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Group Theory I

by: Donny Graham

Group Theory I MTH 912

Donny Graham
GPA 3.57

U. Meierfrankenfeld

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U. Meierfrankenfeld
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Date Created: 09/19/15
Group Theory Lecture Notes for MTH 912913 0405 Ulrich Meierfrankenfeld April 26 2007 Contents O CO 4 G ro up Theory 11 Group Action 12 Balanced Products of G sets 13 Central by nite groups 14 Finite p Groups 15 A p complement Theorem General Representation Theory 21 Basic De nitions 22 Krull Schmidt Theorem 23 Maschke s Theorem 24 Jacobson Radical 25 Simple modules for algebras 26 Tensor Products 27 lnduced and Coinduced Modules 28 Tensor Induction and Transfer 29 Clifford Theory Character Theory 31 Semisimple Group Algebra 32 Characters 33 Burnside s paqb Theorem 34 An hermitian form 35 F robenius7 Theorem 36 Quaternion Groups 37 Groups with quaternion Sylow 2 subgr0up Linear Algebra 41 Bilinear Forms 5 d3 CONTENTS Representations of the Symmetric Groups 101 51 The Symmetric Groups 101 52 Diagrams7TableauX and Tabloids 102 53 The Specht Module 108 54 Standard basis for the Specht module 112 55 The number of simple modules 116 56 p regular partitions 119 57 Series of R modules 123 58 The Branching Theorem 124 59 SW23 126 510 The dual of a Specht module 132 Brauer Characters 135 61 Brauer Characters 135 62 Algebraic integers 137 63 The Jacobson Radical ll 143 64 A basis for CE 145 65 Blocks 150 66 Brauer s Frist Main Theorem 156 67 Brauer s Second Main Theorem 165 Chapter 4 Linear Algebra 41 Bilinear Forms De nition 411 de bilinear form Let R be a ring V an R module and W a right R module and s V X W gt RUw gt U i w a function Let A Q V and B Q W Suppose that s is R bilinear that is 221 riU ijj 21 rU39 w si for all vi 6 V wj E W and rhsi E R Then a a s is called a bilinear form b b s is called symmetric ifV W and U L w w L U for allUw E V c z s is called symplectic ifV W and U L U 0 for allU E V d C Let U E V and w E W we say that U and w are perpendicular and write U L w if U w 0 e d We say that A and B are perpendicular and write A L B ifa L b for all a E A b e B 8 AL w E W i A L w and LB U E V i U L B AL is called the right perp of A and LB the left perjn of B 9 f IfA is an R submodule of V de ne 3A W gt A by sAwa a i w for all a E Aw E W h g If B is an R submodule of W de ne sB V gt B by sBUb U i b for all 1 e Vb e B i h s is called non degenerate if VL 0 and LW 0 If V is free with basis V and W is free with basis W then the V X W matrix U i w vgv gw is called the Gram Matrix of s with respect to V and W ObserUe that the Gram Matrix is just the restriction of s to V X W 91 92 Chapter 4 Linear Algebra Let I be a set R a ring W 1R and V DIR De ne s V x W a R 1 10 261 0110 Note that this is well de ned since almost all 1 are zero Note also that if we View 1 and 10 as I x 1 matrices we have 1 10 0T10 As a second example let V be any R module and W V and de ne 1 10 101 If V is a free R module this example is essentially the same as the previous Lemma 412 dual basis Let V be a free R module with basis V For 1i 6 V de ne if E V by 1i1 6W De ne by V a 111 H101wgv V Vak 3 VT H R7a H 04Ullvev a a Both by and inn are R isomomhisms b b Let 10 E V and 1 E V and put gtV0 and 10 gtV110 Then 101 17T10 Proof a Since V is free with basis V the map eBVR a V n a Eva n1 is an R isomorphism Clearly by is the inverse of this map and so by is an R isomorphism To check that inn is an R linear map of right R modules recall rst that V is a right R module via wr0 101r Also VR is a right R module via T1UT TETM We compute WWW WUv WWW 1001 and so 1121 is R linear Given n v E V R then 10 V a R Zvev 811 a Eva sun is the unique element of V with 101 m for all 1 E V that is with gtV110 nu So inn is a bijection b For 1i 6 V let su if 1 and Tu Then 1 2761 sun and so 101 2761 su101i 2761 sum 17T10 D VA De nition 413 dual map Let R be a ring and 0 V a W an R linear map Then the R linear map 04 W gt V gt a j o a is called the dual of 04 Lemma 414 matrix of dual Let R be a ring and V and W free R modules with basis V and W respectively Let 0 V gt W be an R linear map and M its matrix with respect to V dndW Let6 E W Then vltalt6gtgt MT Wlt6 Proof Let 1 E V Then the 1 coordinate of 1121 oz6 is oz61 6oa0 6a0 By de nition of M mwvwewvev 041 Zwew mwvw and so Section 41 Bilinear Forms 93 vsa5 5avv E mwv5w MUM45 wEW D Lemma 415 associated nondeg form Let R be a ring and s V x W a R an R bilinear form Let A be an R subspaoe of V and B an R subspaoe of W Then EABAA iB x BBmAiaAmiebBmAi sally is a well de ned non degenerate R bilinear form Proof Readily veri ed D Lemma 416 basic bilinear Let R be a ring and let 3 V x W a R be an R bilinear form a a Let A be an R subspaoe of V then AL ker SA b b Let B be an R subspaoe of W then LB ker SB 0 C s is non degenerate if and only if 3V and SW are 1 Proof a and b are obvious and 6 follows from a and D Lemma 417 nite dim nondeg Let lF be a diuision ring and s V x W a lF a non degenerate lF bilinear form Suppose that one of V or W is nite dimensional Then both V and W are nite dimensional both 3V and SW are isomoijnhisms and dimF V dimF W Proof Without loss dimF V lt 00 and so dim V dim V By 41667 3V and sW are 1 1 and so dim W dim V dim V So also dim W is nite and dim V dim W dim W Hence dim V dim W dim W dim V Since 3V and sW are 1 1 this implies that 3V and sW are isomorphisms D Corollary 418 dual sbasis LetlF be a diuision ring 3 V x W a lF a non degenerate lF bilineanform B a basis for V Suppose that B is nite Thenfor each b E B there exists a unique b E W with sab lab for all ab E B Moreouer b l b E B is an lF basis for W Proof By 417 sV W a V is an isomorphism Let b E V with b a ab and de ne b s1b D 94 Chapter 4 Linear Algebra De nition 419 defzs dual basis Let F be a diuision ring 3 V x W a F a non degenerate IF bilinear form 3 a basis for V A tuple i b E B such that for all ab E B b E W a i ab and i b E B is basis for W is called the basis for W dual to B with respect to 3 De nition 4110 de adjoint Let R be ring 33 x Wl a R 12 R bilinear forms and 04 V1 gt V2 and W2 gt W1 R linear maps We say that 04 and B are adjoint with respect to 31 and 32 or that B is an adjoint ofa prouided that WA iw22 U1 13w21 for all U1 6 V1 w2 6 W2 Lemma 4111 basic adjoint Let R be a ring 3i W x Wl a B7 127w a v wi i 17 2 R bilinear forms and 04 V1 gt V2 and W2 gt W1 R linear maps Then 04 and B are adjoint i 31V1 o B 04 o s2V2 Proof Let U1 6 V1 and w2 6 W2 Then 6W1 1 w22 82V2w2avl a82V2w2U1 of O 82V2w2vl and U1 i 3w21 81V15w2vl 81V1 O 5w2vl and the lemma holds D Lemma 4112 kernel of adjoint Let R be a ring 3i V x Wl a R 12 R bilinear forms and 04 V1 gt V2 and W2 gt W1 R linear maps Suppose oz and B are adjoint Then kera L1m B with equality if LW2 0 Proof Let U1 6 V1 Then U1 6 kera ltgt 04011 0 ltgt if Wzi 0au1 102 0va 6 W2 ltgt U1 3w2 0Vw2 6 W2 ltgt U1 6 Lime D Lemma 4113 unique adjoint Let R be a diuision ring 3139 V x Wl a R 12 R bilinear forms and 04 V1 gt V2 and W2 gt W1 R linear maps Suppose 31 is non degenerate and V1 is nite dimensional ouer R a a There exists a unique adjoint Glad of 04 with respect to 31 and s2 Section 41 Bilinear Forms 95 b b Suppose that also 32 is non degenerate and V2 is nite dimensional Let Vi be a basis for V and U 1 U 6 Vi the basis W dual to V with respect to 8 IfM is the matrix of 04 with respect to V1 and V2 then MT is the matrix for Glad with respect to V2 and V1 Proof a By 417 31V1 is an isomorphism and so by 4111 31quot1 o 04 o 32V2 is the unique adjoint of 04 b Let U 6 Vi Then the U1U2COe 1Clth of M is aU1 1 U22 By de nition of the adjoint am 1 am U1 1 oadag1 and so b holds Corollary 4114 dual basis for subspace LetlF be a eld V a nite dimensional lF space and s V X V gt lF an non degenerate symmetric lF bilinear form on V Let W be an s non degenerate lF subspace of V Let V be an lF basis for V and W an W basis for W Let V U 1 U E V and W a 1 w E W be the corresponding dual basis for W and V respectiUely Let M mvw be the V X W matrix 0UerlF de ned by UWL Z r7imw 1WL wEW for all U E V Then a 2 mm UEV Proof Since W is non degenerate V W Wi Let Oz V a W be the orthogonal projection onto W that is if U w 1 y with w E W and y E Wt then w aU Observe that the matrix of 04 with respect to V and W is MT Let B W a Vw a w be the inclusion map Then for all U E V w E W av1wv1wvl w and so 3 is the adjoint of 04 Thus by 4113b the matrix for B with respect to W and V is MTT M So a 3a Emma UEV D Lemma 4115 gram matrix Let R be a ring V a free R module with basis V and W a free right R module with basis W Let in V gt V R gtW V gt WR gtV1V gt VR and W1W gt VR be the associated isomorphisms Let s V X W gt R be bilinear form and M its Gram Matrix with respect to V and W Let U E V w E W U tnU and I Ww 96 Chapter 4 Linear Algebra a a lt7 1w WM b b Mti Nu11M the Null space of M c a new WM 1 d1 WSWU MTfJI 6 e1 V3Vw Mw Proof a We have U Zea7771a w Zbew 5131 and M a 1 bab Since 3 is R bilinear7 u 1 w Z 71aa 12ml aTMw aEVbEW b By a w 6 Vi iff aTMw 0 for a117171ff M72 0 and iff w e Nu11M c u e LW mam10 iEMTa 0 1M eNullMT 1 Let 71 sWU and 71 W U Then by right module version of 412 On the other hand 71w sWUw U 1 w UTM w Thus 7T 5T M and so 71 MT u and d holds e Let 71 sVw and 71 lt1gtV17i Then by 412 71U UT 71 On the otherhand 71U sVwU U 1 w 71TMw S0 71 M71 and e holds D Lemma 4116 gram matrix of dual basis LetIF be a division ring and s Vx W a F a non degenerate IF bilinear form Let V and W be IF basis for V and W respectively and V and W the corresponding dual basis for W and V Let M be the Gram matrix for s with respect to V and W Let N the Gram matrix for s with respect to W and Then a a MT is the matrix for My with respect to V and b b N is the matrix for idW with respect to W and V c C M and N are inverse to each other Section 41 Bilinear Forms 97 71 Proof a We have idV V 91K W 53 V By 4115d7 the matrix of sW with respect to V and W is M By de niton of W the matrix of 3 with respect to W and W is the identity matrix So a holds 10 Similar to a7 use 3V and 4115e c By 10 N 1 is the matrix of idW with respect to V and W Note that idV is the adjoint of idW So by a and 411300 N l MTT M D Lemma 4117 circ and bilinear Let R be a commutatiue ring G a group and let V and W be RG modules Let s V X W gt R be R bilinear form a a s is G inuariant i aou l w u l aw for all a E inRG b b Leta 6 RG Then AWa etaH with equality if Vi 0 Proof a Recall rst for a 2960 agg E Rg7 aquot 2960 agg l Thus 3 is G invariant ltgt gulgwulw VgEG7u V7wEW uauguisabijectionltgtulgwg 1ulw V96G7 Uel7 LUEW sistilinear ltgt ulawa ulw VaERG7u V7w W b By a a and aquot are adjoints So In follows from 4112 D Lemma 4118 extending scalars and bilinear Let R g 13 be an extensions of Tings and s V X W gt R an R bilinear form There exists a unique R bilinear form R RVX W RRHRa uw b al ewb for allab Ru E Vw E V Proof Observe that the map 1 X V X W X 1 to7 a7u7b7w gt am uwb is R balanced in at7 u and b7 w The universal property of the tensor product now shows the existence of the map 5 A simple calculation shows that s is R bilinear Lemma 4119 extending scalars and intersections Let lF K be an extension of diuision rings and V on lF space a a LetW be a set ole subspaces of V Then K WK W WSW WSW 98 Chapter 4 Linear Algebra b b Lets V X W gt IF be an IF bilinear form and extend s to a bilinear form 5 K 811 VX W 1FIK AK see 4118 LetX an IF subspace ofV Then IK 1FXL IK XL Proof a Suppose rst that W W17 W2 Then there exists IF subspaces X1 of W1 with W1 XiW1 W2 Observe that W1W2 Wl W2X1X2 For X an IF subspace of V let X IK 1FX IK 1FV ThenWl W1 W2 Yi and W1 W2 W1 W2Y1 Y2 and so W1 OWZ W1 W2 So a holds if 2 By induction it holds ifW is nite In the general case let U E V Then there exists a nite dimensional U V with U E U Moreover7 there exists a nite subset X of W with W O XeYX W O XeW X By the nite case7 V O XeYX U XeY X and so a is proved b Note that XL meX xi So by a we may assume that X IFz for some z E X If X 1 V7 then also X 1 V and we are done Otherwise dim VXL 1 and so also dimVXL 1 From Fth ltVwe conclude that Xi D Lemma 4120 symmetric form for p2 LetIF be a eld with charIF 2 De ne a IF gt IF7 f gt f2 and let IF f by the IF space with IF f IF as abelian group scalar multiplication f I k le Lets a symmetric form on V and de ne oz V gt IF f U gt U I U Then 04 is IF linear W kera U E V I U I U 0 is an IF subspace s 1W is a symplectic form and dim VW dimF r0 dim zz r Proof Since Uw I Uw U IUU I ww I Uw I w U I U2U I wwl w U l U w l w and fU l fU f2 U l U f v U l U conclude that 04 is IF linear Thus W kera is an IF subspace of V and VW E lma Also dimF lma dimF IF The map 77 idF IF x IF a IF2 x IF7 f7 k a f27 k provides an isomorphism of the IF space IF and the IFZ space IF So dim1w IF dim1z2 IF Cleary s 1W is a symplectic form D Lemma 4121 symplectic forms are even dimensional Let IF be a eld V a nite dimensional IF space and s a non degenerate symplectic IF form on V Then there exists an IF basis Ul397l E i1i2i n for V with Ui I 1 6139jsgni In particular dimF V is euen Proof Let 0 7 U1 6 V Since U1 0 Vt7 there exists U E V with U1 1 U 7 0 Let U71 U1 1 U 1U Then U1 1 U71 1 U71 1 U1 Let W IFU1U1 The Gram i i i 1 i i Matrix of s on W With respect to U17 U1 is gt So the Gram matrix has determinant 71 0 1 7 0 Thus W is non degenerate and so V W Q Wt Hence also WL is non degenerate and the theorem follows by induction on dimF V D Lemma 4122 selfdual and forms LetIF be eld G a group and V simple IFG module Suppose that V is self dual that is V g V as IFG module Section 41 Bilinear Forms 99 a a There exists a non degenerate G inuariant symplectic or symmetric form 3 on V b b Suppose that charlF 2 and lF is perfect Then either V lFG or s is symplectic a Let Oz V a V be an lFG isomorphism and t V x V a IF 1110 a ozvw7 the corresponding G invariant lF bilinear form Since V is a simple lFG module any non zero G invariant bilinear form on V is non degenerate De ne rvw tvw twv Then T is a symmetric form If r 7 07 then a holds with s r If r 0 then tvw itwv for all mu 6 V lf charlF 27 then t is symmetric and a holds with s t lf charlF 7 27 then tvv itvv implies that t is symplectic So again a holds with s t 10 Let s be as in a and observe that in either case of a7 s is symmetric Let Oz V a lFa be as in 4120 View lFquot as an lFG module with G acting trivially Then by 4120 04 is lF linear and since Sis G invariant also lFG linear Since lF is perfect7 dimF F So lFquot E lFG has lFG modulo and either 04 0 or 04 is onto If 04 07 s is symplectic If 04 is onto kera 7 V is an lFG submodule of V Since V is simple7 kera 0 and so VglmozF ElFG 100 Chapter 4 Linear Algebra Chapter 5 Representations of the Symmetric Groups 51 The Symmetric Groups For n 6 2 let 0 1 23n and Symn SymQn Let g E Symn and let 09 01 0k be the sets of orbits for g on 9 Let ni and choose notation such that n1 2 n2 2 n3 2 De ne n 0 for all t gt 1 Then the sequence is called the cycle type of 9 Pick aio E O and de ne 17 97ai0 for all j E Z Then aij am if and only ifj E k mod The denote the element 9 by 9 01117 01127 a1n1a217a227 7a2n2 ak17ak27 aknk Lemma 511 conjugacy classes in symn Two elements in Symn are conjugate if and only if they have the same cycle type Proof Let g be as above and h E Symn Then hgh 1 ha11 hlta12 ha1mha21 hlta22 ha2n2 Ham Ham and the lemma is now easily proved D De nition 512 defzpartition of n A partition ofn E N is a non decreasing sequence A0221 of non negative lntergers with n Elil i Note that if is a partion of n the necessarily A 0 for almost all i For example 4 4 4 3 3 1 1 1 1 0 O O is a partition of 22 We denote such a partition by 433214 Observe that the cycle type ofg E Sym is a partition of n Together with 313f we conclude 101 102 Chapter 5 Representations of the Symmetric Groups Lemma 513 number of partitions Let n E Z1 The follwing numbers are equal a a The numbers of partitions of n b b The numbers of conjugacy classes of Symn c C The number of isomorphism classes of simple CSymn modules B Our goal now is to nd an explicit 1 1 correspondence between the set of partions of n and the simple CSymn modules We start by associating a Symn module MA to each partition A of n But this modules is not simple In later section we will determine a simple section of MA De nition 514 de lambda partition Let I be a set of size n and A a partition of n A A partition of is a sequence A A0221 of subsets of A such that a lal IU1Ai b b Ai Aj0foral11 iltjltoo 0 Cl lAil i39 For example 1 3 5 2 4 6 0 we will write such a partition as 0 is a 3 2 1 partition of16 where In 1 2 3 HMH HgtOJ U The lines in this array are a remainder that the order of the elements in the row does not matter On the otherhand since sequences are ordered 3 5 T 7g Let MA be the set of all A partions of In Note that Symn acts on A Via WA 7rA1 Let r be a xed eld and let Mquot M111 rMo Then Mquot is an lFSymn module Note that for MW U 1N Let l the unique bilinear form on MA with orthonormal basis MA Then by l is Symn invariant and non degenerate H to a 35 to a H 52 DiagramsTableaux and Tabloids De nition 521 de diagram Let D Q 2 x 2 a z Let ij kl 6 2 X 2 Then hi provided thati k andj 1 Section 52 DiagramsTableauX and Tabloids 103 b a D is called a diagram i iffor alld E D and e 6 2 X 2 with e g at one has 5 E D c b The elements of diagram are called the nodes of the diagram dc rZXZXijgtiandcZXZXijgtj e e The i th row ofD is D14 D O x 2 and the j column ofD is D7 2 x f ldl DlDil1 and NW WU De nition 522 defzdiagram2 A 6 23 de ne ll 334 6 2 X Zl1 j M Lemma 523 basic diagram Let n E N Then the map D a AD is a bijection between the Diagram of size n and the partitions of n The inverse is is by A gt Proof Let D be a diagram of size n and put A AD Let i E N and let j be maximal with i7j E D By maximality ofj and the de nition of a diagram7 E D iff k g j Thus j Ai and D Let k g i Since i7 A 6 D7 the de ntion of a diagram implies k A and so Ai Ak Thus A is non increasing Clearly Ai lDl n and so A is a partition of n Conversely suppose that A is a partition of n Let i7j E D and a7 1 6 2 x 2 with a g i and b g j Then a g i AJ Ab and so a7 1 E Thus A is a diagram Clearly Ai that is AA A D We draw diagams as in the following example De nition 524 defzdominates Let A and a be partitions of n 6 2 We say that A dominates Ia and write A E a if j j Z i Z Z M i1 i1 for allj 6 2 104 Chapter 5 Representations of the Symmetric Groups Note that dominates is a partial orderigg but not a total ordering For n 6 we have 571 16 On rare occasions it will be useful to have a total ordering on the partition De nition 525 de lexiographic ordering Let A and a be partitions ofn 6 2 We write A gt Ia provided that there eristsi 6 2 with Ai gt m and Aj uj for all 1 g j lt i Observe that H lt is a total ordering on the partitions of n7 called the leriographic ordering If A D a and i is minimal With Ai 7 mui7 then Aj al and 221 Ag 2 221 lulu Thus Ai 2 Ia and so A gt a De nition 526 defzconjugate partition a a Let D Q 2 x 2 Then D j7i i7j E D D is called the conjugate of D b b Let A be a partition of n Then A is the number of nodes in the i th column of Lemma 527 basic conjugate a a The conjugate of a diagram is a diagram b b Let D be a diagram Then the rows of D are the conjugates of the columns of D D Di 0 C Let A be a partition of n Then A is a partition of n and A X Section 52 DiagramsTableauX and Tabloids 105 Proof a follows immediately from the de nition of a diagram b is obvious c By b Ag Thus A AA So c follows from 523 B Lemma 528 reverse ordering Let A and u be partitions of n Then A E In if and only if A 1 u Proof Let j 6 2 and put i u jDe ne the following subsets of 2 x 2 Top a7b l a g Bottom a7b l a gt Leftab bgj Rightablbgti Since A dominates In 1 lTop All 2 lTop Ml By de nition oft Iiij Ai 2 j and A141 gt j Thus Top Left Q In and Bottom Right W p 0 Hence 2 lTop Left All lTop Left M and 3 lBottom Right Z lBottom Right O In From 1 and 2 we conclude 4 lTop Right Z lTop Right 3 and 4 imply lRight Z lBottom O 1 Since n we conclude lLeft o All 2 Left o Ml Thus 1 A g 1 p and X g 1 D 106 Chapter 5 Representations of the Symmetric Groups De nition 529 defztableau Let A be a partition of n A A tableau is a function t A Hgt In We denote tableaux as in the following example 514 23 denotes the 32 tableau t 11 H 4 1 2 a 1 13 a 4 21 a 2 2 2 a 3 De nition 5210 defzpartition 0f tableau Let t D H In be a tableau Then At tDi1 and A t At is called the row partition oft and A t the column partition oft Note that if t is a A tableau then At is a A partition of In and A t is a A partition of In For example 243 2 3 ift61 thenAt 1 5 5 De nition 5211 de tabloids Let st be A tableaur a a s andt are called row equivalent if At As An equivalence class of this relations is called a tabloid and the tabloid containing t is denoted by b b s andt are called column equivalent if A t As The equivalence class of this relations containingt is denoted by For example ift then i5 q H H q q H Lemma 5212 action on tableaux Let A be partition of n Let 7r 6 Symn and st be A tableaux to to DJ OJ to DJ to a a Symn acts transitively on the set of A tableaur via 7rt 7r 0 t b b 7rAt A7rt c C s andt are row equivalent i jrs d 7rt are row equivalent In particular Symn acts on the set of A tabloids via 7rt Lt Section 52 DiagramsTableauX and Tabloids 107 Proof a Clearly 7rt 7r 0 t de nes an action of Symn on the set of A tableaux Since st a bijections from A a In p s o t 1 E Symn Then p o t s and so the action is transitive b Let D Then At and so 7rAt MIX3021WtDi1rtDi1 AW c s is row equivalent to t iff As At and so iff 7rAs 7rAt So by b iff A7rs A7rt and iff 7rt and 7139s are row equivalent B Let A Am1 be A partition of In Let 7r 6 Symn Recall that 7139 E 00A means WA A and so 7rAi Al for all i CsymnA NsymnAl SymAl So CsymnA has order A 00 AI i1 r De nition 5213 def row stabilizer Lett be a tableau The Rt CsymnAt and C CsymtA t R is called the row stabilzer and C the column stablizer oft Lemma 5214 Char row equiv Lets andt be A tableaur The s andt are row equiva lent l s 7rt for some 7139 E Rt Proof Then by 5212a7 s 7rt for some 7139 E Symn Then 8 is row equivalent to t if and only if At A7rt By 5212b7 A7rt 7rAt and so 8 and t are row equivalent iff 7r 6 Rt D Lemma 5215 basic combinatorical lemma Let A and a be partlons of n t a A tableau and s a a tableau Suppose that for all l7j Ati Asj 1 That is no two entrees from the same row oft lie in the same column of 3 Then Aill Moreouer lfA a then there exists A tableau r such that r is row equlualent to t and r is column equlualent to s Proof Fix a column C of Changing the order the entrees of C neither effects the assump tions nor the conclusions of the lemma So we may assume that ifl appears before j in C7 then i also lies earlier row than j in the tableau t We do this for all the columns of s It follows that an entree in the k row of t must lie in one of the rst k rows of 3 Thus 251 Ai 2171 M and la dominates A Suppose now that A Ia Since A1 al and the rs row oft is contained in the rst row of 37 the rst row of At1 As1 Proceeding by induction we see that Atk Ast for all s and t So 8 and t are row equivalent D 108 Chapter 5 Representations of the Symmetric Groups 53 The Specht Module De nition 531 defth Let G be a group H Q G R a ring and f 6 RC Then fH ZheH fhh Lemma 532 basic fh Let G be a group R a ring and f 6 RC Suppose that f view as a function is a multiplicative homomorphism a a LetAB Q G such that the maps AX B gt G a b gt G is 171 then fAB fAfB b b Let A g B G and T a left transuersal to A in B Then f3 foA c C Let A1A2An G and A A l 1 g i g n Suppose A LlAi then fA fAlng fix d 1 Suppose f is a class function then for allg E G and H Q G ngg 1 ngga Proof a Since the map a b a ab is 171 every element in AB can be uniquely written has ab with a E A and b E B Thus fAfB ZaeAfaa39ZbgB fbb ZaEAJJE Bfafbab Egg11763 fabab Zoe13 fcc fAB b is a special case of a c follows from a and induction on n d Readily veri ed Since the map E a At is a well de ned bijection between the tabloids and the the partitions of In we will often identify t with At In particular we have t 6 MA De nition 533 polytabloid Lett be tableau a a kt sgnCt Ema sgnmr E FSymn b b et kti 260 sgmr e Mquot et is called a polytabloid c C S is the F subspace of MA spanned by the polytabloids SA is called a Specht module d 1 FA is the left ideal in FSymn generated by the kt t a tableau 325 14 The C Sym13 x Sym24 kt 11L1 7 13 7 24 1324 and i 325 125 345 145 14 34 12 32 As a rst example consider t Section 53 The Specht Module 109 As a second example consider n 7 11 and t Then Cl Symij 17013 k4 1i7j and ForielnputziIni 12quot39271z1quot39n Then MW U is the IF space with basis xi7i E In and et xi 7 mi Thus Sltn71 1Fltmjimilifj lngt2fiilft6FlZfi0ml2HWL i1 11 The reader should convince herself that if charlF l n then SW44 is a simple lFSymn module and if charlF l n then x ELI xi 6 SW44 and SWANFm is a simple lFSymn module Lemma 534 transitive 0n polytabloids Let 7r 6 Symn andt a tableau a lzl Wkt39n39il km 17 a wet e c b Symn acts transitively on the set of A polytabloz39ds d C SA is a FSymn submodule of MA e d If 7139 6 Ct then km kt sgnwkt and em sgnn39et Proof a We have Cm 7rCt7r 1 and so by 532d applied to the class function sgn on Symn7 km sgncm sgnwctrl 39n39sgnogr 1 wk ril b USing b7 57 th Wktwilwi Wkti 7H5t c and d follow from e Since 7139 6 Ct Cm Ct Ct39n39 Thus kt km and kt Zagct Sgna 04 Z ect sgnww 57quot sgn7r Z ecn sgn 3 sgn7rk 7r The second statement follows from the rst and mi u D 110 Chapter 5 Representations of the Symmetric Groups Lemma 535 action of es 011 ml Let A and ILL be partitions of n a a If WMA 7g 0 then A g p b b ft and s are A tableau with hst 7 0 then then hi ies Proof Let s be a ILL tableau and t and A tableau with list 7 0 Suppose rst that there exists ai 7 j E In such that i and j are on the same row oft and in the same column of 3 Let H Symij 17 Then sgnHi i sgniji7ji i E 0 Since ij are in the same column of s H g O9 and we can choose a transversal T to H in Cs Then list sgnTsgnHt O7 contrary to our assumption Thus no such ij exists So by 5215 A 1 ILL Moreover7 if A Iii there exists a A tableau T which is row equivalent to t an columns equivalent to 3 Hence k7 kg and f 5 Moreover 7139s r for some 7139 6 CS and so by 534e7 list e7 sgn7reS D Lemma 536 es self dual Let A and ILL be partitions of n and s an u tableau Then a lal ks kg 5 bl kSMAL AM ks39 c C hth Fes and AMnUss esi d 1 hi U l eses for allv 6 Mt Proof a If 7139 6 CS then also 7r 1 6 Cs Moreover sgn7r sgnnquot1 and a holds b Follows from a and 4117 0 By 535 55MA F55 and so by b AMMS a 1 By c ksu fes for some 1 E F Hence UlEsUlksEksU lif5t lif D Lemma 537 and ml FAMquot sA and AMAFA SM Section 53 The Specht Module 111 Proof This follows immediately from 536b and 536c D Lemma 538 submodules of ml Supp F is a eld and let A be a partition ofn and V be an FSymn submodule of MA Then either FAV SW and 539 g V orFAV 0 and SA V Proof If FAV 0 then by 537 V SM So suppose FAV 7 0 Then ksV 7 0 for some A tableau 3 So 536 implies ksV FeS ksMquot Since by 5345 implies ksV ksMquot for all A tableaux 8 Thus FAV FAMquot SA and SA V D If F K is a eld extensions we View MA M has a subset of SH Note also that Mfg is canonically isomorphic to 1K 1F MA Put DA SASA SAL Lemma 539 dl dl Let A be a partition ofn IfF is a eld then FADquot Dquot Proof By 538 either FASA S or SA SAL In the rst case FADA DA and in the second DA 0 and again FADA DA Proposition 5310 dldu Let A and a be partitions of n with DA 0 Suppose F is a eld If DA is isomorphic to an FSymn section of Mf then A 1 la In particular Dquot a D then A n Proof By 539 FADquot Dquot 7g 0 Hence also FADt 7g 0 and FAMM 7g 0 So by 5355 A 1 a If DA E D 7 the D is a section of MA and so a 1 A and la A D Lemma 5311 scalar extensions of ml Let A be a partition of n and F K a eld extension a a Si KSquot K FS b b S gi iqSM gnome SM e d S gms ginsms n y SAmSM d n 1an Dquot Proof a is obvious b follows from a and 411900 a follows from a7 b and 4119a d follows from a and D Lemma 5312 11 absolutely simple Let A be a partition of n and suppose DA 7 0 Then DA is an absolutely simple lFSymn module 112 Chapter 5 Representations of the Symmetric Groups Proof By 5311d it suffices to show that DA is simple So let V be an lFSymn submodule of SA with SA SAL V By 538 either SA V or V SAL In the rst case V sA and in the second V sA m SM and V s o SM Thus Dquot sAsA o SM is simple D 54 Standard basis for the Specht module Proposition 541 garnir relations Lett be a A tableau i lt j E Z1 X Q A ti and Y Q Atj Let T be any transversal to SymX X SymY in SymX U Y a a sgnT 5 is independent from the choice of the tmnuersal T b b If lXUYl gt Ag Then sgnTet 0 Proof a Let 7r 6 SymX U Y and p E SymX x SymY 0 Then 534 sgn7rp 7rp 61 sgn7r7r SewOW 5 and so a holds b Since lX O Y gt A 2 Xi there exists i E X andj in Y such that i and j are in the same row of t So 1 7 0 If 7139 E SymX U Y7 then 7139 and 7139 lie in differen cosets of SymX x SymY Hence we can choose R Q SymX U Y such that R O R i7j Q and R U R is a transversal to SymX U SymY By a we may assume T R U R and so sgn7r7ret sgnT sgnngn1Zj sgnR 17 sgnTet sgnR 17 ijet O D De nition 542 de garnir Lett be a A tableau i lt j E Z1 X Q A t and Y Q A t a a TXY is the set of all 7139 E SymX U Sme such that the restrictions of 7139 ot to 7r 1X and 71quot Y are increasing b b GXYt sgnTxy GXYt is called a Garnir element in FSymn Lemma 543 basic garnir Lett be a A tableau i lt j E Z1 X Q A ti andY Q A tj a a TXY is a tmnsusersal to SymX X SymY in SymX U Y Section 54 Standard basis for the Specht module 113 b b If XUY gt A Then GXYtet 0 Proof a Just observe that if 7139 E SymX U SyinY7 then there exists a unique element p E SymXUSymY such that the restriction of 7rp to t 1X and to t 1Y are increasing b follows from a and 541b D Consider it 5 A 3 2 t X 2 5Y 3 Then GXyet 0 gives De nition 544 defincreasing tableau Let A be a partion ofn andt a A tableau a a rt r o t 1 and C s o til So i E In lies in row rti and column cti oft b b We say thatt is row increasing 0 is increasing on each row Allt oft c C We say thatt is column increasing if rt is increasing on column A t Note that rt only depends on i and so we will also write ri for rt lndeed E 3 iff Tt T5 Lemma 545 basic increasing Let A be a partion ofn andt a A tableau a a t contains a unique row increasing tableau b b t contains a unique column increasing tableau c C Let 7r 6 Symn andi E I Then rti rm7ri Proof a and b are readily ver ed Cr7rto7rro7rot o7rrot 73 D De nition 546 de standart tableau Let A be a partition ofn andt a A tableau A standard tableau is row and column increasing tableau A tabloid is called standard if it contains a standard tableau ft is a standard tableau then 5 is called standard polytabloid By 545a7 a standard tabloid contains a unique standard tableau We will show that the standard polytabloids form a basis of SA for any ring F For this we need to introduce a total order on the tabloids De nition 547 defzorder tabloids Lett and 3 be the distinct A tabloids Leti E In be maximal with rii 74 732 Then i lt E provided that r i lt r i 114 Chapter 5 Representations of the Symmetric Groups Lemma 548 basic order tabloids lt is a total ordering on the set of tabloids Proof Any tabloid t is uniquely determined by the tuple rii1 Moreover the ordering is just a lexiographic ordering in terms of it associated tuple D Lemma 549 proving maximal I Let A and B be totally ordered sets amd f A a B be a function Suppose A is nite and 7139 E SymA with f 74 1 Opt Let a E A be maximal with fa 74 f7ra Iff is non decreasing then fa gt f7ra and is non increasing then fa lt f7ra Proof Reversing the ordering on F if necessary we may assume that f is non decreasing Let J j E J fj gt fa and let j E J Since 1 is non decreasing7 j gt a and so by maximality of f f7rj fj gt fa Hence 7rJ Q J Since J is nite this implies 7rJ J andso since 7139 is 1717 7rIJ Q IJ Thus 7ra J7 f7ra at and since fWWDt JWWDlt a The above lemma is false if I is not nite even if there exists a maximal at De ne f 2 a 01 by 0 ifi 0 and 1 otherwise De ne 7r ZZJ a i 1 Then 1 is non decreasing and a 0 is the unique element with at 7 f7ra But 0lt1 i Allthough the lemma stays true if there exists a maximal a and f is increasing de creasing lndeed in thus case J 017139 and so 7rI J I J Lemma 5410 proving maximal Lett be a x tableau and X Q In a a Suppose that rt is non decreasing on X Then E g t for all 7139 E SymX b b Suppose that rt is non increasing on X Then E gt t for all 7139 E SymX Proof a Suppose that E 7 i Let i be maximal in In with rti 7 rmi Note that rmi rt7r 1i Since rt is non decreasing 549 gives rti lt rt7r 1i rmi Thus i lt it b Similar to a D Lemma 5411 maximal in et Lett be column increasing tableau Thent is the mar imal tabloid inuolued in et Proof Any tabloid involved in et is of the form E with 7139 6 0 Since rt is increasing on each column7 we can apply 5410 to the restriction of 7139 to each of the columns So the result holds D Section 54 Standard basis for the Specht module 115 Lemma 5412 linear independent and order LetlF be ring V a uector space with a totally ordered basis 3 and E a subset of V Let b E B and U E V We say that b is inuolued in U if the b coordinate of U is non zero Let by be maximal element of V inuolued in U Suppose that the bhl E E are pair wise distinct and the coe icient fl of bl inl is not a left zero diUisor a a E is linearly independent b b Suppose in addition that each fhl E E is a unit and E is nite PutC bl 1 l E E and D B C a a E U is an R basis for M b b Suppose R is commutatiue and 1 be the unique R bilinar form on M with orthormal basis 3 Then a a For each d E 7 there exists a unique ed 6 d 1 BC with ed 6 EL b b ed 1 d 6 1D is an R basis for EL 0 a nil Br Proof a Let 0 71 fl 6 L F Choose l E E with bl maximal with respect to fl 71 0 Then bl gt bk for l 71 k E E with fk 71 0 So bl is involved in fll7 but in not other fkk Thus Zle fll 71 0 and E is linearly independent 10 We assume without loss that fl 1 for all l E E 10a Let m 21765 mbb E M We need to show that m 6 RH U E If mb 0 for all b 6 35 this is obvious Otherwise pick b 6 35 maximal with mb 71 0 and let l E E with b bl Then by induction on b7 m 7 mbl 6 RH U E 1010 We will rst show that ROCOELO Let 0 71 m Zle mlbl and choosel with ml 71 0 and bl minimal Then m 1 l ml 71 0 and m EL 1010a This is just the Gram Schmidt process For completeness here are the details Let E l17l2ln and bi bli with b1 lt b2 lt Put e0 d and suppose inductively that we have found el 6 d 1 Rbl 1 1 Rei with el T lj for all 1 g j ei If i lt it put ei1 el 7 el 1 li1bl1 Then eHl 1 li1 1 and since bi T lj for all j g i Put ed en By quot7 ed is unique 101010 Clearly ed 1 d E D is R linearly independent Moreover if m ZbewB mbb 6 cl then m m 7 Edd mded 6 RC o ri So implies m 0 and 101010 holds 10100 m 2176M mbb 6 ch By 10a there exists m 6 M with m m 6 RD and so we may assume that me 0 for all c E C Then 0 m 1 ed md for all at E D and so m 0


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