on clifford theory with galois action.pdf

8/10/2019 On Clifford theory with Galois action.pdf

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ON CLIFFORD THEORY WITH GALOIS ACTION

FRIEDER LADISCH

A b s t r a c t . LetG be a finite group, Na normal subgroup ofGand IrrN.Let F be a subfield of the complex numbers and assume that the Galois orbit of

over F is invariant inG. We show that there is another triple (G1, N1, 1) ofthe same form, such that the character theories ofGover and ofG1 over1 areessentially the same over the fieldF and such that the following holds:G1 hasa cyclic normal subgroup C contained in N1, such that 1 =

N1 for some linearcharacter ofC, and such thatN1/Cis isomorphic to the (abelian) Galois group of

the field extension F()/F(1). More precisely, the same means that both triples

yield the same element of the Brauer-Clifford group BrCliff(G,F()) defined by

A. Turull.

1. I n t r o d u c t i o n

1.1. Motivation. Clifford theory is concerned with the characters of a finite grouplying over one fixed character of a normal subgroup. So letG be a finite group andNG a normal subgroup. LetIrr N, where Irr Ndenotes the set of irreduciblecomplex valued characters of the groupN, as usual. We write Irr(G|) for the setof irreducible characters of

G which lie above in the sense that their restriction to

N has as constituent.In studying Irr(G| ), it is usually no loss of generality to assume that is

invariant inG, using the well knownClifford correspondence[8,Theorem 6.11]. In thissituation,(G, N, )is often called a character triple. Then a well known theorem tellsus that there is an isomorphic character triple(G1, N1, 1)such thatN1 Z(G1)[8,Theorem 11.28]. Questions about Irr(G|) can often be reduced to questions aboutIrr(G1|1), which are usually easier to handle. This result is extremely useful, forexample in reducing questions about characters of finite groups to questions aboutcharacters of finite simplegroups.

Some such questions involve Galois automorphisms or even Schur indices [12,20].

Unfortunately, both of the above reductions are not well behaved with respect toGalois action on characters and other rationality questions (like Schur indices of theinvolved characters). The first reduction (Clifford correspondence) can be replacedby a reduction to the case where is semi-invariant over a given field F C [14,Theorem 1]. (This means that the Galois orbit of is invariant in the groupG.)

Date: September 11, 2014.

2010 Mathematics Subject Classification. 20C15.Key words and phrases. Brauer-Clifford group, Clifford theory, Character theory of finite groups,

Galois theory, Schur indices.1

arX

iv:1409.3559v1[math.GR]11Sep

2014


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2 FRIEDER LADISCH

Now assuming that the character triple (G, N, ) is such that is semi-invariantinG over some field F, usually we can not find a character triple (G1, N1, 1) withN1 Z(G1), and such that these character triples are isomorphic over the fieldF. We will give an exact definition of isomorphic over F below, using machinery

developed by Alexandre Turull [22,23]. For the moment, it suffices to say that acorrect definition should imply thatG/N=G1/N1 and that there is a bijectionbetween

Irr(

G|) andIrr(G1|1 ) (unions over a Galois group) commutingwith field automorphisms over F and preserving Schur indices over F. Now if, forexample, Q() =Q(

5)(say), then it is clear that we can not find a triple isomorphic

over F and such that 1 is linear. The main result of this paper, as described inthe abstract, provides a possible substitute: At least we can find an isomorphiccharacter triple(G1, N1, 1), where the properties ofN1 and 1 are somewhat undercontrol. This result is probably the best one can hope for, if one wants to take intoaccount Galois action and Schur indices.

1.2. Notation. To state the main result precisely, we need some notation. Instead ofcharacter triples, we find it more convenient to use Clifford pairs as introduced in [11].

LetGand G be finite groups and let :GG be a surjective group homomorphismwith kernelKer = N. Thus

1 N G G 1

is an exact sequence, andG/N=G via . We say that (, )is a Clifford pairoverG. (Note thatG, Gand Nare determined by as the domain, the image and thekernel of, respectively.) We usually want to compare different Clifford pairs overthe same group G, but with different groupsG and N.

Let F C be a field. (For simplicity of notation, we work with subfields ofC,the complex numbers, but of course one can replace C by any algebraically closedfield of characteristic 0 and assume that all characters take values in this field.)Then Irr N is called semi-invariant inG over F (where NG), if for everygG, there is a field automorphism =g Gal(F()/F) such that g =. Inthis situation, the map gg actually defines an action ofG on the field F() [7,Lemma 2.1].

To handle Clifford theory over small fields, Turull [22, 23] has introduced theBrauer-Clifford group. For the moment, it is enough to know that the Brauer-Clifford group is a certain set BrCliff(G,E) for any group G and any field (or, moregenerally, ring) E on whichG acts. Given a Clifford pair (, )and a fieldF such that is semi-invariant over F, the groupG acts on F() and the Brauer-Clifford groupBrCliff(G,F())is defined. Turull [22, Definition 7.7] shows how to associate a certainelement,,FBrCliff(G,F())with(, )and F. Moreover, if(, )and (1, 1)are two pairs over G such that and 1 are semi-invariant over F and induce thesame action ofG on F() =F(1), and if,,F= 1, 1,F, then the charactertheories ofGover and ofG1 over1are essentially the same, including rationality


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ON CLIFFORD THEORY WITH GALOIS ACTION 3

properties over the field F. (See [22, Theorem 7.12] for the exact statement.) Thisjustifies it to view to such Clifford pairs as isomorphic overF.

1.3. Main result. The following is the main result of this paper. We state it forsubfields of the complex numbers C, but it should be clear that in fact C can stand

for any algebraically closed field of characteristic 0, if all characters are assumed totake values in that fixed field C.

Theorem A. Let FC be a field and1 N G G 1

be exact andIrr N be semi-invariant inG over F. Then there is another exactsequence

1 N1

G1 G 1

1

and1Irr N1, such that F() = F(1) asG-fields, such that,,F= 1, 1,F in BrCliff(G,F()),

and such that the following hold:

(a)G1 has a cyclic normal subgroup CG1 withC N1,(b) there is faithfulLin C with1=N,(c) is semi-invariant inG1 over F and(d) N1/C=Gal(F()/F()).

As mentioned before, Turulls result [22, Theorem 7.12] yields that there arecorrespondences of characters with good compatibility properties. We mention a fewproperties here, and refer the reader to Turulls paper for more:

Corollary B. In the situation of TheoremA, for each subgroup H G there is abijection between

Z[Irr(1(H)|)] and Z[Irr(11 (H)|1)].The bijections can be chosen such that their union commutes with restriction andinduction of characters, with field automorphisms over the field F, and with multi-plications of characters ofH, and such that it preserves the inner product of classfunctions and fields of values and Schur indices (even elements in the Brauer group)overF.

We can say something more about the groupG1 in the main theorem.Corollary C. In the situation of TheoremAand forU1= (G1)1 andV = (G1) (theinertia groups of1 andinG1), we haveU1 = V N1 andN1 V =C(cf. Figure1),and

1, (1)|U1,F()= , (1)|V,F().For every subgroup X withN1 XU1, induction yields a bijection

Irr(X V|)X Irr(X|)


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4 FRIEDER LADISCH

commuting with field automorphisms overF() and preserving Schur indices overF().

U1G1

N1N =1

V

C

= Gal(F()/F())

(

Gal(F()/F))

F i g u re 1 . Subgroups of

G1 in TheoremA

Observe that since is faithful, we actually have V = CG1(C) andC Z(V). Soover the bigger field F() and in the smaller groupU1, we can replace the Cliffordpair (1, (1)|U1) by the even simpler pair (, (1)|V). This is, of course, just theclassical result mentioned before, which is usually proved using the theory of projectiverepresentations and covering groups.

1.4. Relation to earlier results. A number of people have studied Clifford theoryover small fields, including Dade [2,3,4], Isaacs [7], Schmid[15,16] and Riese [13],cf.[14]. While we use some ideas of these authors, most important for our paper isthe theory of the Brauer-Clifford group as developed by Turull [22,21,23], whichsupersedes in some sense his earlier theory of Clifford classes [18]. In particular, it isessential for our proof that the Brauer-Clifford group and certain subsets of it areindeed groups, a fact which, it seems to me, has not been important in the applicationsof the Brauer-Clifford group[19,25]so far.

A paper of Dade[2] contains, between the lines, a result similar to our main theorem(but in a more narrow situation): Dade studies the situation where (in our notation)

is invariant inG and has values in F. Then a cohomology class [] H2(G,F)is defined. This class determines part of the character theory ofG over , but notcompletely, since it does not take into account the Schur index of itself. Afterproving some properties of this cohomology class, Dade shows that all cohomologyclasses with these properties occur, by constructing examples. When examining thisconstruction, one will find that the examples have almost the same properties as thegroupG1 in TheoremA. One can deduce that all such cohomology classes come fromcharacters of some cyclic by abelian group.

TheoremAcontains the classical result that every simple direct summand of thegroup algebra ENof a finite groupNover a field E of characteristic zero is equivalentto a cyclotomic algebra[26]. (This is the case G= 1 of TheoremA.) Our proof ofTheoremAuses ideas from the proof of the result that Schur algebras are equivalentto cyclotomic algebras, as presented by Yamada [26].


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2. R e v i e w o f t h e B r a u e r - C l i f f o r d g r o u p

Let us briefly recall the relevant definitions. Details can be found in the papers ofTurull[22,23], see also [5]. LetG be a group. A G-ringis a ring Z(with1) togetherwith an action ofGonZvia (unital) ring automorphisms. We use exponential notation

zzg to denote such an action. Let Zbe a commutativeG-ring. AG-algebra overZ is aG ringA together with a homomorphism ofG-rings : Z Z(A). This meansthat A is an algebra over Zand that the algebra unit : Z A has the property(zg) =(z)g. We usually suppress mention of the algebra unit and simply write z afor (z)a.

We only need G-algebras over fields in this paper. Let Zbe a field on which Gacts. A G-algebra over Z is called central simple, if it is central simple as algebraoverZ. The Brauer-Clifford groupBrCliff(G, Z) is the set of equivalence classesof central simple G-algebras over Zunder a certain equivalence relation. To definethe equivalence relation, we need the skew group ring ZGofG over Z (also called

the crossed product), which is the set of formal sumsgG gcg, (cg Z) withmultiplication defined by

gG

gcg

hG

hdh

=g,h

ghchgdh.

If V is a right ZG-module, then EndZV is a G-algebra over Z, called a trivialG-algebra. Two G-algebras S and T over Z are called equivalent, if there areZG-modules V and Wsuch that

SZEndZV=TZEndZWasG-algebras overZ. The Brauer-Clifford groupBrCliff(G, Z)is the set of equivalenceclasses of central simple G-algebras over Z, with multiplication induced by tensoringoverZ.

The Brauer-Clifford group is an abelian torsion group [22, Theorem 3.10,5,Theo-rem 5].

Next let (, :G G) be a Clifford pair and F C a field. Assume that issemi-invariant inG over F. (The last assumption is not necessary in Turulls theory,but we only need this case, and the notation can be simplified somewhat in thiscase.) So for every gG, there is a field automorphismgGal(F()/F) such thatgg =. The map gg defines an action ofG on the field F() [7, Lemma 2.1]and thus the Brauer-Clifford group BrCliff(G,F()) is defined. Turull showed [22,Definition 7.7] how to associate an element

,,FBrCliff(G,F())to (,,F). We recall the construction. Let e = e(,F) Z(FN) be the centralidempotent ofFN corresponding to. Note thate is invariant inGsince is semi-invariant. An FG-moduleV is called -quasihomogeneous ifV e = V. Let V bea nonzero quasihomogeneous FG-module. Then S = EndFNV is a G-algebra. Forz Z(FN e), the map v vz is in S, and this defines a canonical isomorphismZ(FNe)= Z(S) ofG-algebras. The central character belonging to defines an


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6 FRIEDER LADISCH

isomorphism Z(FNe)= F(), which commutes with the action ofG (namely, we have(z

g) =(z)g). Thus we can view Sas a central simple G-algebra over F(). The

equivalence class ofSdoes not depend on the choice of the quasihomogeneous moduleV [22,Theorem 7.6]. Following Turull, we write ,,F for this equivalence class.

(Actually, Turull defines,,Fas an element ofBrCliff(G, Z), whereZ= Z(FN e).)Now suppose that (, ) and (1, 1)are two semi-invariant Clifford pairs over G

such that F() = F(1). Assume that both Clifford pairs induce the same action ofGon F(), that is, for each gG there is g Gal(F()/F) such that gg = andgg1 =1 for all gG. If

,,F= 1, 1,F,

then the character theory ofGover the Galois conjugates ofand the character theoryofG1 over the Galois conjugates of1 are essentially the same[22, Theorem 7.12],as we mentioned in the introduction.

3. S u b e x t e n s i o n s

Let

1 K G G 1be an exact sequence and Irr K. (Thus(, ) is a Clifford pair.) Suppose thatHis a supplement ofKinG, so thatG=HK. SetL =HK. ThenH/L=G/K=G.Suppose that Irr L. We want to compare the elements ,,F and ,

|H,F.We do this under additional assumptions, which for convenient reference we collecthere:

3.1. Hypothesis. Let

1 L H G 1

1 K G G 1

|H

be a commutative diagram of finite groups with exact rows, let Irr KandIrr Lbe irreducible characters and let F be a field of characteristic zero such that the

following conditions hold:(a) n= [L, ]> 0.(b) F() = F() =:E.

(c) For every hHthere is =hGal(E/F) such that h = and h =.To reduce visual clutter, we write

e= e(,F)=

Gal(E/F)

e and f=e(,F)=

Gal(E/F)

e

for the corresponding central primitive idempotents in FKand FL, respectively.


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3.2. Lemma ([9, Lemma 6.3]).

i:=

Gal(E/F)

ee

is aH-stable nonzero idempotent inFKe, and we haveei= i = ie andf i= i = if.3.3. Lemma ([9, Lemma 6.4]).

Z(iFKi)= Z(FKe)= F() =F()= Z(FLf)asG-rings.

3.4. Lemma ([9, Lemma 6.5]). SetZ= Z(iFKi) and letS= (iFKi)L. ThenS is acentral simpleG-algebra overZ=E = F() with dimensionn2 overZ, and

CiFKi(S) =FLi= FLf.Thus the equivalence class ofSdefines an element in BrCliff(G,E). The main

result of this section is:3.5. Theorem. Assume Hypothesis3.1. Then

,,F [S] =, |H,F (inBrCliff(G,E)).

This result is of course related to the results in[9], but we didnt use the languageof the Brauer-Clifford group in our previous paper. Thus we give a translation here.Theorem3.5is also related to results in [24].

The precise result depends on the following conventions: All modules over FG andother group algebras are right modules, and endomorphism rings also operate fromthe right. Thus for anyF-algebraA operating on a moduleVwe get a homomorphism

A EndF(V), both operating from the right. ThenEndA(V)is simply the centralizerof the image ofAin EndF(V). With other conventions, one may have to exchange and in the formula of Theorem3.5.

Proof of Theorem3.5. Let V be an FG-module with V e =V. Then the G-algebraEndFK(V) is in ,,F. Moreover, set W = V i. Then W is an FH-module withW f=V if=V i= W, and thus EndFL(W) is a G-algebra in , |H,F.

We may also consider W = V i as a module over iFKi. Consider the algebraA= EndiFKi(V i). SinceHacts onV i and elements ofL commute with elements ofA, we see that G=H/Lacts on A. We claim that

A=EndFK(V)as G-algebras over Z. In fact, it is well known that EndR(V)= EndiRi(V i) foridempotentsi in a ring R withR= RiR. We apply this with R= FKe. It remainsto verify that the isomorphism commutes with the action ofG. The isomorphism isgiven by sending EndR(V) to iEndiRi(V i)defined by (vi)(i) =vii. (Wemention that the inverse sends EndiRi(V i) to defined by v =vaib,where a, b R are such that 1R =aib. It is easy to check that these areinverse ring homomorphisms.) Now

vhi= vh1hi= vh1ih= v(i)h


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8 FRIEDER LADISCH

forhH, so the isomorphism i commutes with the action ofG=H/L. Theclaim is proved.

Now identify iFKi with a subalgebra of T := EndE(W). Then we have A =CT(iFKi). Since all these algebras are central simple over E, we have

T =A(iFKi)=A E(iFKi).In the same way, identify FLf= FLi with a subalgebra of iFKi T, and setB= EndFL(W) = CT(FLi). As above, we have

T =B(FLi)=BE(FLi).Clearly, we have

A, SB.Since

T=A

E(iFKi)

=A

E(S

E(FLi))

=(A

ES)

E(FLi),

it follows that

B=AS=A ES.Since B,

|H,Fand A,,F, the theorem follows. 3.6. Corollary. Assume Hypothesis 3.1 with n = [L, ] = 1. Then ,,F =,

|H,F.Proof. Clear since then S= E yields the trivial element ofBrCliff(G,E).

The assumptions of the corollary hold in particular when L = Irr L andF() = F(). To see another situation where Hypothesis 3.1holds with n= 1, weneed a simple lemma, which is a minor extension of a result of Riese and Schmid[14,Theorem 1]. (The bijection in the following lemma also preserves Schur indices overF, but we will not need this fact.)

3.7. Lemma. LetA G be a normal subgroup and Irr A. LetF be a field and := Gal(F()/F). Define

T={gG| : g =}.Then induction defines a bijection between

Irr(T|

) and Irr(G|

)

commuting with field automorphisms overF. In particular, F(G) = F() forIrr(T| ).Proof. WriteGto denote the inertia group of. Clifford correspondence [8, Theo-rem 6.11] yields that induction defines bijections from Irr(G| )ontoIrr(T| ), andfrom Irr(G| ) onto Irr(G|). Thus induction defines a bijection from Irr(T|)ontoIrr(G|). The same statement holds withreplaced by. (Of course, we haveG =G.)


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It follows that induction yields a surjective map

Irr(T| )

Irr(G|)

which is injective when restricted to some Irr(T

|). It remains to show that there

occurs no collapsing between two different sets Irr(T | 1) and Irr(T | 2). Soassume that G =G = for Irr(T| 1) and Irr(T|2). Then 1 and2 are both irreducible constituents ofA. By Cliffords theorem there isgG with1g =2 or g1

12 =. By the very definition ofTwe see that gT. But then

= g Irr(T| 2). Since induction is bijective when restricted to Irr(T| 2), itfollows that = as wanted.

Obviously, induction commutes with field automorphisms: We have ()G = (G).We always have F(G)F(). IfGal(F()/F(G)), thenG = (G) = ()Gand thus = by bijectivity. It follows = 1 and thus F() =F(G) as claimed.The proof is finished.

3.8. Corollary. Let

1 K G G 1be an exact sequence of finite groups andIrr K be semi-invariant overF inG.Assume there isA G withAKand letbe an irreducible constituent ofA. SetH={gG| Gal(F()/F) :g =}.ThenG=HK, and forL=H K there is a uniqueIrr(L|) with = K,and we haveF() =F() and,,F= ,

|

H

,F for this.

Proof. We first show thatG=HK. Let gG. Since is semi-invariant inG, thereis a field automorphismsuch that g =. We may extendto an automorphismofF(, ). Then g is a constituent ofA. By Cliffords theorem, we have

gk =

for some kK. ThusgkH.Lemma3.7yields the existence of a with = K.We verify that Hypothesis3.1holds. We haven = [L, ]= [,

K]= 1. Lemma3.7

yields that F() = F(). Finally, let hH. Then there is Gal(F()/F) withh = . Then (h)K = = K. Since h Irr(L|h) and h = for somefield automorphism, it follows from Lemma3.7that h =.

We have now shown that Hypothesis 3.1holds with n = 1. Corollary3.6 yields the

result. 3.9. Remark. It follows directly from Lemma 3.7 that for each subgroup U withKUG there is a correspondence between Irr(U| ) and Irr(UH| ). It isalso elementary to prove that these correspondences commute with restriction andinduction of characters. In fact, every property of the correspondence that followsfrom the equality ,,F= ,

|H,Fand the results of Turull [22]can be provedelementarily, without using the Brauer-Clifford group. But we will need Corollary3.8in inductive arguments to come later, and there we can not do without the languageof the Brauer-Clifford group.


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10 FRIEDER LADISCH

CorollaryCis a special case of Corollary 3.8, applied over a bigger field:

Proof of CorollaryC. Assume the situation of TheoremA.Recall thatC, N1 G1with C N1, and that 1 = N1 with Irr C. By definition in Corollary C,

U1 = (G1)1 is the inertia group of1 inG1, andV = (G1) is the inertia group ofinG1. Clearly, N1 =1Irr N1 implies that V N1 = C.We apply Corollary3.8to the exact sequence

1 N1 U1 U 1with C instead ofA, instead ofand the field F()instead ofF. Note thatH=V.We get thatU1 = V N1 and 1, (1)|U1,F() = , (1)|V,F(). The statementabout induction inducing a bijection Irr(XV | ) Irr(X| 1) follows fromLemma3.7.

4. A s u b g r o u p o f t h e S c h u r - C l i f f o r d g r o u p

Let E be a field and G a group which acts on E and fixes the subfield F. Recallthat in[11] we defined the Schur-Clifford groupSC(F)(G,E) to be the subset ofBrCliff(G,E)of all,,Fsuch that (, )is a Clifford pair over G that induces the

given action ofG on F() =E. We also showed thatSC (F)(G,E)is a subgroup of theBrauer-Clifford group, ifE is contained in a cyclotomic extension ofF. Further, inthat case we have

SC(F)(G,E) =SC(EG)(G,E) =:SC(G,E).Fix a finite group G and a field F. We consider two subclasses of Clifford pairs

(, ) where, as usual, the homomorphism is part of an exact sequence

1 N G G 1andIrr N. The main result of this section is that both subclasses yield the samesubgroup of the Schur-Clifford group.

4.1. Definition. LetMbe the class of all Clifford pairs (, ) such that(a) there is A G with AN,(b) there is a linear character Lin Asuch that = N.

LetC be the subclass ofMcontaining the pairs inMsuch that(c) the Aabove is cyclic and is faithful,

(d) is semi-invariant inG over F, and(e) N/A=Gal(F()/F()).

Recall that for a classXof Clifford pairs, we definedSCX(G,E) ={,,FBrCliff(G,E)|(, ) X}.

In this terminology, TheoremAsays thatSC(G,E) =SCC(G,E).There is some redundancy in these conditions:


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4.2.Lemma. If(a), (b)and(d) above hold for some Clifford pair(, ), so does(e).(Thus Condition(d) in TheoremAfollows from the other conditions in TheoremA.)

Proof. Since is semi-invariant in

G over F, there is for every n N an n

Gal(F()/F) such that nn = . Since (n)N = n = , we must have nGal(F()/F()). This defines an homomorphism from N/A into Gal(F()/F()).

SinceN Irr N, this homomorphism is injective. Now letGal(F()/F()). Then andare constituents of = and thus conjugate inN, so thatn = for somenN. It follows that = nGal(F()/F()). Thus nnGal(F()/F())issurjective. Thus Condition(e)holds.

The following result follows directly from the definitions.

4.3.Lemma. Let(, :GG)be a Clifford pair andKKer a normal subgroupof

G. Then ,,F = ,,F, where :

G/K G is the map induced by and

Irr(N/K) is the character viewed as character ofN/K (whereN= Ker ).Note that when Conditions(a),(b)and(d)above hold, then K=Ker is normal

inG, and also K= Ker . Thus we can factor out Kand get a Clifford pair suchthat(c)is true, too.

4.4.Proposition. LetE be a field extension ofF on whichG acts asF-algebra. Then

SCM(G,E) =SCC(G,E).Proof. By definition, we have SCC(G,E) SCM(G,E). To show the converse inclusion,we begin with a Clifford pair (, )

Mthat induces the given action on E= F().

Here :G G and Irr N with N = ker as usual. Let A Nbe a normalsubgroup ofGand Lin Aa linear character with = N.

By assumption, the character is semi-invariant inG. SetG0={gG| there isGal(F()/F) such that g =},N0=NG0, 0 = |G0 , and 0=N0.

Note thatN0 =N =. By Corollary3.8, we have thatG=G0N, that F() = F(0),

and that ,,F= 0, 0,F. Condition(d)(and thus(e)) holds for the Cliffordpair(0, 0). By Lemma4.3 and the remark following it, we can factor out the kernel

of and we get a Clifford pair inC that yields the same element of the Brauer-Cliffordgroup as(, ).

4.5. Corollary.SCC(G,E) is a subgroup ofSC(G,E) andBrCliff(G,E).Proof. It suffices to show thatSCM(G,E)is a subgroup. By Theorem 5.7 from[11]we have to show:

(a)SCM(G,E)=,(b) When (, ) M, then (, ) M,(c) When (1, 1) and (2, 2) M, then (1 2, 1G2) M.


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12 FRIEDER LADISCH

The first assertion follows from Corollary 5.2 in [11], and the second is clear. For the

third, assume AiNi is the abelian normal subgroup ofGi, where1 Ni

Gi G 1

i

is the exact sequence belonging to i, and let iLin Ai withi = Nii fori= 1,2.Recall that the Clifford pair (1 2, 1G2) is defined as follows: Let

G1GG2={(g1, g2)G1G2|g11=g22}be the pullback and 1G2 :G1GG2G the canonical homomorphism. ThenKer(1G2) = N1N2, and 12 Irr(N1N2) has the usual meaning. Itfollows that A1A2 G1GG2 is an abelian normal subgroup and 12 =N11 N22 = (1 2)N1N2. This shows the last item.

5. R e d u c t i o n t o p r i m e p o w e r g r o u p s5.1. Lemma. Leta SC(G,E). Thena SCC(G,E) if and only if thep-partsap ofa are inSCC(G,E) for all primesp.Proof. Recall thatBrCliff(G,E)is torsion. Thusa=

p apis the product of itsp-parts

ap, and ap a. The result follows sinceSC(G,E) andSCC(G,E) are subgroups(Corollary4.5).

For any subgroup H G, there is a group homomorphism

ResGH: BrCliff(G,E)BrCliff(H,E)which is induced by viewing a G-algebra as an H-algebra. This restriction homo-morphism sends,,Fto ,,F, where is the restriction ofto the preimage1(H) ofH [11, Proposition 7.1].

In the proof of the next result, we also need the corestriction map

CoresGH: BrCliff(H,E)BrCliff(G,E)defined in [10].

5.2. Lemma. Leta SC(G,E) havep-power order for the primep and letP Gbe a Sylowp-subgroup ofG. Thena SCC(G,E) if and only ifResGP(a) SCC(P, E).Proof. The only if part is clear and does not depend on a SC(G,E) havingp-power order.

Now assume ResGP(a) SCC(P,E). Apply the corestriction mapCoresGP : BrCliff(P,E)BrCliff(G,E).

It follows from [11, Corollary 9.2] thatCoresGPmapsSCM(P, E)intoSCM(G,E). SinceSCM(G,E) =SCC(G,E)by Proposition4.4, we have CoresGP(ResGP(a)) SCC(G,E).By[10,Theorem 6.3], we get CoresGP(Res

GP(a)) =a

|G:P|. Sinceahasp-power order, itfollows that a a|G:P|. Thusa SCC(G,E) as claimed.


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By the last two results, to show that an arbitrary element a SC(G,E) is in factcontained inSCC(G,E), we can assume that a has p-power order and that G is ap-group. Note that it was essential in the above proofs thatSC(G,E) andSCC(G,E)are groups.

5.3.Remark. By similar arguments, one can show: An element aBrCliff(G,E)is inSCC(G,E)if and only ifResGP(a) SCC(P,E) for all Sylow subgroups P ofG. Note,however, that even when P is a p-group, the exponent ofBrCliff(P, E) orSC(P,E)is in general not a p-power. For example, BrCliff(1,E) =Br(E), andSC(1,E) is ingeneral not the trivial group.

6. R e d u c ti o n t o a l a rg e r f i e l d

In this section, we show that when Gis ap-group and,,Fhasp-power-order,then we can replace the fields E and F by certain larger fields to prove the main

theorem (TheoremA).First, we assume the following situation: Let E be a field on which a group G acts,

and let F be a field contained in the fixed field EG. Suppose that there are two otherfields K L F such that K= EL and E L= F. Assume that L/F is Galois andlet =Gal(L/F)=Gal(K/E). The situation is summarized in the following picture:

F

L

E

K

In this situation, G acts on K= EF L. IfS is a central simple G-algebraover E, then SF L= SE K is a central simple (G)-algebra over K. Thisdefines a group homomorphismBrCliff(G,E)BrCliff(G,K). In fact, this grouphomomorphism is the composition of

BrCliff(G,E) BrCliff(G ,E) BrCliff(G ,K),Inf

where the first map is the inflation map induced by the epimorphism G G,and the second map is induced by scalar extension from E to K. Our first goal is to

show that the above group homomorphism is actually an isomorphism:

6.1. Proposition. The maps

SSE K and TT := CT()define inverse maps between the isomorphism classes of central simpleG-algebrasoverE and central simple(G )-algebras overK. Moreover, they induce mutuallyinverse isomorphisms

BrCliff(G,E)=BrCliff(G ,K).


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14 FRIEDER LADISCH

It is clear thatS=(SE K), and it is a result of Hochschild [6, Lemma 1.2] thatT=T E K. But to see that the inverse map sending T toT respects equivalenceclasses, we need some more general arguments, and it will be more convenient for usto reprove Hochschilds result.

Let K denote the skew group ring with respect to the action of on K (seeSection2). The ring K acts on K from the right by

x

c =

xc.

This makes K into a right K-module. So ifV is a vector space over E, thenVE Kis a right K-module. For a right K-moduleW, we still write W ={w W|w =w for all }.

The next lemma basically follows from the fact that K= M||(E) which is wellknown from Galois cohomology.

6.2. Lemma. For everyE-vector spaceV and every rightK-moduleW, we have

V=(VE K) and W=W E Knaturally, andEndK(W)=EndE(W). (In fact, we have a category equivalencebetween the category of vector spaces overE and the category of modules overK.)

Proof. It is clear that (VE K)=Vnaturally for any E-vector spaceV.Conversely, let W be a K-module. We have to show that wk wk is an

isomorphism W E K=W. Let aK. Then

a

=

,

a =

TrKE(a)

in K. Letb1,. . . ,bn be a basis ofK overE and leta1,. . . ,an be the dual basis withrespect to the form (x, y)TrKE (xy), so that TrKE (biaj) =ij. Then for all kK wehave

k=n

i=1

TrKE (kai)bi=n

i=1

TrKE (kbi)ai.

SetEij :=ai

bj K.It follows that

EijErs =ai

TrKE (bjar) bs= jrEis.

In particular, theEijs are linearly independent over Eand so, by counting dimensions,form a basis ofK. Since (

i Eii)Ers = Ers, it follows that

i Eii = 1. (We have

now proved that theEijs form acomplete set of matrix units.) With all this notationin place and equations proved, it is routine to verify that

W wn

i=1

wai

biW E K

is the inverse of the natural map W E K=W.The third isomorphism in the lemma is a consequence of the second.


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6.3.Lemma. LetW1 andW2 be two K-modules. Then(W1K W2)=W1E W2 .Proof. We have a natural injection ofW1 EW2 into(W1KW2). It follows fromLemma6.2that both spaces have dimension (dimK W1)(dimK W2)over E. Thus theinjection is an isomorphism.

Proof of Proposition6.1. A(G )-algebra over K can be viewed as a K-module.It is easy to check that the isomorphisms of Lemma 6.2are isomorphisms of(G )-algebras.

IfS1 andS2 are equivalent, thenS1 EK and S2 EK are equivalent, as explainedat the begin of this section.

Now assume that [T1] = [T2] in BrCliff(G ,K). To finish the proof, we need toshow thatT1 and T

2 are equivalent. By assumption, there areK[G ]-modules

P1 and P2 such that

T1KEndK(P1)=T2KEndK(P2).Taking centralizers of and using Lemma6.3, it follows that

T1 EEndK P1=T2 EEndK P2.(Note that (EndK Pi)

= EndK Pi.) From Lemma6.2, it follows that EndK Pi=EndE(Pi ). The centralizer P

i is an EG-submodule ofPi, and the isomorphisms

above are isomorphisms ofG-algebras. Thus T1 and T2 are equivalent.

We write

C(): BrCliff(G ,K)BrCliff(G,E), C()[T] = [T].6.4. Lemma. We have

C()(SC(G ,K)) SC(G,E) andC()(SCM(G ,K)) SCM(G,E).

Proof. Let

1 N G G 1be an exact sequence and V an FG-module with S=EndFNV, and such that theclass ofSis inSC(G ,K). Then for M=1() we have an exact sequence

1 M

G G 1,

where is G G G. The equality EndFM(V) = S proves thefirst assertion.

Let Irr Nbe an irreducible constituent of the character ofVN. Then F() =K= Z(S). Since acts faithfully on K, it follows that m = for all mM\ N.ThusM is irreducible, and it follows that [S] =M, , F. This shows, in particular,the second assertion.

6.5. Lemma. Let [S] BrCliff(G,E). If [SF L] SCM(G,K), then [S]|| SCM(G,E).


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16 FRIEDER LADISCH

Proof. Note that we can viewSFLas a(G )-algebra. Applying the corestrictionmap with G G , we get[10, Theorem 6.3]

CoresGG ResGG [SF L] = [SF L]||.

Since we assume Res

G

G [SF L] SCM(G,K) it follows[11, Corollary 9.2] that[SF L]|| SCM(G ,K).

ApplyingC() yields, by Lemma6.4, that [S]|| SCM(G,E). Recall that we want to prove that an arbitrary element a SC(G,E) is in fact

an element ofSCC(G,E), and that we have already reduced to the case where ahasp-power order andG is ap-group. The results of this section yield a further reduction:

6.6. Corollary. LetG be ap-group and let

a= , :

GG, F SC(G,E)

havep-power order. (So is semi-invariant overF inG andF() = E.) Let bea primitive|G|-th root of unity. Then there is a unique field L such that|L : F|is a p-number and|F() : L| is a p-power. If ,,L SCC(G,L()), then a =,,F SCC(G,E).Proof. The field L is uniquely determined as the fixed field of a Sylowp-subgroup ofthe abelian Galois group Gal(F()/F).

By assumption, E = F(). Since ,,F = ,,EG, we may assume withoutloss of generality that EG = F. Then|E : F| is a power ofp, since G is a p-group.Thus E L= F. With K= EL=E F L, the assumptions from the beginning of thesection hold.Now L() = LF() = LE = K and ,,L = ,,FF L. If ,,LSCM(G,K) then ,,F|L:F| SCM(G,E) by Lemma6.5. Since a = ,,F hasp-power order, it follows that a a|L:F| SCM(G,E) =SCC(G,E). This proves thecorollary.

7. R e d u c t i o n t o e l e m e n t a r y g r o u p s

Let L be a field of characteristic 0. Recall that a group His called L-elementaryfor the prime por L-p-elementary, if the following two conditions hold:

(a) H=P Cis the semidirect product of a normal cyclic p-groupCand ap-group

P.(b) The linear characters ofCare semi-invariant over L inH.

The second condition is sometimes expressed differently. We explain the connection.LetLin Cbe a faithful character ofC. Then is semi-invariant inH=P Cif foreveryyP there isGal(L()/L) such that y =. Now L() =L(), where is a primitive|C|-th root of unity. Then for Gal(L()/L), there is a uniquek = k() (Z/|C|Z) with = k(). The second condition is equivalent to: ForeveryyP there is Gal(L()/L) such that cy =ck() for allcC.

We need the following part of the generalized induction theorem:


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7.1. Proposition (Brauer, Berman, Witt[17, 12.6]). LetG be a finite group with|G| = bpk, where the primep does not divide b, and letL be a field. Then we canwrite

b1

G

=

X,aX,

G

with integers aX,, where (X, ) runs through the set of pairs such that X is an

L-p-elementary subgroup ofG and is the character of anLX-module.The next result is similar to results of Dade[1, cf.8, Theorem 8.24] and Schmid[16,

Lemma 5.1]. See also [26, Proposition 3.3].

7.2. Theorem. LetG be a finite group with normal subgroup NG such thatG/Nis ap-group for some primep. LetIrr N beL-semi-invariant, whereL is somefield of characteristic0 such that|L() : L| is ap-power, where is an exp(G)-throot of unity. Then there exists anL-p-elementary group X

G andIrr(N X)

such that the following hold:(a)G= X N.(b) [, N]0 modp.(c) L() =L().(d) For each x X there is = x Gal(L()/L) such that x = and

x =.

Proof. Write

b1G= X,

aX,G,

as in Proposition7.1.Set = Gal(L()/L) and

= TrL()L () =

.

Then

b||= [,b1G]N=X,

aX,[, (G)N]

=X,

aX,[,

t[

G:XN]

tXtN

N] (Mackey)

= X,aX,|G: X N|[, (XN)N

] ( is invariant inG)=X,

aX,|G: X N|[, (XN)N]=||

X,

aX,|G: X N|[, (XN)N],since (XN)

N is a character with values in L. Since b0 mod p, it follows thatthere is an L-elementary subgroupXand a character of an LX-module, such that

aX,|G: X N|[,NXN]0 mod p.


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18 FRIEDER LADISCH

Fix such anX andand set Y =X N. First we note that sinceG/N is a p-groupand|G: X N| 0 modp, it follows thatG= X N.

Let= Gal(L()/L) and set P= (X/Y) . This group acts on the charactersofNand ofY. The character Y is invariant under the action ofP, since it is the

restriction from a character ofXwhich has values in L. Sinceis semi-invariant inG, it follows that everyP-conjugate of is of the form with =Gal(L()/L)and thus isP-invariant, as is Y, of course. Thus YY is invariant under P.

For eachIrr Y, letS()be the sum of the characters in the P-orbit of. SinceYY is invariant, we may write

YY =

cS(), (c= [YY, ]Y)

where the sum runs over a set of representatives of the P-orbits ofIrr Y. It followsfrom

0

[,NY ] = [Y, YY] = c[Y, S()]

that there is Irr Y such that c[Y, S()] 0. After replacing by anothercharacter in its orbit (if necessary), we may assume that [Y, ]0 mod p. For therest of the proof, we fix a Irr Y such that

[YY, ][Y, S()][Y, ]0 mod p,and we show that this has the desired properties. Of course we already have that[, N] = [Y, ]0 modp. It remains to show the last two properties.

The groupP = (X/Y) Gal(L()/L) is a p-group by our assumption. Write Pand P for the stabilizers of and in P. We claim that P =P. First we show

PP. Since is semi-invariant in X, we have P= P. Thus P is normal in PandP /P=X/X= is abelian. Thus every P-orbit contained in the P-orbit ofhas the same length|P :P P|. Since [Y, xy]= [Y, x] for y P, it followsthat|P : P P| divides

[Y, S()] =

x[P:P]

[Y, x]0 mod p.

Thus|P:P P|= 1 and PP.As we mentioned before, is theP-orbit sum of and Y isP-invariant. It follows

that

[YY, ] = x[P:P][(x)YY, ]

0 mod p

is divisible by the p-power|P:P|. Therefore,P=P as claimed.From P= P it follows that a field automorphism over L fixes if and

only if it fixes . Thus L() =L(). Since is semi-invariant inG, there is, for everyxX, an x such that xx =. If is any extension ofx to L(), then(xY,)P = P. It follows that also xx =. The proof is finished. 7.3. Corollary. In the situation of Theorem7.2, let :GG be an epimorphismwith kernelN. Then,,L= , |X,L.


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Proof. It follows from Theorem7.2that Hypothesis3.1holds for

1 Y X G 1

1 N G G 1

|X

over the field L. Thus Theorem3.5applies and yields that

, |X,L= [S] ,,L,where [S]is the equivalence class of the algebra S= (iLN i)Y, with

i=

Gal(L()/L)

(ee).

It remains to show that [S] = 1, in other words, Sis a trivial algebra G-algebra over

L().Pick G-algebras A,,Land B, |X,L with A L()S=B , as in theproof of Theorem3.5.By Lemma3.4,we know that Sis central simple of dimensionn2 over L(), where n = [Y, ] 0 mod p. Since L() is a splitting field of allgroups involved, it follows that A LL() is a direct product of matrix rings overL(), and A L() L() is a matrix ring over L(). The same holds for B, and thusSL() L()= Mn(L()). Since|L() : L()|is a power ofpandnis prime to p, itfollows that S= Mn(L()).

The action ofG on L() extends naturally to an action ofG on Mn(L()). Viasome fixed isomorphism S= Mn(L()), we get an action of G on S. We write : G

Aut Sfor this action, that is we write s(g) to denote the effect of this action.

This action is different from the G-algebra action onS= (iLN i)Y we already have,but both actions agree on Z(S)= L().

It follows that for gG, the mapSss(g1)g is an L()-algebra automorphismofS. Thus there is (g) S such that s(g1)g = (g)1s(g) for all s S, orsg =s(g)(g). By comparing left and right hand side of

s(xy)(xy) =sxy =s(x)(x)(y)(y) =s(xy)(x)(y)(y),

we see that

(x)(y)(y) =(x, y)(xy) for some (x, y)L().Thus : G S= Mn(L()) is a twisted projective representation and Z2(G,L()).

Taking determinants in the above equation yields that n is a coboundary. Weknow also that |G| is a coboundary. Since|G| is a power ofpandn is prime to p, itfollows that itself is a coboundary. So after multiplying with a coboundary, wemay assume that (x)(y)(y) =(xy) for all x,yG.

Now we can show that Sis a trivial G-algebra. Let V = L()n and view V as arightS-module via our fixed isomorphism S= Mn(L()). We have to define a rightG-module structure on V such that S=EndL()(V) as G-algebras over L().


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20 FRIEDER LADISCH

The action ofG on L() defines an action ofG on Vwhich we denote by vvg.It has the property

(vs)g =vgs(g) for sS .Define v

g:= vg(g). Then

(v x) y = (vx(x))y(y) =vxy(x)(y)(y) =vxy(xy) =v (xy)for v V and x, y G, and (v)g = (vg)g for v V, L() and g G.Thus V is a right module over the crossed product L()G. The computation

(v g1)s

g=

vg1

(g1)sg

(g) =v(g1)s(g)(g) =vsg

shows that S=EndL()(V) as G-algebras. This finishes the proof that [S] = 1 inBrCliff(G,L()).

8. P r o o f o f t h e t h e o r e m f o r e l e m e n t a r y g r o u p s

Let F be a field of characteristic 0. We need the following lemma which states thatsemi-invariance is in some sense transitive:

8.1. Lemma. Let M, N be normal subgroups ofG with M N. Suppose that Irr M isF-semi-invariant inNand thatIrr(N| ) isF-semi-invariant inG.Then isF-semi-invariant inG.Proof. Let gG. There is Gal(F()/F) such that g = . We may extend to F(, ). For simplicity, we denote this element ofGal(F(, )/F) by , too. Theng is a constituent ofgM =M, and so there is nNsuch thatg =n. Since is semi-invariant in N, there is Gal(F()/F) such that n =. Thusg =.Since gG was arbitrary, is semi-invariant inG over F.

LetG be an F-elementary group with respect to some prime p. ThenG = P CwhereP is a Sylowp-subgroup andCis a normal cyclic subgroup ofp-order. In thenext result, we will only need thatGhas a normal abelian subgroupCsuch thatG/Cis nilpotent. In this case, every irreducible character ofG is induced from a linearcharacter of some subgroup containing the normal subgroup C [8, Theorem 6.22].(This is even true ifG/C is supersolvable, but we will also need the nilpotence ofG/C.)8.2. Theorem. Let

1 N G G 1be an exact sequence of groups. Assume that there is a normal abelian subgroup C Nsuch thatG/C is nilpotent. Suppose thatIrr N isF-semi-invariant inG. Thenthere are subgroupsA M H ofG and a linear characterLin A such that

(a) A H,G= H N andC A M=H N,(b) = N,(c) F() =F(), where= M,(d) ,,F= , |H,F.


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GN

N =

H

M= M

A

C

Our proof of this result follows closely the proof of Proposition 3.6 in Yamadasbook[26]. The latter is the caseN=G (that is, G = 1) of Theorem8.2, except forcondition(d).

Proof of Theorem8.2. Consider the set of all subgroups T Nsuch that(1) C T

G,

(2) there is Irr Tsuch that = N,(3) for all nN, there is Gal(F()/F) such that n =.

LetTbe minimal among the subgroups having Properties(1)(3). (Note that thereare such subgroups, for example, N.) If the correspondingis linear, we set A= T,= , H=G and M = N. Then = M = and(a)(d)of the proposition aretrivially true. So in the case where (1) = 1, the proof is finished.

Assume(1)> 1. Our first goal is to find a proper subgroup T0< Twhich is normal

inGand such thatis induced from a character 0Irr T0. The non-linear characteris induced from a linear character of some subgroup B such thatC B < T. LetSbe a maximal subgroup ofT containingB , and thus C. Then is induced from some

character ofS. Since S T (because T /C is nilpotent), we see that T\S= 0. Set

U=gG S

g.

Then C UG. We claim thatvanishes on T\ U. Since vanishes outside S, itfollows that g vanishes outside Sg forgG. But g andare Galois conjugate (byLemma8.1), and thus vanishes outside Sg. Since gG was arbitrary, it followsthat vanishes outside Uas claimed.

Since U, T G andG/U is nilpotent, there exists T0 G such that UT0 T

G and

|T :T0

|is a prime p. Since vanishes outside U, it vanishes outside T0.

Since|T :T0|= p, we see that =T0 for some 0 Irr T0. We have found a T0 aswanted.

In the following, set

G0={gG| Gal(F(0)/F) : g0 =0},N0 =G0 N and 0 =N00 . Then N0 =N0 =N = and 0 Irr(N0| 0) is theunique element with N0 = (Lemma3.7). Corollary3.8, applied to the extension

1 N G G 1


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22 FRIEDER LADISCH

and the normal subgroup T0N, yields that the following hold:(a)G=G0N,(b) = N0,(c) F() =F(0),

(d) ,,F= 0, |G0 ,F.(We can also show thatG=G0Tby applying Corollary3.8to the extension TGG/T, but actually we will not need this for the proof.)

GG0

HN

N0

M

T

T0

A

C

F i g u re 2 . Proof of Theorem8.2

Note that N0 = (T0)

N =N =, so(1)and(2)hold for T0. It follows from theminimality ofT that(3)does not hold forT0, which means thatN0 < N. Thus we mayapply induction (on|N/C|) to conclude that there are subgroups A M HG0and Lin A such that Properties(a)(d)hold withG, N, replaced byG0, N0,0. It follows thatG=G0N=H N and M=H N0 = HG0 N=H N (seeFigure2), that = N0 =

N, and that F() = F(0) = F(). Finally, we have that

,,F= 0, |G0,F= , |H,F. 8.3. Corollary. In the situation of Theorem8.2, ,,F SCC(G,F()).Proof. By Proposition 4.4, it suffices to show that ,,F SCM(G,F()). ByTheorem8.2, we have ,,F= , |H,F, and it is obvious that the conditions in

Theorem8.2yield that , |H,Fis inSCM(G,F()). (See Definition4.1.) 8.4. Remark. In fact, the construction from the proof of Theorem8.2itself yields anMthat also has the following properties:

(e) M={nN| Gal(F()/F) : n =},(f) M/A=Gal(F()/F()).

Proof. By Lemma4.2,(e)together with M Irr M implies(f).To see (e), let N0 be as in the proof above and writeM ={n N |

Gal(F()/F()) : n =}. By induction we may assume thatM N0 =M. Since


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REFERENCES 23

N =Irr N and N0 =0Irr N0, we know from(f) respective Lemma4.2thatM/A=Gal(F()/F()) and M/A=Gal(F()/F(0)). But since F() = F(0), itfollows that M=M. The proof is finished.

9. P r o o f o f t h e m a i n t h e or e m

For the convenience of the reader, we summarize here how Theorem Afrom theintroduction follows from the various results proved so far. We restate Theorem A,using the notation introduced in Section4.

9.1. Theorem. LetE be a field on which the finite group G acts. ThenSC(G,E) =SCC(G,E).Proof. Let a = ,,F SC(G,E). We have to show that a SCC(G,E). ByLemmas5.1and5.2, we may assume thata hasp-power order and thatG is ap-group.By Corollary6.6, it suffices to show that ,,L SCC(G,L()), where L is suchthat a splitting field ofG and all its subgroups has p-power degree over L. ThenCorollary7.3yields that we may assume thatG is L-elementary for the prime p. Inthis case, Corollary8.3yields that ,,L SCC(G,L()).

R e f e r e n c e s

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3. Everett C. Dade, Extending irreducible modules, J. Algebra 72, no. 2 (1981), 374403,d o i: 10.1016/0021-8693(81)90300-8,MR641332 (84g:20009)(cit. on p.4).

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779799, d o i: 10.1016/j.jalgebra.2005.12.034,MR2381809 (2008j:20015)(cit. onp.4).

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6. Gerhard Hochschild, Automorphisms of simple algebras, Trans. Amer. Math. Soc. 69,no. 2 (1950), 292301, J s t o r: 1990359,MR0037834 (12,315b)(cit. on p.14).

7. I. Martin Isaacs,Extensions of group representations over arbitrary fields, J. Algebra 68,no. 1 (1981), 5474, doi: 10.1016/0021-8693(81)90284-2, MR604293 (82f:20025)(cit. on pp.2,4,5).

8. I. Martin Isaacs,Character Theory of Finite Groups, New York: Dover Publications,Inc., 1994, xii+303, corrected reprint,MR1280461(cit. on pp.1,8,17,20).

9. Frieder Ladisch, Character correspondences induced by magic representations, (Aug.

2011), arXiv: 1004.4538v2 [math.RT] (cit. on p.7).10. Frieder Ladisch, Corestriction for algebras with group action, (Sept. 10, 2014), arXiv:

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U n i v e r s i t t R o s t o c k, I n s t i t u t f r M at h e m at i k , U l m e n s t r . 6 9 , H a u s 3 ,

1 8 0 5 7 R o s t o c k , G e r m a n y

E-mail address: [email protected]
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