod p),so that σj ≡ σj−1 ≡ −2 (mod p), by our indu tive hypothesis. Consequently, dim(Dλ) ≡−2 6≡ 0 (mod p), for n ≡ 0 (mod p2), and Dλ has then vertex Pn. This �nally proves theproposition.Proposition 4.18. With the assumptions of Remark 4.16, let r > 1 or n 6≡ 0 (mod p). Thenthe assertion of Theorem 4.1 holds, for λ = (n−m,m) = (n− kp− r, kp + r).Proof. By Proposition 4.4, there exists some j ∈ {0, 1} su h that Dλ and the FSn−j−1-module D(n−kp−r−j,kp+r−1) have ommon verti es. Furthermore, Proposition 4.5 ensures thatSλ ∼= Dλ if and only if S(n−kp−r−j,kp+r−1) ∼= D(n−kp−r−j,kp+r−1). Arguing by indu tion onr, and applying Propositions 4.6 and 4.8, we dedu e that the verti es of Dλ di�er from thedefe t groups of its blo k if and only if ondition (2) of Theorem 4.1 is satis�ed. From thisthe proposition follows.Remark 4.19. To summarize, we have now ompleted the indu tion, also for the ase m 6≡ 0(mod p). This �nally ompletes the proof of Theorem 4.1.5 Further resultsLet p be odd, and onsider a simple FSn-module D(n−m,m). Provided m ≤ p+1

2 p − 1, theverti es of D(n−m,m) are now known, by Theorem 4.1. On the other hand, as mentionedin Remark 3.3, D(n−m,m) is ompletely splittable, for m ≥ n−p+22 . As a dire t onsequen eof these results, the verti es of all simple FSn-modules labelled by two part partitions areknown, for n ≤ p2 + 2p− 2. Indeed, we an do better. In the following, we will show that theabove methods enable us to determine the verti es of all simple FSn-modules labelled by twopart partitions, for n < 2p2. In ase p = 3, we an even a hieve n < 27, by arrying out someadditional omputer al ulations. 16

Remark 5.1. (a) First of all we should emphasize that the dimension arguments used through-out the previous se tion generally fail, for m ≥ p(p+1)2 . For instan e, assuming that k = p+1

2 ,n = 2p2, m = kp, and λ = (n−m,m), we have

s + 1 = 2p2 − (p + 1)p + 1 = (p − 1)p + 1,

t1 = 0, d1 = 1, δ = p− 1, t = t2 = 1, d2 = p− 1,⌊

m

p2

⌋

=

⌊kp

p2

⌋

= 0,∞∑

i=0

ciqi = Γs(q) = 1,in the notation of Theorem 2.11 and Remark 2.10. Consequently, c0 = 1, ci = 0 for i ≥ 1, andhen e

dim(Dλ) =

(n

kp

)

−

(n

kp− 1

)

−

(n

kp − p + 1

)

+

(n

kp− p

)

≡ 0 (mod p).(b) Note that Propositions 4.3 and 4.4 do also hold, for k > p−12 . Consequently, in order todetermine the verti es of all simple FSn-modules labelled by two part partitions (n−m,m),it su� es to onsider the ases where m ≡ 0 (mod p) and n ≡ 0 (mod p), m ≡ 0 (mod p)and n ≡ −1 (mod p), and the ase where m ≡ 1 (mod p) and n ≡ 0 (mod p).With this in mind, we obtain the following:Proposition 5.2. Let k := p+1

2 , r ∈ {0, . . . , p − 1}, and m = kp + r. Furthermore, letp2 + p ≤ 2m ≤ n < 2p2 and λ := (n−m,m). Then one of the following ases o urs:(i) The verti es of Dλ are the defe t groups of the blo k of FSn ontaining Dλ.(ii) n ≡ bp + 2r− 1 (mod p2), for some b ∈ {1, . . . , p−1

2 }. In this ase, Dλ ∼= Sλ with vertexP := Pn−2kp−2r × (Pkp)

2, and P is not a defe t group of the blo k ontaining Dλ.Proof. The proof is similar to that of Theorem 4.1. Therefore we omit the details in the al ulations. Firstly, by Theorem 2.6, we obtain that Sλ ∼= Dλ if and only if n ≡ bp + 2r − 1(mod p2), for some b ∈ {1, . . . , p−1

2 }. In all these ases, Dλ has vertex P := Pn−2kp−2r ×(Pkp+r)

2 =SnPn−p2−p−2r × (Pp)

p+1, by Theorem 2.4. Moreover, Dλ is then ontained inthe blo k of FSn with p- ore (p + r − 1, r) and p-weight w = n−2r+1−pp . This shows that

P =SnPn−p2−p−2r × (Pp)

p+1 <SnPn−p−2r+1 = Ppw so that P is not a defe t group of theblo k ontaining Dλ.Next, we onsider the ase where r = 0 and n ≡ bp−1 (mod p2), for some b ∈ {0, p+1

2 , . . . , p−

1}. Then, we again apply Theorem 2.11, in order to determine dim(Dλ). If n = ap2+bp+p−1,for some a ∈ N and some b ∈ {p−12 , . . . , p− 1}, then Theorem 2.11 yields

dim(Dλ) =

(n

kp

)

−

(n

kp − 1

)

−

(n

(k − p + b)p

)

+

(n

(k − p + b)p− 1

)

.This shows that dim(Dλ) ≡ −2 (mod p), for b = p−12 . Furthermore, for b > p−1

2 , we havedim(Dλ) =

n− 2kp + 1

(kp)!

kp−2∏

i=0

(n− i)−n− 2(k − p + b)p + 1

((k − p + b)p)!

(k−p+b)p−2∏

i=0

(n− i) =z

(kp)!,17

where z ≡ 2pk((p − 1)!)k b2(b − 1) · · · (b − (k − 2)) 6≡ 0 (mod pk+1). Sin e νp((kp)!) = k, thisimplies dim(Dλ) 6≡ 0 (mod p). Hen e Dλ has vertex Pn, in this ase.Now, let r = 0 and n ≡ 0 (mod p). Then ResSn

Sn−1(Dλ) ∼= D(n−kp−1,kp) ⊕ D(n−kp,kp−1),by Theorem 2.8. Moreover, D(n−kp,kp−1) has vertex Pn−p, by Theorem 4.1. Sin e n 6≡ 0

(mod p2), we have |Pn : Pn−p| = p, and Dλ has thus vertex Pn, by Lemma 2.3. If r = 0 and−1 6≡ n 6≡ 0 (mod p), then Dλ and D(n−kp−1,kp) have ommon verti es, and Dλ 6∼= Sλ andD(n−kp−1,kp) 6∼= S(n−kp−1,kp). This follows from Proposition 4.3. Consequently, in these asesDλ has vertex Pn, by what we just proved.We may now assume that r = 1. Assume further that n ≡ 0 (mod p). Then D(n−kp−2,kp) |ResSn

Sn−2(Dλ), by Theorem 2.8. Sin e n−2 6≡ −1 (mod p), we already know that D(n−kp−2,kp)has vertex Pn−p. Sin e n 6≡ 0 (mod p2), we again have |Pn : Pn−p| = p, and Dλ has vertex

Pn, by Lemma 2.3. If n ≡ 1 (mod p) then Dλ and D(n−kp−2,kp) have ommon verti es, byProposition 4.4, and Dλ ∼= Sλ if and only if D(n−kp−2,kp) ∼= S(n−kp−2,kp), by Proposition 4.5.Thus the verti es of Dλ are as laimed, by what we have proved above. If 0 6≡ n 6≡ 1 (mod p)then Dλ and D(n−kp−1,kp) have ommon verti es, by Proposition 4.4, and Dλ ∼= Sλ if andonly if D(n−kp−1,kp) ∼= S(n−kp−1,kp), by Proposition 4.5. By our previous onsiderations, thisyields the laimed assertion on the verti es of Dλ.Finally, assume that r > 1. Then Dλ and Dµ := D(n−kp−r−j,kp+r−1) have ommon verti es,for some j ∈ {0, 1}, by Proposition 4.4. Furthermore, Dλ ∼= Sλ if and only if Dµ ∼= Sµ, byProposition 4.5. We now argue by indu tion on r, and �nally obtain the verti es of Dλ as laimed. This ompletes the proof of the proposition.Proposition 5.3. Let 1 ≤ n ≤ 2p2− 1. Then the verti es of all simple FSn-modules labelledby two part partitions are known. If m ∈ {0, . . . , ⌊n2 ⌋} then D(n−m,m) has the defe t groups ofits blo k as its verti es, unless S(n−m,m) ∼= D(n−m,m). In the latter ase D(n−m,m) has vertexPn−2m × (Pm)2.Proof. For p = 3, the assertion follows immediately from Theorem 4.1 and Proposition 5.2.Therefore, we may now assume that p > 3. By Theorem 4.1 and Proposition 5.2, the verti esof D(n−m,m) are known to be as laimed, for m ≤ p+1

2 p + p− 1. Moreover, for m ≥ n−p+22 , wehave n − 2m ≤ p − 2. Consequently, D(n−m,m) is then ompletely splittable, and the defe tgroups of its blo k are verti es of D(n−m,m), by Theorem 3.4. In parti ular, the assertion ofthe proposition follows, for n ≤ p2 + 4p− 2.It thus remains to onsider the ases where p2+4p−1 ≤ n ≤ 2p2−1 and m ∈ {p+3

2 p, . . . , ⌊n−p+12 ⌋}.For this, we write m = kp + r, for appropriate k ∈ {p+3

2 , . . . , p− 1} and r ∈ {0, . . . , p− 1}. Inanalogy to our proof of Theorem 4.1, we will argue by indu tion on m, and will distinguishbetween three ases.Case 1: m ≡ 0 (mod p) and n ≡ −1 (mod p). Then m = kp, for some k ∈ {p+32 , . . . , p− 1},and n = p2 + xp− 1, for some x ∈ {4, . . . , p}. In the notation of Theorem 2.11, we thus have

18

s + 1 = p(p + x− 2k), t = t1 = 1, d1 = p + x− 2k, δ = (2k − x)p,⌊

m

pt+1

⌋

= 0, Γs(q) = 1.Consequently,dim(D(n−m,m)) =

(n

kp

)

−

(n

kp− 1

)

−

(n

p(x− k)

)

+

(n

p(x− k)− 1

)

,and, in parti ular, D(n−m,m) ∼= S(n−m,m) if and only if k > x. Hen e, for k > x, the moduleD(n−m,m) has vertex Pn−2m × (Pm)2, by Theorem 2.4. Now suppose that k < x. Then weobtain

dim(D(n−m,m)) =n− 2kp + 1

(kp)!

kp−2∏

i=0

(n− i)−n− 2(x− k)p + 1

(p(x− k))!

p(x−k)−2∏

i=0

(n− i) =z

(xp)!,where

z ≡

{

2px(x− 2k)((p − 1)!)x(x− 1) · · · (x− (k − 1))(k + 1) · · · x (mod px+1), for x < p

−2px+12k((p − 1)!)x(p− 1) · · · (p− (k − 1))(k + 1) · · · (p − 1) (mod px+2), for x = p.Sin e νp((xp)!) = x for x < p, and νp((xp)!)) = x+1 for x = p, this shows that dim(D(n−m,m)) 6≡0 (mod p). Finally, suppose that k = x < p. Then we have

dim(D(n−m,m)) =

(n

xp

)

−

(n

xp− 1

)

− 1 ≡ −2 (mod p).Hen e, also in this ase, dim(D(n−m,m)) 6≡ 0 (mod p). Consequently, D(n−m,m) has vertexPn, provided n = p2 +xp− 1, for some x ∈ {4, . . . , p}, and m = kp, for some k ∈ {p+3

2 , . . . , x}with k ≤ p− 1. This ompletes ase 1.Case 2: m ≡ 0 (mod p) and n 6≡ −1 (mod p). If n ≡ 0 (mod p) then ResSn

Sn−1(D(n−m,m)) ∼=

D(n−m,m−1) ⊕D(n−m−1,m), by Theorem 2.8. The simple FSn−1-module D(n−m,m−1) belongsto the blo k of FSn−1 with p- ore (p − 2, 1) and p-weight n−pp . Moreover, D(n−m,m−1) 6∼=

S(n−m,m−1), by Theorem 2.6. Thus, by indu tion, D(n−m,m−1) has vertex Pn−p. Sin e n 6≡ 0(mod p2), this implies |Pn : Pn−p| = p so that D(n−m,m) has then vertex Pn, by Lemma 2.3.If n 6≡ 0 (mod p) then D(n−m,m) and D(n−m−1,m) have ommon verti es, by Proposition 4.3.Moreover, D(n−m,m) 6∼= S(n−m,m) and D(n−m−1,m) 6∼= S(n−m−1,m), by Proposition 4.3. There-fore we may then argue by indu tion on n, and obtain that D(n−m,m) has vertex Pn, by ourprevious onsiderations.Case 3: m 6≡ 0 (mod p). Then m = kp + r, for some k ∈ {p+3

2 , . . . , p − 1} and somer ∈ {1, . . . , p − 1}. If r = 1 and n ≡ 0 (mod p) then ResSn

Sn−1(D(n−m,m)) ∼= D(n−m−1,m) =

D(n−kp−2,kp+1), by Theorem 2.8. Sin e 2r − 1 6≡ n − 1 6≡ 2r − 2 (mod p), the modulesD(n−kp−2,kp+1) and D(n−kp−2,kp) have ommon verti es, by Proposition 4.4. By Theorem 2.6,19

D(n−kp−2,kp) 6∼= S(n−kp−2,kp) so that D(n−kp−2,kp) has vertex Pn−p, by our indu tive hypoth-esis. Sin e n 6≡ 0 (mod p2), we have |Pn : Pn−p| = p, and D(n−m,m) has thus vertex Pn, byLemma 2.3.Finally, let r > 1 or n 6≡ 0 (mod p). By Proposition 4.4, there is an appropriate j ∈ {0, 1}su h that D(n−m,m) and the simple FSn−1−j-module D(n−m−j,m−1) have ommon verti es.Furthermore, D(n−m,m) ∼= S(n−m,m) if and only if D(n−m−j,m−1) ∼= S(n−m−j,m−1), by Proposi-tion 4.5. Hen e, by indu tion, D(n−m,m) has the defe t groups of its blo k as verti es, unlessD(n−m,m) ∼= S(n−m,m). In the latter ase, D(n−m,m) has learly vertex Pn−2m × (Pm)2, byTheorem 2.4. This ompletes the proof of the proposition.Remark 5.4. (a) Hen e, for p = 3, the �rst module whose verti es annot be determined viathe above methods is the simple FS18-module D(12,6) of dimension 7752 ≡ 0 (mod 3). Wehave investigated this module with the omputer. By Theorem 2.8, we have ResS18

S17(D(12,6)) ∼=

D(12,5) ⊕D(11,6). Furthermore, by Theorem 4.1, the simple FS17-module D(12,5) has vertexP15. Thus

P9 × (P3)2 = P15 <S18 P ≤ P18 = P9 × P9,for some vertex P of D(12,6), by Lemma 2.1. From Lemma 2.3 we now dedu e that either P =

P18, or P = P9× (P3)3. Next, onsider the y li group C = 〈(1, . . . , 9)(10, . . . , 18)〉 ≤S18 P18.Our omputations show that

ResS18C (D(12,6)) ∼= F ⊕ 5M1 ⊕M2 ⊕ proj,where M1 and M2 are inde omposable of dimension 3 and 5, respe tively. The inde ompos-able proje tive dire t summands of ResS18

C (D(12,6)) are easily dete ted using the algorithm�ProjSummands� presented in [4℄. The trivial module F has learly vertex C, and hen e Chas to be ontained in some vertex of D(12,6), by Lemma 2.1. Sin e C 6≤S18 P9 × (P3)3, we�nally on lude that D(12,6) has vertex P18.(b) For p = 3, the next module whose verti es annot be derived from Theorem 4.1, or fromProposition 5.2, is the simple FS18-module D(11,7) of dimension 4845 ≡ 0 (mod 3) whi h wehave also treated with the omputer. Here we have ResS18

S16(D(11,7)) ∼= D(10,6)⊕D(9,7), by Theo-rem 2.8. Moreover, D(10,6) has vertex P15, by Proposition 5.2. Thus, with the same argumentsas above, we obtain that D(11,7) has either vertex P18 or vertex P9 × (P3)

3. When restri tingD(11,7) to the y li group C of order 9 mentioned in (a), we also get F | ResS18

C (D(11,7)), sothat D(11,7) in fa t has vertex P18.( ) In fa t, we an now determine the verti es of all simple FSn-modules labelled by twopart partitions, for p = 3 and n ≤ 26. The alternating FSn-module D(n−⌊n2⌋,⌊n

2⌋) has learlyvertex Pn, for all n. Furthermore, we observe the following:

n = 19: D(13,6) ←→0 D(12,6), D(12,7) ←→2 D(11,6), D(11,8) ←→0 D(11,7). Hen e these mod-ules have the defe t groups of their blo ks as verti es.n = 20: D(14,6) ∼= S(14,6) with vertex (P3)

6 whi h is not a defe t group of the blo k on-taining D(14,6). Moreover, we have D(13,7) ←→2 D(13,6), D(12,8) ←→2 D(11,8), dim(D(11,9)) =20

(209

)−

(208

)−

(203

)+

(202

)+ 1 = 41041 ≡ 1 (mod 3). Thus all these modules have vertex P18.

n = 21: ResS21S20

(D(15,6)) ∼= D(15,5) ⊕D(14,6), ResS21S20

(D(14,7)) ∼= D(13,7) and ResS21S20

(D(12,9)) ∼=

D(12,8) ⊕D(11,9). Thus these modules have vertex P21 = P18 × P3, by Lemma 2.3. Moreover,D(13,8) ←→0 D(12,7) with vertex P15.n = 22: D(16,6) ←→0 D(15,6) with vertex P21, and D(15,7) ∼= S(15,7) with vertex (P3)

6 whi h isnot a defe t group of the blo k ontaining D(15,7). Moreover, D(14,8) ←→0 D(14,7) with vertexP21. Finally, D(12,10) ←→2 D(11,9) and D(13,9) ←→0 D(12,9). Hen e D(12,10) has vertex P18,and D(13,9) has vertex P21.n = 23: Sin e dim(D(17,6)) =

(236

)−

(235

)− 1 = 67297 ≡ 1 (mod 3), the module D(17,6)has vertex P21. Furthermore, D(16,7) ←→2 D(16,6) with vertex P21 and D(15,8) ←→2 D(14,8),

D(13,10) ←→2 D(13,9), and dim(D(14,9)) =(239

)−

(238

)−

(236

)+

(255

)+ 1 = 259579 ≡ 1 (mod 3).Also these modules have thus vertex P21.

n = 24: Here we have ResS24S23

(D(18,6)) ∼= D(18,5) ⊕ D(17,6) and ResS24S23

(D(17,7)) ∼= D(16,7) sothat both modules have vertex P24, by Lemma 2.3. Furthermore, D(16,8) ∼= S(16,8) with vertex(P3)

6 whi h not a defe t group of the blo k ontaining D(16,8), and D(13,11) ←→0 D(12,10)has vertex P18. Finally, ResS24S23

(D(14,10)) ∼= D(13,10) and ResS24S23

(D(15,9)) ∼= D(15,8) ⊕ D(14,9).Consequently, both modules have vertex P24 = P21 × P3, by Lemma 2.3.n = 25: D(19,6) ←→0 D(18,6), D(18,7) ←→2 D(17,6), D(17,8) ←→0 D(17,7), D(14,11) ←→0

D(14,10), D(15,10) ←→2 D(14,9), and D(16,9) ←→0 D(15,9). Hen e all these modules have thedefe t groups of their blo ks as verti es.n = 26: dim(D(20,6)) =

(266

)−

(265

)−

(263

)+

(262

)= 162175 ≡ 1 (mod 3) so that D(20,6) has ver-tex P24. We also have D(19,7) ←→2 D(19,6) and D(18,8) ←→2 D(17,8) so that both modules havevertex P24. Furthermore, dim(D(14,12)) =

(2612

)−

(2611

)−

(266

)+

(265

)+

(263

)−

(262

)= 1769365 ≡ 1

(mod 3), D(15,11) ←→2 D(14,11), D(16,10) ←→2 D(16,9), and D(17,9) ∼= S(17,9). Thus all thesemodules have vertex P24.The simple FS27-module D(21,6) of dimension 200655 ≡ 0 (mod 9) is the �rst module whoseverti es annot be determined with the above methods, and whi h we urrently annot treatwith the omputer either. We suspe t that the verti es of D(21,6) are the Sylow 3-subgroupsof S27.To summarize, Proposition 5.3 and our additional onsiderations in hara teristi 3 lead tothe following:Corollary 5.5. Let p > 2 and 1 ≤ n ≤ f(p) where f(3) = 26, and f(p) := 2p2 − 1 forp > 3. Then the verti es of all simple FSn-modules labelled by two part partitions are known.If m ∈ {0, . . . , ⌊n2 ⌋} then D(n−m,m) has the defe t groups of its blo k as its verti es, unlessS(n−m,m) ∼= D(n−m,m). In the latter ase D(n−m,m) has vertex Pn−2m × (Pm)2.

21

6 Closing remarksWe lose with some remarks and questions arising from our results in the previous se tions. Inview of Theorem 4.1 and Corollary 5.5, we are led to the following general question on erningsimple FSn-modules labelled by two part partitions:Question 6.1. Let p > 2 and n ∈ N. Consider the simple FSn-module D(n−m,m), form ∈ {0, . . . , ⌊n2 ⌋}. Provided D(n−m,m) 6∼= S(n−m,m), are the verti es of D(n−m,m) then pre iselythe defe t groups of the blo k ontaining D(n−m,m)? Of ourse, by Theorem 2.4, D(n−m,m)has vertex Pn−2m × (Pm)2 if D(n−m,m) ∼= S(n−m,m).Remark 6.2. Suppose that p > 2, and let µ be any p-regular partition. Then the simpleFSn-modules Dµ and Dµ ⊗ S(1n) =: Dm(µ) have ommon verti es. Here, m(µ) denotes theMullineux onjugate partition of µ. In this way, on e knowing the verti es of simple FSn-modules labelled by two part partitions, we also have the verti es of the simple FSn-moduleslabelled by the respe tive Mullineux onjugate partitions. The rather te hni al pro edure for onstru ting the Mullineux onjugate of a given p-regular partition of n, is the ontent of theMullineux onje ture ( f. [19℄), proved by Ford and Klesh hev in [6℄.The tables below display the verti es of all simple FSn-modules labelled by two part partitionsand of those labelled by their Mullineux onjugate partitions, for p = 3 and n ≤ 26. Blo ks areparametrized by their 3- ores, and simple modules by their labelling partitions; neighbouringpartitions are always Mullineux onjugate to ea h other. Moreover, olumn �def.� ontains thedefe t groups of the blo ks under onsideration. Whenever a simple FSn-module is isomorphi to some Spe ht module, the orresponding partition is displayed in olumn �Spe ht�.Remark 6.3. Suppose that p = 2, and �x n ∈ N. As stated earlier, in this ase the verti es ofsimple FSn-modules parametrized by two part partitions are rather poorly understood. How-ever, some partial results on verti es of the simple FSn-modules D(n−m,m), for m ∈ {2, 3}and m ∈ {4, 5}, respe tively, an be found in [23℄ and [7℄, respe tively. Some further om-putational results on erning verti es of simple FSn-modules D(n−m,m) are given in [2℄, [3℄and [23℄. Building on our omputational data already mentioned in the introdu tion, wesuspe t that the simple FSn-module D(n−2,2) has either vertex Pn, or n = 5, or n ≡ 3(mod 4). As explained earlier, the Sylow 2-subgroup of A4 is a vertex of D(3,2). Moreover, forn ≡ 3 (mod 4), we obtain S(n−2,2) ∼= D(n−2), by Theorem 2.6, and D(n−2,2) has then vertexPn−4 × (P2)

2 /∈ Syl2(Sn), by Theorem 2.4.Referen es[1℄ W. Bosma, J. Cannon, C. Playoust, The Magma algebra system. I. The user language, J.Symboli Comput., 24 (3-4) (1997), 235�265[2℄ S. Danz, Theoretis he und algorithmis he Methoden zur Bere hnung von Vertizes irredu-zibler Moduln symmetris her Gruppen, PhD thesis, Jena (2007)(http://www.minet.uni-jena.de/algebra/preprints/preprints.html)[3℄ S. Danz, Verti es of low dimensional simple modules for symmetri groups, to appear inComm. Algebra 22

Simple FSn-modules in hara teristi 3

n blo ks def. parts. vtx. Spe ht1 (1) 1 (1) 1 (1)2 (2) (12) 1 (2) (12) 1 (2) (12)3 ∅ P3 (3) (2,1) P3 (3) (13)4 (1) P3 (4) (22) P3 (4) (14)

(3, 1) (2, 12) 1 (3,1) (2, 12) 1 (3, 1) (2, 12)5 (2) (12) P3 (5) (3,2) P3 (5) (15)

(22, 1) (4,1) P3 (2, 13) (4, 1)6 ∅ P 2

3(6) (32) P 2

3(6) (16)

(5,1) (3, 2, 1) P 23 -

(4, 2) (22, 12) 1 (4,2) (22, 12) 1 (4, 2) (22, 12)7 (1) P 23 (7) (4,3) P 2

3 (7) (17)

(5,2) (3, 2, 12) P 2

3-

(3, 1) (2, 12) P3 (6,1) (32, 1) P3 (6, 1) (2, 15)8 (2) (12) P 2

3(8) (42) P 2

3(8) (18)

(5,3) (3, 22, 1) P 23 (5, 3) (23, 12)

(4, 3, 1) (7,1) P 2

3(2, 16) (7, 1)

(32, 12) (6,2) P 23 - -9 ∅ P9 (9) (5,4) P9 (9) (19)

(8,1) (42, 1) P9 -(6,3) (32, 2, 1) P9 -

(4, 2) (22, 12) P3 (7,2) (4, 3, 12) P3 (7, 2) (22, 15)10 (1) P9 (10) (52) P9 (10) (110)

(8,2) (42, 12) P9 -(7,3) (4, 3, 2, 1) P9 -

(3, 1) (2, 12) P 2

3(9,1) (5, 4, 1) P 2

3(9, 1) (2, 18)

(6,4) (32, 22) P 2

3 (6, 4) (24, 12)11 (2) (12) P9 (11) (6,5) P9 (11) (111)

(8,3) (42, 2, 1) P9 - -(52, 1) (10,1) P9 (2, 19) (10, 1)

(5, 4, 12) (9,2) P9 - -(4, 3, 22) (7,4) P9 - -Table 1: simple FSn-modules in hara teristi 3; n = 1, . . . , 11

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n blo ks def. parts. vtx. Spe ht12 ∅ P12 (12) (62) P12 (12) (112)

(11,1) (6, 5, 1) P12 -(9,3) (5, 4, 2, 1) P12 -(8,4) (42, 22) P12 -

(4, 2) (22, 12) P 23 (10,2) (52, 12) P 2

3 (10, 2) (22, 18)

(7,5) (4, 32, 2) P 2

3(7, 5) (25, 12)13 (1) P12 (13) (7,6) P12 (13) (113)

(11,2) (6, 5, 12) P12 -(10,3) (52, 2, 1) P12 -(8,5) (42, 3, 2) P12 -

(3, 1) (2, 12) P9 (12,1) (62, 1) P9 (12, 1) (2, 111)

(9,4) (5, 4, 22) P9 - -14 (2) (12) P12 (14) (72) P12 (14) (114)

(11,3) (6, 5, 2, 1) P 4

3(11, 3) (23, 18)

(8,6) (42, 32) P12 - -(7, 6, 1) (13,1) P12 (2, 112) (13, 1)

(62, 12) (12,2) P12 - -(52, 22) (10,4) P12 - -

(5, 4, 3, 2) (9,5) P12 - -15 ∅ P15 (15) (8,7) P15 (15) (115)

(14,1) (72, 1) P15 -(12,3) (62, 2, 1) P15 -(11,4) (6, 5, 22) P15 -(9,6) (5, 4, 32) P15 -

(4, 2) (22, 12) P9 (13,2) (7, 6, 12) P9 (13, 2) (22, 111)

(10,5) (52, 3, 2) P9 - -16 (1) P15 (16) (82) P15 (16) (116)

(14,2) (72, 12) P15 -(13,3) (7, 6, 2, 1) P15 -(11,5) (6, 5, 3, 2) P15 -(10,6) (52, 32) P15 -

(3, 1) (2, 12) P12 (15,1) (8, 7, 1) P12 (15, 1) (2, 114)

(12,4) (62, 22) P 4

3(12, 4) (24, 18)

(9,7) (5, 42, 3) P12 - -Table 2: simple FSn-modules in hara teristi 3; n = 12, . . . , 16

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n blo ks def. parts. vtx. Spe ht17 (2) (12) P15 (17) (9,8) P15 (17) (117)

(14,3) (72, 2, 1) P15 (14, 3) (23, 111)

(11,6) (6, 5, 32) P15 - -(82, 1) (16,1) P15 (2, 115) (16, 1)

(8, 7, 12) (15,2) P15 - -(7, 6, 22) (13,4) P15 - -(62, 3, 2) (12,5) P15 - -(52, 4, 3) (10,7) P15 - -18 ∅ P18 (18) (92) P18 (18) (118)

(17,1) (9, 8, 1) P18 -(15,3) (8, 7, 2, 1) P18 -(14,4) (72, 22) P18 -(12,6) (62, 32) P18 -(11,7) (6, 5, 4, 3) P18 -

(4, 2) (22, 12) P12 (16,2) (82, 12) P12 (16, 2) (22, 114)

(13,5) (7, 6, 3, 2) P 43 (13, 5) (25, 18)

(10,8) (52, 42) P12 - -19 (1) P18 (19) (10,9) P18 (19) (119)

(17,2) (9, 8, 12) P18 -(16,3) (82, 2, 1) P18 -(14,5) (72, 3, 2) P18 -(13,6) (7, 6, 32) P18 -(11,8) (6, 5, 42) P18 -

(3, 1) (2, 12) P15 (18,1) (92, 1) P15 (18, 1) (2, 117)

(15,4) (8, 7, 22) P15 (15, 4) (24, 111)

(12,7) (62, 4, 3) P15 - -20 (2) (12) P18 (20) (102) P18 (20) (120)

(17,3) (9, 8, 2, 1) P18 - -(14,6) (72, 32) P 6

3 (14, 6) (26, 18)

(11,9) (6, 52, 4) P18 - -(10, 9, 1) (19,1) P18 (2, 118) (19, 1)

(92, 12) (18,2) P18 - -(82, 22) (16,4) P18 - -

(8, 7, 3, 2) (15,5) P18 - -(7, 6, 4, 3) (13,7) P18 - -(62, 42) (12,8) P18 - -Table 3: simple FSn-modules in hara teristi 3; n = 17, . . . , 20

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n blo ks def. parts. vtx. Spe ht21 ∅ P21 (21) (102) P21 (21) (121)

(20,1) (102, 1) P21 -(18,3) (92, 2, 1) P21 -(17,4) (9, 8, 22) P21 -(15,6) (8, 7, 32) P21 -(14,7) (72, 4, 3) P21 -(12,9) (62, 5, 4) P21 -

(4, 2) (22, 12) P15 (19,2) (10, 9, 12) P15 (19, 2) (22, 117)

(16,5) (82, 3, 2) P15 (16, 5) (25, 111)

(13,8) (7, 6, 42) P15 - -22 (1) P21 (22) (112) P21 (22) (122)

(20,2) (102, 12) P21 -(19,3) (10, 9, 2, 1) P21 -(17,5) (9, 8, 3, 2) P21 -(16,6) (82, 32) P21 -(14,8) (72, 42) P21 -(13,9) (7, 6, 5, 4) P21 -

(3, 1) (2, 12) P18 (21,1) (11, 10, 1) P18 (21, 1) (2, 120)

(18,4) (92, 22) P18 - -(15,7) (8, 7, 4, 3) P 6

3(15, 7) (27, 18)

(12,10) (62, 52) P18 - -23 (2) (12) P21 (23) (12,11) P21 (23) (123)

(20,3) (102, 2, 1) P9 × P 4

3 (20, 3) (23, 117)

(17,6) (9, 8, 32) P21 - -(14,9) (72, 5, 4) P21 - -(112, 1) (22,1) P21 (2, 121) (22, 1)

(11, 10, 12) (21,2) P21 - -(10, 9, 22) (19,4) P21 - -(92, 3, 2) (18,5) P21 - -(82, 4, 3) (16,7) P21 - -(8, 7, 42) (15,8) P21 - -(7, 6, 52) (13,10) P21 - -Table 4: simple FSn-modules in hara teristi 3; n = 21, . . . , 23

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n blo ks def. parts. vtx. Spe ht24 ∅ P24 (24) (122) P24 (24) (124)

(23,1) (12, 11, 1) P24 -(21,3) (11, 10, 2, 1) P24 -(20,4) (102, 22) P24 -(18,6) (92, 32) P24 -(17,7) (9, 8, 4, 3) P24 -(15,9) (8, 7, 5, 4) P24 -(14,10) (72, 52) P24 -

(4, 2) (22, 12) P18 (22,2) (112, 12) P18 (22, 2) (22, 120)

(19,5) (10, 9, 3, 2) P18 - -(16,8) (82, 42) P 6

3(16, 8) (28, 18)

(13,11) (7, 62, 5) P18 - -25 (1) P24 (25) (13,12) P24 (25) (125)

(23,2) (12, 11, 12) P24 -(22,3) (112, 2, 1) P24 -(20,5) (102, 3, 2) P24 -(19,6) (10, 9, 32) P24 -(17,8) (9, 8, 42) P24 -(16,9) (82, 5, 4) P24 -(14,11) (72, 6, 5) P24 -

(3, 1) (2, 12) P21 (24,1) (122, 1) P21 (24, 1) (2, 123)

(21,4) (11, 10, 22) P9 × P 43 (21, 4) (24, 117)

(18,7) (92, 4, 3) P21 - -(15,10) (8, 7, 52) P21 - -26 (2) (12) P24 (26) (13

2) P24 (26) (126)

(23,3) (12, 11, 2, 1) P24 (23, 3) (23, 120)

(20,6) (102, 32) P24 - -(17,9) (9, 8, 5, 4) P24 (17, 9) (29, 18)

(14,12) (72, 62) P24 - -(13, 12, 1) (25,1) P24 (2, 124) (25, 1)

(122, 12) (24,2) P24 - -(112, 22) (22,4) P24 - -

(11, 10, 3, 2) (21,5) P24 - -(10, 9, 4, 3) (19,7) P24 - -

(92, 42) (18,8) P24 - -(82, 52) (16,10) P24 - -

(8, 7, 6, 5) (15,11) P24 - -Table 5: simple FSn-modules in hara teristi 3; n = 24, 25, 2627

[4℄ S. Danz, B. Külshammer, R. Zimmermann, On verti es of simple modules for symmetri groups of small degrees, to appear in J. Algebra[5℄ K. Erdmann, Tensor produ ts and dimensions of simple modules for symmetri groups,Manus ripta Math. 88, 357�386 (1995)[6℄ B. Ford, A. Klesh hev, A proof of the Mullineux onje ture, Math. Z. 226 (1997), 267�308[7℄ B. Fotsing, Symmetris he Gruppen, einfa he Moduln und V ertizes, PhD thesis, Jena,published by Shaker-Verlag, Aa hen (2007)[8℄ J. Grabmeier, Unzerlegbare Moduln mit trivialer Youngquelle und Darstellungstheorie derS huralgebra, Bayreuth. Math. S hr. 20, 9�152 (1985)[9℄ , A. J. Green, On inde omposable representations of a �nite group, Math. Z. 70 (1959),430�445[10℄ D. J. Hemmer, The Ext1-quiver for ompletely splittable representations of the symmetri groups, J. Group Theory 4 (2001), 401�416[11℄ D. J. Hemmer, D. K. Nakano, Support varieties for modules over symmetri groups, J.Algebra 254 (2002), 422�440[12℄ G. D. James, The representation theory of the symmetri groups, Berlin, Heidelberg, NewYork, Springer (1978)[13℄ G. D. James, A. Kerber, The representation theory of the symmetri group, En y l. Math.Appl. 16, Addison-Wesley, Reading (1981)[14℄ A. Klesh hev, Completely splittable representations of symmetri groups, J. Algebra 181(1996), 584�592[15℄ A. Klesh hev, Linear and proje tive representations of symmetri groups, Cambridge Uni-versity Press (2005)[16℄ R. Knörr, On the verti es of irredu ible modules, Ann. Math. (2) 110 (1979), 487�499[17℄ S. Martin, S hur algebras and representation theory, Cambridge Tra ts in Mathemati s,vol. 112, Cambridge University Press (1993)[18℄ J. Müller, R. Zimmermann, Green verti es and sour es of simple modules of the symmetri group, Ar h. Math. 89 (2007), 97�108[19℄ G. Mullineux, Bije tions of p-regular partitions and p-modular irredu ibles of the sym-metri groups, J. London Math. So . (2) 20 (1979), 60�66[20℄ H. Nagao, Y. Tsushima, Representations of �nite groups, A ademi Press, San Diego(1989)[21℄ J. B. Olsson, Lower defe t groups in symmetri groups, J. Algebra 104 (1986), 37�56[22℄ A. J. E. Ryba, Fibona i representations of the symmetri groups, J. Algebra 170 (1994),678�686 28

[23℄ R. Zimmermann, Vertizes einfa her Moduln der symmetris hen Gruppen, PhD thesis,Jena (2004) (http://www.minet.uni-jena.de/algebra/preprints/preprints.html)Susanne Danz, Mathematis hes Institut, Friedri h-S hiller-Universität Jena,07737 Jena, Germanyemail: susanned�minet.uni-jena.de

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