GROUPSandREPS LW 1139 Pmath745notes

8/11/2019 GROUPSandREPS LW 1139 Pmath745notes

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mlbaker.org presents

PMATH 445/745Groups and Representations

Dr. David McKinnonFall 2013 (1139)University of Waterloowww.math.uwaterloo.ca/dmckinno

Disclaimer: These notes are providedas-is, and may be incomplete or contain errors.

Contents

1 Representations 21.1 Direct sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Decomposition of representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3 Tensor products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2 Characters 62.1 Class functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.2 Symmetric groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.3 Almost abelian groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.4 Induced representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

3 Modules 143.1 C[G] and its modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153.2 Tensor products of modules over commutative rings . . . . . . . . . . . . . . . . . . . . . . . . . . . 173.3 Integrality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193.4 Frobenius divisibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203.5 Frobenius reciprocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213.6 Mackey . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

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1 Representations

1.1 Definition. Let Gbe a group. ArepresentationofGin a vector space Vis a homomorphism: G GL(V).Vis always taken to be a finite-dimensional C-vector space. There are such things as infinite-dimensional represen-tations; theyre not that useful since they dont end up being matrices. WhyC? Because other fields are harder.In practice, it will happen a lot that the matrices entries lie in R or Q.

Here, GL(V) is the group of invertible linear transformations V V . IfV = Cn, this is just the group ofn nmatrices. We want invertible matrices, because group elements are invertible. There is always a finite-dimensionalsubspace ofV that captures all the information of, so this restriction is usually superfluous.

1.2 Example.The trivial homomorphism G GL(C1) is called the trivial (or unit)representationofG.1.3 Example. Let G = D4, the group of symmetries of the square. There are 8 elements in this group. Note thatrotations and reflections are linear maps. Hence there is a representation ofG in C2 corresponding to its descriptionas symmetries of a square.

1.4 Example. Let G = Sn, and V = Cn. The permutation representation ofSn is the homomorphism that

takes a permutation to the corresponding permutation matrix.

The last two examples above are injective homomorphisms. We have a special name 1 for these.

1.5 Definition. A representation is called faithful iff it is injective.

1.6 Example. Let Gbe a finite group acting on a finite set X. There is a permutation representation correspondingto this action, on the vector space

V = xX

Cvx.

1.7 Example. Every group G acts on itself by left multiplication. IfG is finite, the corresponding permutationrepresentation is called the left regular representation.

It turns out that this representation, for a finite group, is the most important representation of them all. Basically,all the representation-theoretic information about G can be obtained from this one.

Sometimes, emphasis is placed on the vector space Vrather than the homomorphism . Hence a representationcan also be thought of as a vector space with an extra piece of information (the action ofG).

1.8 Definition. Let : G GL(V)be a representation ofG. A subspaceW V isG-stable, orG-invariant, ora subrepresentation ofV iff(g)( w) W for all g G, w W. In other words, iff(g)(W) W for all g G.In other other words, iff (g)

|W

GL(W) for all g

G. In other other other words, iff (G)

|W

GL(W).

1.9 Example.Let G = Z2, V = C2, and : G GL2(C) be given by

(0) =

1 00 1

, (1) =

0 11 0

.

Let W= span{(1, 1)}. Then W is G-invariant. Also,{0} and V are (always) G-invariant. W = span{(1, 1)} isalsoG-invariant.

1.10 Fact. It is easy to see that to check the G-invariance of W, it suffices to check that (gi)( wj) W forgeneratorsgi ofG and a basis{wj} ofW.1.11 Definition. Let : G GL(V)and :G GL(W)be representations ofG. Amorphism(orintertwiner)from to(or fromV toW) is a linear transformation T :V Wsuch that for allg G, the following naturalitysquare commutes:

V

(g

) VT TW

(g) Wthat is, T((g)(v)) = (g)(T(v)) for all g G, v V . In other words,T commutes with the G-actions on V andW.

1Since a group G can be thought of as a (locally small!) category in which every morphism is invertible (i.e. a groupoid) with onlyone object, and representations can be thought of as functors from G to the category of vector spaces over some field, the use of thisterminology is unsurprising (although historically, the terminology for representations may have preceded the terminology for functors).If you know what Im talking about, then by now you can probably guess what the morphisms between representations will turn out tobe.

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1.25 Definition. A morphismT from to is an isomorphismiff there is a morphism T from to such thatT T= id and T T= id.Proof. By a simple induction it suffices to prove that ifV is not irreducible then there are two proper subrepresen-tationsW, W such that V=W W. Thus, assume V is reducible. Then there is a proper, nontrivial subrepre-sentation W V. LetW= W. We just need to show thatW= W is G-invariant. Thus, choose any w W,and anyg G. (From now on, I will omit from the notation). We want to show that g( w) W. To do this, wemust take anyv Wand show that v, g( w) = 0. Since is unitary, g1 is unitary, so v, g( w) = g1(v), w = 0.This means that g( w) W so W is G-invariant, so W, W are both representations and V= W W, asdesired.

1.26 Theorem. Let G be a finite group, : G GL(V) be a finite dimensional representation. Then there issome complex inner product, on V such that(g) is unitary for all g G.Proof. Let, dumb be any old inner product on V . Define a new pairing on V :

u, vawesome=gG

g(u), g(v)dumb

The pairing easily satisfiesu1+ u2, va = u1, va+ u2, va andcu, va = cu, va andu, va = v, ua. Ifu= vthenu, ua R0 because each summand is, and so the sum is 0 iffg(u) = 0, which happens iffu= 0, as desired.Now(g)is unitary with respect to, a for all g G:

g(u), g(v)a =hG

h(g(u)), h(g(v))d =hG

(hg)(u), (hg)(v)d =hG

h(u), h(v)d = u, va.

Note that these two theorems immediately imply that every finite dimensional representation of a finite group G isisomorphic to a direct sum of irreducible representations.

1.27 Theorem.Let : G GL(V) and : G GL(W) be irreducible representations of a group G. LetT :V Wbe a morphism. Then Tis either an isomorphism or 0.Proof. The kernel ofT is a subrepresentation ofV , which by the irreducibility of must be either V or0. Ifker Tis V , then T = 0. Ifker T = 0, then Tis one-to-one, so since Im T = 0 or W, we have either T = 0 or T onto. SoeitherT = 0 or T is an isomorphism.

1.28 Theorem (Schurs Lemma).Let T :V Vbe a morphism of irreducible representations. Then T = idfor some scalar C.

Proof. Let be an eigenvalue ofT. Then T Iis a morphism from V toV (easy to check). Since T I is notan isomorphism, it must be 0. SoT =I.1.29 Theorem.Let G be a finite abelian group, : G GL(V)an irreducible representation. Then dim V = 1.Proof. It turns out (check it!) that ifG is abelian, then for everyg G, (g) : V Vis a morphism, so by Schur(g) =gI for some g C. But this means every subspace is G-invariant, so since is irreducible, V cant haveany nontrivial subspaces and so dim V = 1.

In fact, G is abelian iff all its irreducible representations are one-dimensional.

1.3 Tensor products

1.30 Definition. Let V andWbe complex vector spaces. LetHbe the vector space whose basis is

{v

w

|v

V, w W}. H is very large.Define a subspace R ofHto be the span of all vectors in Hof the following forms:

v1 w + v2 w (v1+ v2) w v w1+ v w2 v ( w1+w2) (v) w (v w) v ( w) (v w)DefineV W=H/R.1.31 Example. IfV =W ={0} then V W ={0} because H= span{0 0}, and (0 0) = (0) 0 =0 0for all .

1.32 Example. V ={0}, W any vector space. ThenV W ={0} because for any w W, 0 w+ 0 w =(0 + 0) w= 0 w, so0 w= 0.

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1.33 Example.Say V= span{v}, W= span{w}, withv, w nonzero. Then V Wis spanned by elements of theform(v) ( w). But(v) ( w) = (v w). SoV Wis spanned by{v w}. To see thatv w is not zero,we use a very useful trick.

Letq: H V Wbe the reduction mod R linear transformation. Thenqis onto and His nonzero, so it sufficesto show thatq is not identically 0. Define T :H Cby T( ai(iv) (i w)) = aiii which is clearly a lineartransformation. The image ofT is C, becauseT(v w) = 1. SoT is onto. Moreover,R ker T, so by the UniversalProperty of Quotients (UPQ), T :V W C is well-defined.1.34 Theorem (UPQ).Let Ube a vector space, K Uany subspace, q: U U/Kthe reduce mod K lineartransformation. Let T : U

V be a linear transformation. Then there is a linear transformationT : U /K VsatisfyingT =T q iffK ker T.

U V

U/K

T

q T

Also,Im T = ImT andkerT=q(ker T).In particular, T :V W C is well defined and onto, so dim V W = 1.1.35 Theorem. Let V andWbe finite-dimensional vector spaces. Let

{v1, . . . , vn

}and

{w1, . . . , wm

}be bases for

V andW, respectively. Then{vi wj} is a basis ofV W. In particular, dim(V W) = (dim V)(dim W).Proof. Span: Let

k xk yk be an arbitrary element ofV W. Then

xk yk =

k

i

aikvi

j

bjk wj

= aikbjk(vi wj) span{vi wj}.Linear independence: define a linear transformation T from H Cnm by

T

ck

aikvi

bjk wj

=

ckaikbjkeij

where {eij} is the standard unit basis vector in Cnm, Cnm being viewed asn mmatrices, andeij the matrix withall zeroes except for a one in the(i, j) entry. Since T(vi

wj) = eij, Tis onto. It is easy to check that R

ker T.

For example,

T(v1 w+ v2 w (v1+ v2) w) = T

aivi

bj wj

+

aivi

bj wj

(ai+ ai)vi

bj wj

=

aibjeij+

aibjeij

(ai+ ai)bjeij

= 0.

By the UPQ, T :H/R Cnm is well defined soT :V W Cnm is well defined and onto. Therefore, since{eij}is linearly independent in Cnm, their preimages{vi wj} are linearly independent in V W.

SayT :U V, S: W Xare linear transformations. We can define T S: U W V Xby(T S)

ui wi

=

T(ui) S( wi).

1.36 Example. M1 =

a bc d

, M2 =

e fg h

, inM2(C). Then the matrix ofM1 M2 with respect to the basis

{ei ej} ofC2 C2 is aM2 bM2cM2 dM2

=

ae af be bf ag ah bg bhce cf de df cg ch dg dh

.

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1.37 Definition. Let : GGL(V) and : GGL(W) be representations. Then : GGL(V W) isgiven by

[( )(g)]

vi wi

=

[(g)](vi) [(g)]( wi).1.38 Example. Let G = S3, : GGL(C) trivial, : G GL(C) the sign representation. Then : GGL(CC) =GL(C)is ( )(g) = (g)(g) = (g).1.39 Example. If : G GL(C) is one-dimensional and :G GL(V) is any representation, then = 1,where1(g) = (g)(g).

1.40 Question. If, are irreducible, is

irreducible?

Answer. No. See Assignment 2.

1.41 Question. What if : G GL(V)is irreducible? Is irreducible then?Answer. Never, ifdim V 2.1.42 Theorem.Let : G GL(V) be a representation with dim V 2. Then is reducible.Proof. Define : H V V (where His the free vector space on V V) by( aivi wi) = ai wi vi. ThenR ker (easy check) so : V V V V is well-defined. LetSym2 Vbe the 1-eigenspace of, Alt2 V be the(1)-eigenspace of. We claim thatSym2 V and Alt2 V are G-invariant subspaces ofV V. Note that Sym2 Vand Alt2 V have zero intersection and Sym2 V+ Alt2 V = V V, so V V=Sym2 V Alt2 V as vector spaces.So if we show the claim, well be done. Say

ai(vi wi) Sym2 V. Then

((

)(g))( ai(vi wi)) = ( ai([(g)](vi) [(g)](wi)))=

ai([(g)](vi) [(g)](wi))=

ai[(g)](wi) [(g)](vi)= [( )(g)](

aiwi vi)

= [( )(g)]((

aivi wi))= [( )(g)](

aivi wi)

So [( )(g)]( ai(vi wi)) Sym2 V. So Sym2 V isG-invariant. (Alt2 V is similar).Note: ifdim V =n, then dim V V =n2, and dimSym2 V =n +

n2 =

n(n+1)2 anddimAlt

2 V = n2 =

n(n1)2 .

2 Characters

2.1 Definition. Let : G GL(V) be a representation. The characterof is the function : G Cgiven by

(g) = Tr((g)).

Ifdim = 1, then = . If is irreducible, then p is called an irreducible character.

If =, then = . This is because trace is invariant under linear isomorphisms.(1) = dim V and(g1hg) = (h).

AssumeG is finite, dim V is finite. Then: =+ (because the matrices have and as blocks) =. (g1)is the sum of the eigenvalues of(g1), which is the sum of the reciprocals of the eigenvalues of(g),

which is the sum of the complex conjugates of the eigenvalues of(g). Since the eigenvalues are roots of unity.This is equal to (g).

2.2 Theorem. Let , be representations ofG in V , W respectively. LetT :V Wbe any linear transformation.ThenT =

g(g

1) T (g)is a morphism . In particular, if and are irreducible and not isomorphic,thenT= 0.

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Proof. On p. 20 of notes, or as Cor. 1 in section 2.2 of the textbook.

2.3 Definition. C[G] = {f :G C} is the complex group ring (or group algebra).C[G] has a natural inner product

, = 1|G|gG

(g)(g).

This corresponds to the standard inner product on Cn (divided by|G|) by

f (f(g1), . . . , f (gn)).

2.4 Theorem. Irreducible characters are an orthonormal subset ofC[G].

Proof. We show that if, are irreducible representations, then

, =

1 if =0 if =.

: G GL(V), :G GL(W). Choose bases for V , W. For each g G, write

(g) = (rij(g)), (g) = (tij(g))

as matrices. Then

,

=

1

|G| g,i,j rii(g)tjj (g) = 1

|G| g,i,j rii(g)tjj (g1)LetT :V W be any linear transformation. Then

T =g

(g1) T (g)

is a morphism . If =, then T= 0, so ifT = (Tij):g,i,j

tki(g1)Tijrj(g) = 0

for any k, . Set Tij = 0 except Tk = 1. Then

g

tkk(g1)r(g) = 0

so summing over k gives, = 0. If= , then = , so we assume = . If = , then T = I bytheorem, and

=|G| (Tr T)

dim V .

Exactly the same argument as before gives, = 1.This means there are only finitely many irreducible representations ofG, up to isomorphism, because the corre-sponding characters are an orthonormal set in a finite dimensional vector space.

Better: If = (m11) . . . (mrr) for mi Z and i pairwise nonisomorphic irreducible representations thenmi=

, i

.

Even better, irreducible decompositions of representations are unique up to isomorphism.

Even more better, non-isomorphic representations have different characters, because the character determines theirreducible decomposition.

Even more more better, with = (m11) . . . (mrr)as above, =m21+ . . . + m2r, so, = 1iff isirreducible.

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2.5 Example.Representation ofS3 in C2:

1

1 00 1

2

(12)

0 11 0

0

(13) 1 0

1 1

0

(23) 1 10 1 0

(123)

0 11 1

1

(321) 1 1

1 0

1

, = 16 (22 + 02 + 02 + 02 + (1)2 + (1)2) = 1so is irreducible.

Begin shameless copying from Ehsaans notes.

2.6 Example. Let : S3 GL(C3)be the permutation representation. Whats the character of? Well, ()isthe number of fixed points:

() =

3 if = 11 if is a 2-cycle

0 if is a 3-cycle.

Therefore, = 2, meaning that is not irreducible. Actually, since, is the sum of the squaresof thedimensions of the irreducible subrepresentations, the only way we could have, = 2is if

= (2-dimensional) (1-dimensional).

Let be the representation from the above example. Then

, =16

(3 2 + 1 0 + 1 0 + 1 0 + 0 (1) + 0 (1)) = 1

implying that is (isomorphic to) a summand of. The other summand has to be 1-dimensional. But note thatC(1, 1, 1)is a 1-dimensional invariant subspace on which acts trivially, therefore

= (trivial).

2.7 Example.Let G be a finite group ofn elements g1, . . . , gn. LetV :=

i Cgi be the vector space with G as abasis. ThenG has a representation on V:

(g)gi := ggi.

This is called the left regular representationofG. Just like in the above example, the character is obtainedby nothing that (g) is the number of fixed points of(g). But in this case, ifg= 1 then g cant have any fixedpoints!

(g) = |G| ifg = 10 otherwise.

Therefore,, = |G|, so is not irreducible unless G = {1}. If is any other representation ofG, then

, = 1|G| (|G| dim() + 0 + . . . + 0) = dim()

(because(1) = dim()). So, if =d11 . . . drr is a sum of irreducible representations i anddi:= dim(i),then

|G| = , =d21+ . . . + d2r.Perhaps a useful fact which follows from the above: ifis an irreducible representation ofG, thendim()

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2.1 Class functions

Recall that characters are constant on conjugacy classes (some proposition above). In general, a function f :G Cis called a class function on G if it is constant on conjugacy classes; that is, f(ghg1) =h for all g , hG. Theset of class functions onG is a subspace ofC[G].

2.8 Proposition.The irreducible characters form an orthonormal basis for the space of class functions on G.

In particular, the number of irreducible representations ofG is equal to the number of distinct conjugacy classes ofG.

Proof. Let V be the space of class functions on G and let W be the span of the irreducible characters ofG (inC[G]), so W V. We already know that irreducible characters are orthonormal, so it just needs to be shown thatW V = 0.

End shameless copying from Ehsaans notes.

Proof. We will show that ifWis the span of the irreducible characters then WV = {0}. This will implyW =V(sinceW V). Note that W =Wand thus W = W, so f W ifff W. So say f W. Then f W,so, f = 0for any (irreducible) character. Define

Tf(v) =g

f(g)[(g)](v) for any representation ofG.

( : G GL(V), Tf : V V .) It is straightforward to check thatT

f is a morphism . If is irreducible,

thenTf =Ifor some C. But 0 = , f =

g(g)f(g) =

gf(g) Tr((g)) = Tr(Tf), so = 0. SoT

f = 0

for any irreducible representation , and so by linearity, Tf = 0 for any representation .

Let be the left regular representation, v = gi:

0 = Tf(v) =g

f(g)[(g)](v) =g

f(g)ggi

so since{ggi} = {g} is linearly independent, it follows that f(g) = 0for all g G. This means W V = {0}, asdesired, so W =Vand the irreducible characters span the space of class functions.

LetG = D4. Well write down a list of all irreducible characters ofG. This is called acharacter table for G.

D4 = {x, y| yx = x1y, x4 =y2 = 1} = {1, x , x2, x3,y,xy,x2y, x3y}.We can think of x as a 90 degree clockwise rotation, and y as reflection in the vertical axis of symmetry. Theconjugacy classes are:{1},{x2},{x, x3},{xy,x3y},{y, x2y}.Aside: x1yx = x2y andy1xy= yxy= x1.

The trivial character is irreducible, so we can start the table:

{1} {x2} {x, x3} {xy,x3y} {y, x2y}triv 1 1 1 1 1sgn 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1

square 2 2 0 0 0Have a homomorphism : D4 Z2, (xayb) = bmod 2. This gives a sign representation on D4 by (g) =(1)(g). Have a homomorphism : D4 (Z2)2, (xayb) = (amod 2, bmod 2). This gives two more irreduciblecharacters:

1(xayb) = (1)a mod 2 2(xayb) = (1)a+b mod 2.

There are lots of ways to find the last character ofD4. The easiest is to use the fact that, with the other 4 chars,it is an orthonormal basis for the space of class functions. Or you can just guess it; its the realisation ofD4 as thesymmetries of a square in C2.

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2.9 Example.Compute a character table for A5.

[STUFF]

Need 5 representations; have permutation representation in C5: perm : (5, 2, 1, 0, 0).

perm, perm = (52 1 + 22 20 + 12 15 + 02 12 + 02 12)/60 = 2.

Does perm have a subrepresentation isomorphic to triv?

perm, triv = ((5 1) 1 + (2 1) 20 + (1 1) 15 + (0 1) 12)/60 = 1

so yes! Definept = perm triv. Its irreducible. Need 3 more! TrySym2(pt)and Alt2(pt). We need a formulafor the chars, though.

2.10 Theorem. Let : G GL(V) be a representation of a finite group. Lets and a be Sym2() and Alt2()respectively. Let , s, a be chars for , s, a respectively. Then

s(g) =1

2((g)2 + (g2)) a(g) =

1

2((g)2 (g2))

Proof. SinceG is finite (or since is unitary), (g) is diagonalizable. Let{e1, . . . , en} be an eigenbasis for V, withcorresponding eigenvalues1, . . . , n. Then(g) =

i, and (g

2) =

2i .

Now,

[s(g)](ei ei) = [(g)](ei) [(g)](ei) = 2

i (ei ei), [s(g)] = (ei ej+ ej ei) = ij(ei ej + ej ei).So

s(g) =

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2.2 Symmetric groups

Question: What are all the irreducible representations ofSn? The conjugacy classes ofSn are the sets of permuta-tions with the same cycle structure. Conjugacy classes in Sn are in 1-1 correspondence with Young tableaux. TheYoung tableau associated to a conjugacy class (r1, r2, . . . , rk) [r1+ . . . + rk = n, r1 r2 . . . rk] is[DIAGRAM]

[EXAMPLES]

2.11 Definition. A numbering of a Young tableau is an injective function from {1, . . . , n} to the boxes in thetableau. Sn acts on the set of numberings of a fixed tableau.2.12 Definition. A tabloid is an equivalence class of numberings (of some fixed tableau), where two numberingsare equivalent iff for each row, both numberings contain the same set of numbers (albeit possibly in a differentorder).

Sn also acts on the set of tabloids of a given fixed shape. Lets be a fixed shape (Young tableau). LetMs be

TCT, where Tranges over all tabloids of shape s. There is a permutation representation ofSn in Ms coming

from the Sn-action on the tabloids.

LetTbe a numbering of a tableau, and let C(T)be subgroup ofSn of elements Sn that preserve the columnsofT, i.e. m and (m)are in the same column for all m. DefineR(T)to be the same kind of thing, only with rows.

Let[T] be the tabloid associated to T. Define

vT

= C(T)(1)sgn()[(T)]It is easy to check that vTdoes not depend on T, but only on [T]. Let S

s = spanT{vT}. We will show thatSs isan Sn-invariant subspace ofM

s. For any Sn:

(vT) =

(1)sgn()[(T)]

=

(1)sgn()[ (T)] =

C((T))(1)sgn()[ 1(T)]

because C((T)) = C(T)1

=

C(T)(1)sgn()[(T)] = v(T).

SoSn permutes thevT, meaning thatSs isSn-invariant. S

s is called the Specht module associated to the shape s.

Plan: We will show that Ss is irreducible, and that Ss =St ifs=t. This will imply that{Ss} is a complete listof irreducible representations ofSn up to isomorphism.

2.13 Definition. Let s and t be Young tableaux of the same size (same number of boxes). Writes = (s1, . . . , sk),t= (t1, . . . , tk) (setsi orti to 0, if necessary, to give t and s the same number of rows). We say that s dominatestiff for all m,

s1+ s2+ . . . + sm t1+ . . . + tm.We say that s strictly dominates t iffs dominates t and s =t.[EXAMPLES]

2.14 Theorem. Let T, T be numberings of shapes s and s of the same size, with s not strictly dominating s.Then either:

1. There are 2 different numbers in the same row ofT and the same column ofT, or2. s= s and there is some p R(T), q C(T)such that p(T) = q(T).

Proof. Assume (1) is not true. We will show that (2) holds. We want to findp R(T) and q C(T) such thatp(T) =q(T). Choose q1 C(T) so that all the numbers in the first row ofT are in the first row ofq1(T) whichis possible because all those numbers are in different columns ofTby (not (1)). Choose q2 C(T) to fix first rownumbers from T and permute second row numbers from T to the second row (or higher) ofq2q1(T). Keep goinguntil you get a q= qk q1 C(T) such that for every i, all the numbers in the ith row ofT are in the ith row ofq(T), or higher.

This means thats dominatess. By hypothesis, this meanss = s, and each row ofT has the same set of numbersas the corresponding row ofq(T). So there is some p R(T)such that p(T) = q(T).

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Proof. Let : G GL(V)be an irreducible representation ofG. Let |A. LetW V be an A-invariant subspaceofV. Thendim W = 1. Let W= span{w}. Let W = span{(g)( w) : g G}. ThenW is G-invariant. But ifg1, g2 are in the same left coset xA ofA in G, then g1 = g2a for some a A, so g1( w) = (g2a)( w) = g2( w) forsome C, so the dimension ofW can be no greater than the number of left cosets ofA in G, which is|G|/|A|.SinceV is irreducible (and if we choose W= 0) then W = V and were done.Since Dn has an abelian (cyclic!) subgroup of order n, it follows that every irreducible representation ofDn hasdimension 1 or 2.

SayG1, G2 are finite groups. What are the irreducible representations ofG1 G2?

Let i : Gi GL(Vi) be two representations. Define : G1 G2 GL(V1 V2) by [(g1, g2)]( vj wj) =[1(g1)](vj) [2(g2)](wj), for any gi Gi, vj V1, wj V2. This is usually written = 1 2. It is easy tosee that is a representation.

It is also true that if1 and2 are irreducible, then is also irreducible. To see this, note that the character of satisfies(g1, g2) = 1(g1)2(g2), so:

, = 1|G1 G2|

(g1,g2)

(g1, g2)(g1, g2) = 1

|G1||G2|

(g1,g2)

1(g1)2(g2)1(g1)2(g2)

= 1

|G1|

g1

1(g1)1(g1)

1

|G2|

g2

2(g2)2(g2)

= 1 , 12 , 2 = 1.

We know that if Gi has ai conjugacy classes, then G1 G2 has a1a2 conjugacy classes, so G1 G2 has a1a2irreducible representations, up to isomorphism, and weve constructeda1a2 of them. To see that 1 2=1 2if(1, 2) =(1, 2)note that by a calculation similar to the one above,

, = 1 , 12 , 2

so ifi= i for some i, theni , i = 0, giving, = 0. ( = 1 2). Everything below this is up todate.

2.4 Induced representations

Let G be a finite group, H a subgroup. Let: H

GL(V) be a representation. We define = IndG

H

by... well,its complicated. Chooseg1, . . . , gn G so that g1H , . . . , gnHis a complete and distinct set of left cosets ofH inG. Define W=g1V . . . gnV, where giV is just V with a decorative gi prefixed.Define a hom :G GL(W)by

[(g)] (g1v1+ . . . + gnvn) = gj1 [(h1)](v1) + . . . + gjn [(hn)](vn)

where for eachr , g gr =gjrhr. A straightforward check shows that = IndGH is a representation ofG.

2.21 Example. Say G is any group, H is trivial, : H GL(C)trivial. Then IndGH is the left regular represen-tation. This is because H= {1} and trivial means V = C and g gr = gjr for all r.2.22 Example. Let G = H. Then IndGH= .

2.23 Example.Let G = Z4, H =

2. If : H

GL1(C) is trivial then = Ind

GH is the permutation

representation ofG acting on leftH-cosets by left multiplication.

2.24 Example.Take the same G and H, but let : H GL1(C) be the sign representation; (0) = 1 and(2) = 1. Let= IndGH. Note, in general, that dimIndGH= [G: H]dim , where [G: H] is the index ofH inG(the number of left cosets).

In our case,dim = 2. A basis forW=

giC, whereg1 = 0,g2 = 1, is {0(1), 1(1)}. This turns(n)into a matrix:

(0) =

1 00 1

.

[(1)](0(1)) = (1 + 0)(1) = 1(1).

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[(1)](1(1)) = (1 + 1)(1) = (0)[(2)](1) = (0)(1) = 0(1). So we have

(1) =

0 11 0

[(2)](0 (1)) = (2 + 0)(1) = 0([(2)](1)) = 0(1) = 0(1), so[(2)](1 (1)) = (2 + 1)(1) = 1([(2)](1)) = 1(1) = 1(1)

(2) =

1 00

1and now

(1) =

0 11 0

Notice that=1 2, where1 and2 are the 2 extension of to G.2.25 Theorem.If= IndGH, then

(g) =

g1

i ggiH(g

1i ggi) =

1

|H|

x1gxH(x

1gx)

Proof. Choose a basis {givj} for

giV. Pickg G. For eachi,ggi = gjhifor somehi Hand indexj . The traceof(g)is the sum of all the coefficients ofgivj in[(g)](givj). Ifgj=gi, then the coefficient ofgivj in[(g)](givj)is 0. But gi = gj iffhi = g

1

i ggi, so [(g)](givj) = gi[(g1

i ggi)](vj), thereby increasing the trace of (g) by thetrace(g1i ggi) of(g

1i ggi).

This establishes the first formula. The second follows immediately by group theory.

2.26 Example. Let G= D4,H= y whereyhas order 4 (a 90 degree clockwise rotation, say). Let : H GL(C)be (ya) = ia. What is IndGH? (Call it).

W= C xC, where x G is a reflection.y(xv) = (yx)(v) = (xy1)(v) = x(y1(v)) = x(iv) = ixv.

(1) =

1 00 1

, (y) =

i 00 i

, (y2) =

1 00 1

, (y1) =

i 00 i

(x) = 0 11 0 , (xy) = 0 ii 0 , (xy2) = 0 11 0 , (xy1) = 0 ii 0The character of is

(1, y , y2, y1,x,xy,xy2, xy1) = (2, 0, 2, 0, 0, 0, 0, 0).

This is the same answer that the theorem gives, which is nice.

3 Modules

3.1 Definition. Let G be a group. The complex group ring ofG is

C[G] = gGCg,where the multiplication is (

aigi)(

bjgj) =

i,jaibjgigj .

This C[G] is a ring, commutative iffG is abelian.

3.2 Definition. Let R be a ring (with 1, but not necessarily commutative). Aleft R-module is an abelian groupMwith an operation :R M Msatisfying:

1. r1(r2m) = (r1r2)m.

2. 1m= m.

3. (r1+ r2)m= r1m + r2m.

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4. r(m1+ m2) = rm1+ rm2.

For all r, r1, r2 R, m, m1, m2 M.Bad news: there are way too many R-modules.

IfR is a commutative field, then an R-module is the same thing as an R-vector space.

3.3 Example. Zn is a Z-module. There are plenty of linearly independent subsets ofn elements that do not span.

3.4 Example.Zn is a Z-module.

3.5 Example.Any abelian group is a Z-module and vice versa.

3.6 Example. Let R be any ring,I R any left ideal. Then Iis an R-module. In fact, a left ideal ofR is exactlya leftR-submodule ofR.

Say : G GL(V) is a representation ofG. We can make V into a left C[G]-module byaigi

(v) =

ai[(gi)](v)

In other words, to multiply v by

aigi, use to turn the gi into matrices (linear transformations V V) andthen plugv into the corresponding linear combination of those matrices.

Conversely, let V be a left C[G]-module. Then V is a C-vector space, and we can define:

[(g)](v) = gv

for any g G. It is not hard to check (use module axioms) that (g) is linear for all g G, and that is ahomomorphism from G to GL(V).

3.7 Example. Let G = Z2, and : G GL1(C)given by (1) = 1. Then C[G] = C 0 +C 1, and C is a moduleover C[G], via:

(a (0) + b (1))z = a([(0)](z)) + b([(1)](z)) = az bz= z(a b).Notice that(p(t)) = p(1) C[G] is a ring homomorphism from C[t] to C[G]. Note p(1) involves1 Z2 and not1 C. is onto, since (at +b) =b (0) +a (1). The kernel of is (t2 1). Since C[t]/(t2 1) is 2-dimensionalas a C-vector space, and since : C[t]/(t2 1) C[G] is onto, and since dimC[G] = 2 as a C-vector space, weconclude that is an isomorphism, and so in particular C[G] = C[t]/(t2 1).By the Chinese Remainder Theorem, C[t]/(t2 1) = CC, so C[G] = CC, as rings.3.8 Definition.Let R be a ring (with 1, not necessarily commutative) and let M, N be (left) R-modules. AnR-module homomorphismfrom M toNis a function f :M Nsuch that f(m1+ m2) = f(m1) + f(m2)andf(m) = f(m) for all m, m1, m2 M and R.Note that ifR = C[G] then an R-module homomorphism is exactly a morphism of representations.

An isomorphism ofR-modules is a homomorphism ofR-modules that has an inverse homomorphism.

3.1 C[G] and its modules

3.9 Definition. Let R be a ring, ManR-module. ThenEndR(M)is called the endomorphism ring ofM overR:

EndR(M) := {R-module homomorphisms f :M M}

with pointwise function addition and composition as the operations. (An endomorphismofMis a homomorphismM M).Note that Schurs lemma says that EndC[G] V= C ifV is an irreducible representation ofG.3.10 Definition. Let M, Nbe R-modules. ThenHomR(M, N) = {f :M N, f is an R-module homomorphism}.Note thatHomR(M, N) is anR-module via(f1+ f2)(m) = f1(m) + f2(m)and (rf)(m) = rf(m).

Recall also that ifV1=V2 with V1, V2 irreducible, then HomC[G](V1, V2) = 0.If M, N are R-modules, then M N is also an R-module, with (m1, n1) + (m2, n2) = (m1 + m2, n1 + n2) andr(m, n) = (rm,rn). With this definition, direct sum ofC[G]-modules corresponds to direct sum of representations.

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3.11 Definition. An R-moduleM is simple iff its only R-submodules are 0 and M.

Notice that simple C[G]-modules correspond precisely to irreducible representations ofG. Thus, every C[G]-modulethat is a finite dimensional C-vector space is isomorphic to a direct sum of simple C[G]-modules.

3.12 Definition. An R-module M is called semisimple iff for every submodule N M, there is a submoduleN M such thatM=N N.3.13 Theorem (Maschke).Every C[G]-module is semisimple ifG is finite.

Note we have already proven this if the C[G]-module is a finite dimensional C-vector space.

Proof. Let V be a C[G]-module, W

V any submodule. We want to find a submodule W

V such thatV=W W as C[G]-modules.Let f : V W be some C-linear projection (i.e. f( w) = w for all w W). If only f were a C[G]-modulehomomorphism, thenW = ker fwould win us the day. Sadly, fmay not be a C[G]-module homomorphism, so wehave more work to do.

Defineh : V W byh(v) = 1|G|

gg1f(gv). Then h(v) W because f(gv) W andW isG-invariant. Notice

also that h( w) = w for all w W, so h is a projection onto W. Now we can show that h is a C[G]-modulehomomorphism:

h

aigi

v

=

aih(giv) = 1

|G|g,i

aig1f(ggiv)

=

1

|G| g,i aigi(gi)1f(giv) whereg i= ggi=

aigi

h(v).

SoV=Wker has vector spaces, and Wandker hare both C[G]-modules. SoV=Wker has C[G]-modules.Our next goal is to understand the ring C[G] better.

As a left C[G]-module, C[G] corresponds to the left regular representation. So, as a C[G]-module, we have:

C[G] =n1V1 . . . nrVrwhere V1, . . . , V r are the simple C[G]-modules (corresponding to irreducible representations) with ni = dim Vi,wheredim Vi means dimension as a C-vector space.

3.14 Proposition. C[G] is isomorphic to End(C[G]) as a ring, if we make C[G] into a C[G]-module by left multi-plication.

Proof. Let : C[G] EndC[G] be given by

aigi

=

bigi

bigi

aig1i

.

It is easy to see that is well defined and C[G]-linear. Its also clearly 1-1, so we just need to check that isonto. Thus, let f EndC[G] be an arbitrary endomorphism. We want to show that f = ( aigi) for some

aigi C[G]. Let a = f(1) C[G]. Then for all b C[G], we have f(b) = bf(1) = ba, so we conclude thatf Im , as desired.As a left C[G]-module, C[G]= d1V1 . . . drVr, where V1, . . . , V r are the irreducible representations ofG, upto isomorphism, and di = dim Vi. Since Schurs Lemma implies thatHom(Vi, Vj) = 0 if i= j we see that anyendomorphism ofC[G] must be the sum of endomorphisms ofdiVi for each i.

3.15 Theorem. Let Vbe a simple C[G]-module (i.e. a module corresponding to an irreducible representation) andn any positive integer. Then End(nV) =Mn(C), where nV is the direct sum ofn copies ofV .Proof. Define : Mn(C) End(nV) by

(M) = {(v1, . . . , vn) M

v1...vn

nV}.

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So, for example, 1 23 4

v1v2

=

v1+ 2v2

3v1+ 4v2

It is easy to check that is a well defined ring homomorphism. It is also easy to see that the kernel of is zero,so is 1-1. It remains only to show that is onto. Let f End(nV) we want to show that f=(M) for somematrixM.

Write f= (f1, . . . , f n) where each fi is a C[G]-module homomorphism from nV toV . By restricting fi to the j thcoordinate, we get a C[G]-module homomorphism fij : VV . By Schurs Lemma,fij = aijI for some aij C.So

f(v1, . . . , vn) = (f1(v1, . . . ,vn), . . . , f n(v1, . . . , vn))

= (f11(v1) + f12(v2) + . . . + f1n(vn), . . . , f n1(v1) + . . . + fnn(vn))

= (a11v1+ . . . + a1nvn, . . . , an1v1+ . . . + annvn)

=M

v1...vn

whereM= (aij). So is onto, and therefore a bijection, and thus is an isomorphism of rings.

So C[G]= End(C[G])= End(d1V1 . . .drVr)= End(d1V1). . .End(drVr)= Md1(C). . .Mdr(C).What is the isomorphism? Let i : G GL(Vi) be the representation corresponding to Vi. Define : C[G]Md1(C) . . . Mdr(C) by

(

aigi) =

ai1(gi), . . . ,

air(gi)

=

ai(1(gi), . . . , r(gi))

It is clear that is a ring homomorphism. Its 1-1, because if(

aigi) = 0, then the representations i would belinearly dependent, meaning that their characters are linearly dependent too, but theyre orthonormal, so is 1-1.Since we already know that

C[G] =Md1(C) . . . Mdr(C)as C-vector spaces, we conclude that is also onto.

Notice that this means that the ith componenti : C[G] Mdi(C)of is onto. That is, every linear transformationf : Cdi Cdi can be realised as a linear combination of the matrices i(g) for g G. The centre ofC[G] isisomorphic to the centre ofMd1(C)

. . .

Mdr (C) = CI

. . .

CI

= C

. . .

C.

3.16 Proposition. As a subring ofC[G], the centre is a complex vector space of dimension r (the number ofconjugacy classes ofG).

Proof. Say(

aigi)g = g(

aigi) for all g G. This is the same as

aigig =

aiggi. Thus, if

aigi is in thecentre ofC[G], then for every g,i, j such thatg gi= gjg we must have ai= aj . Butggi = gjg g1gjg= gi soin order for

aigi to be in the centre ofC[G], we need ai = aj ifgi andgj are in the same conjugacy class.

Thus, every element of the centre ofC[G] is in the span of the elements

gCg for Ca conjugacy class ofG. Thisspan has dimension r , so it must equal the centre ofC[G].

3.2 Tensor products of modules over commutative rings

Let R be a ring (must be commutative with 1). LetM, N be R-modules. LetB be the free R-module on the set{m n | m M, n N}. So

B =

ai(mi ni) | mi M, ni N, ai R

LetZbe the set ofR-linear combinations of elements ofB of the following form:

(m1 +m2)nm1 nm2 n m(n1 + n2)mn1 mn2 (rm)nr(mn) m(rn)r(mn)

where m, m1, m2 M, n, n1, n2 N, r R are any elements. Define MRN =B/Z, where the R-moduleB /Zis the abelian group B /Zwith the R-actionr(b + Z) = rb + Z.

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3.17 Example. MRR =M, because every element MRR is equal to m 1 for some m M.3.18 Example. Rn RRm=Rnm, same as for vector spaces.3.19 Example. R = Z, M= Q, N= Z[

2] = {a + b2 | a, b Z}. MRN= QZ Z[

2] = Q(

2).

Define : Q Z[2] Q(2)by ( qi (ai+ bi2)) = qi(ai+ bi2).Then is a hom ofZ-modules. To see that is onto note that Q(

2) = {a+b2 | a, b Q} so every element of

Q(

2)can be written as a+b

2c

fora, b, c Z, and( 1c (a + b2)) = a+b

2

c .

To see that is one-to-one, its enough to show that Q Z[2] and Q(2) are both 2-dimensional vector spaces,and that is a Q-linear transformation. All of these facts are easy to see except dimQQZ[2], which isproven as follows. Any element ofQ Z[2] is of the form qi (ai + bi2) = aiqi 1 + biqi 2 =(

aiqi)(1 1) + (

biqi)(1

2). So Q Z[2] is spanned by 1 1and 1 2 as a Q-vector space. Since isonto, and Q(

2) has dimension 2, we see that dim Q Z[2] = 2 and is an isomorphism, as desired.

Let R be a ring (with 1, not necessarily commutative). Note that the previous construction works only if R iscommutative! LetTbe a ring containing R as a subring, and Mbe an R-module.

DefineB to be the free abelian group on{t m|tT , mM}={ ti mi| ti T , mi M}. Let Zbe theset of integer-linear combinations of all elements ofB of the following forms:

(t1+ t2) m t1 m t2 m, t (m1+ m2) t m1 t m2,

(tr) m t (rm)DefineTRM=B/Zas abelian groups with a left T-module structure by

t

ti mi

=

(tti) mi.

3.20 Example. Let G = Z2, H ={0}, V = C, with the trivial H-action. In other words, V is the trivial one-dimensional representation of H, so V is a C[H]-module. Consider C[G] C[H] V . It is a C-vector space. Anarbitrary element ofC[G] C[H]V is

(ai 0 + bi 1) zi=

(aizi 0 + bizi 1) 1 = (A 0 + B 1) 1 spanC{0 1, 1 1}

so C[G] C[H]V= span{0 1, 1 1}.LetW = C[G]

C[H]V= span

{0

1, 1

1

}. ThenW is a C[G]-module, with

0(0 1) = (0 + 0) 1 = 0 10(1 1) = (0 + 1) 1 = 1 11(0 1) = (1 + 0) 1 = 1 11(1 1) = (1 + 1) 1 = 0 1

So W corresponds to the (2-dimensional) left regular representation ofG, as expected.

3.21 Example.Let H = Z4 =y, G = D4, : H GL(C) given by (n) = in, V = C the correspondingC[H]-module. Lets computeW= C[G] C[H]V as a C[G]-module. An arbitrary element ofC[G] C[H]V is of theform

i j aijgj vi= i j Aijyj

zi+ i j Bijxyj

zi

whereBij =aij forj corresponding to xyj , and Aij =aij forj corresponding toy j .

=i

j

Aij (yj)zi

+i

j

Bijx (yj)zi

=A(1 1) + B(x 1)whereA, B C.

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for someri R, soN =f()for some polynomialf(x) R[x]of degree< N. Thus, g(x) = xN f(x)is a monicpolynomial with coefficients inR and g() = 0.

Backwards: We knowf() = 0for some monic polynomial f(x) R[x]. So ifN= deg(f)we getN =rN1N1 +. . . + r0. SoR[] is generated by1, ,

2, . . . , N1, so were done.

3.27 Example. Which elements ofQ(

5)are integral over Z?

Solution. Q(

5) = {ab

+ cd

5 | a,b,c,d Z}.

What is the minimal polynomial for ab

+ cd

5 over Q? Ifc = 0, weve got a rational number, and we already know

which rational numbers are integers. So assume c

= 0. Then the minimal polynomial is

(x ab c

d

5)(x a

b+ c

d

5) = x2 ( 2a

b)x + ( a

2

b2 5c2

d2)

=x2 ( 2ab

)x + a2d25c2b2

b2d2

When does this polynomial have integer coefficients? Assume WLOG thatgcd(a, b) = gcd(c, d) = 1. Then 2ab Z

impliesb | 2 so WLOG b = 1 or 2. Ifb = 1 then the polynomial is:

x2 (2a)x +

a2d2 5c2d2

so d2 | 5, meaning d = 1. Ifb= 2, then the polynomial is x2 ax+ ( a2d220c24d2 ) so 4d2 | a2d2 20c2. This means4 | a2d2, so4 | d2, sinceb = 2, soa is odd. So d is even, meaning that d = 2d for some integerd. Then

a2d2 20c24d2

=4a2(d)2 20c2

16(d)2 = a

2(d)2 5c24(d)2

Now d must be odd, since otherwise the numerator is odd and 4(d)2 is even, so our fraction is not an integer.Furthermore, we must have

(d)2 | a2(d)2 5c2 = (d)2 | 5c2 = (d)2 | 5 = d = 1 = d= 2.

Thus, ab

+ cd

5 = a+c

5

2 . We also need a and c to be odd, so that a2(d)2 5c2 can be even. So in general, any

integral element ofQ(

5) must be of the form x+y

1+

52

. All these elements are integral over Z because the

minimal polynomial ofx + y

1+

52

is

t x + y1 + 52

t x + y1 52

=t2 (2x y)t + (x2 + 2xy+ y2(2))

3.4 Frobenius divisibility

Where were going: to prove that if is an irreducible representation of a finite group G then dim | |G|.3.28 Theorem. Let Tbe a commutative ring, R T a subring. Assume R, T are Noetherian. Then the set ofelements ofTthat are integral over R is a subring ofT.

Proof. The set SofR-integral elements ofT is nonempty, so it suffices to show that it is closed under +,

, and

.

Thus, let x, y Sbe any elements. Thenx+y, x y, and xy are all elements of the ring R[x, y], which is a fin.gen. R-module by{xiyj} where i, j range over the same sets as those needed to ensure that{xi} generates R[x]and{yj} generates R[y]. The proof now follows from the following lemma.3.29 Lemma. IfR is a Noetherian ring and M is a finitely generatedR-module, then every R-submodule ofM isalso finitely generated.

Proof. SayN M is an R-submodule. SinceM is fin. gen., by m1, . . . , mn, there is a surjective R-module hom.: Rn Mgiven by (r1, . . . , rn) =r1m1+ . . . + rnmn. If we can show that 1(N) = {v Rn | (v) N} is afin. gen. R-module, then Nwill also be fin. gen., by the images (under ) of the generators for 1(N). Thus, wemay assume thatM=Rn. Ifn = 1, then Nis an ideal ofR, so since R is Noetherian, Nis fin. gen. Now proceedby induction on n. Define R-module hom. : Rn R by(r1, . . . , rn) = r1. Thenker = Rn1 is a fin. gen.

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Proof. We have

ResKIndGH= (Ind

GH as a C[K] module)

= C[G] C[H]Vconsidered as a C[K]-module. We want to show that C[G] C[H]V=

i

C[K] C[Hi]V

as C[K]-modules.

Define : C[G] C[H]V

i

C[K] C[Hi]V

by

(kgih v) = (0, . . . , k [(h)](v), . . . , 0)where the nonzero entry is in the ith coordinate. To show that is well-defined, we need to show that ifk1gih1 =k2gih2, thenk1 [(h1)](v) = k2 [(h2)](v)for allv V. Well,k1gih1 = k2gih2 = k1 = k2(gih2h11 g1i ). Then

k1 [(h1)](v) = k2(gih2h11 g1i ) [(h1)](v)=k2 [i(gih2h11 g1i )(h1)](v)

because gih2h11 g

1i =k

12 k1 giHg1i K= Hi. So this is

=k2 [(h2h11 )(h1)](v) = k2 [(h2)](v)as desired. Thus, is a well-defined hom. ofC[K]-modules. Define :

i

C[K] C[Hi]V

C[G] C[H]V by(k1 v1, . . . , kn vn) =

i kigi vi. To see that is well defined, I must show that ifgihg1i Hi, then:

(kgihg1i v) = (k [i(gihg1i )](v)

But (0, . . . , k gihg1

i v, . . . , 0) = kgih v = kgi [(h)](v) and (k [i(gihg1

i )](v)) =kgi [(h)](v), whichis good.((kgih v)) = (0, . . . , k [(h)](v), . . . , 0)

=kgi [(h)](v) = kgih vand

((k1 v1, . . . , kn vn)) =

i

kigi vi

= (k1 v1, . . . , kn vn)So and are mutually inverse, and so is an isomorphism, as desired.

3.34 Theorem (Mackeys Irreducibility Criterion). Let G be a finite group, H

G a subgroup,: H GL(V) a representation. Then IndGH is irreducible iff

1. is irreducible, and

2. For allg G H,g , ResHgHg = 0whereHg = gH g

1 H andg :Hg GL(V) is g(t) = (g1tg).Proof. IndGH is irreducible iffIndGH, IndGHG= 1, iff, ResHIndGHH= 1 (choose g1, . . . , gn double cosetreps for(H, H), g1 = 1), iff,

IndHHiiH= 1, iff

ResHi, iHi = 1, iffResH, H= 1 (correspond-ing to H1) andResHi, iHi = 0 for all i = 1, iff, H= 1 ( is irreducible) andResHi, iHi = 0 forall i = 1, iff is irreducible andResHg, gHg = 0 for all g G H.3.35 Corollary. IfH G is a normal subgroup, : H GL(V) a representation, then IndGH is an irreduciblerepresentation ofG iff

1. is irreducible, and

2.g , = 0 for all g G H.Proof. Immediate from MIC.

3.36 Example. G = D4 =x, y| x2 = y4 = 1, xy = y1x, H =y, : H GL(C) given by (ya) = ia. IsIndGH irreducible?

Since dim = 1, its irreducible. Anyg G H is g = xya, so gH g1 = xH x, so Hg = Hx and g = x. Andx(ya) = (xyax) = (ya) = ia, so x =. Thus, sincex is also irreducible, x , = 0 soIndGHis irreducibleby MIC.

GROUPSandREPS LW 1139 Pmath745notes

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