RATIONAL REPRESENTATIONS
OF A METACYCLIC p−GROUP
YoulcHI IIDA AND TosHIHIKo YAMADA
(Received April 8,1993) ABsTRAcT. The irreducible rational representations of a metacychc p. group, whereρis an odd prime, are completely determined. AMS 1991 Mathematics 5!%句ecεClassification, Primary 20C15. Key words and phrases・Rational representation, metacyclic P−group.1.Introduction.
Let G be a finite group and let Irr(G)be the irreducible complex char− acters Of G. For X∈Irr(C), put Ω(x)−mQ(x)Σxσ,σ∈G・1(Q()(1)/Q), σ where Q is>the rationals and mQ()c)is the Schur index of)(over Q. Let {Xl,)(2,… ,)〈8}be a subset of Irr(G)such that Xτis not algebraically conj ugate to)(」(i≠ゴ), and that any)(∈Irr(G)is algebraically conj ugate to some)(i. In order to determine the irreducible rational representations of G, it suHices to obtain a rational representation Xε, whose character is Ω0(∂,(i=1,2,… ,8). F()r. the rest. pf t.he .paper, p is an odd、prime. R)r aρ一group G, C. R)rd has obtained the fQ}lowing fundamental result. Theore.m l.1(R)1d [2]).’Each『』irred’ucible rational:reptesentatfon. of a finite pl・gro砲js jnduced丘om the’ ’irreducible lrational rePtiesentati・n・fdegree p−10n a section of order P. 1 −
Let G be aρ一grbup and X∈『’Irr(G), The procedure fbr obtainihg an irreducible rational representation X of G,with characterΩ(X), is as79
80
A METACYCLIC p−GROUP
follows:Find a subgroup H of G with a linear characterψsuch that(i) )(1=ψG,(ii)Q(x)=Q(ψ).’ThenΩ()c)=Ω(ψ)G. Furthermore, we have an irreducible rational representationΨof H with characterΩ(ψ)(c£{5, Proposition 1D. Consequently, the induced represeritation X=ΨG is an irreducible rational representation of G with characterΩ(j)c). Thus we will call the above subgroup H of G and the linear characterψof H, a required ραir{H,ψ}丘)r)(∈Irr(G). Remark 1.2. Notation being「as above,’let N=kerψ. Let Ho be thesubgroup of H such that」U⊃Ho⊃」V and[Ho:N]=P. Then Fbrd[2]
observed thatΩ(ψ)=Ω(ψo)H, whereψo is a linear character of Ho with kerψo=N. So,Ω(ψo)is the character of the faithful irreducible rational representationΨo of degree p−10n the group Ho/1V of order p. HenceX=ΨG=(Ψξ)G=Ψ8.
Let G be a metacyclic p−group and let X∈Irr(G). The purpose of this paper is to obtain an irreduCible rational representti tioh X with character Ω(X).By the above result, it su伍ces to obtain a required pair fbr X. Since the image of G by a irreducible complex representation is either a metacyclic p−group or a cyclicρ一group, we may assume that X is faithfu1. Then, G is called a∫faithful metαcyclic p−group and the faithful irreducible complex characters of G is denoted by FIrr(c). In Section 2, faithful metacyclic p−groups are classi丘ed into two types of grbups GI and G2.(]1 is not a semi−direct product, but G2 is split. In Section 3, we determipe FIrr(G∂,(i=1,2). In Section 4, a required pair is given f()r each)(∈FIrr(Gi),(i=1,2). Thus, for any metacyclic p−group G, the irreducible rational represen− tations of G are compl6tely determined.2.Classification.
In this section we will classify faithful metacyclicρ一groups. The center Z(G)of a faithful metacyclic p−group G is cyclic(c£Isaacs[4, Theorem (2.32)j).Let G coptaip a cyclic norma1 subgroup♪X=〈x>of ordel’ p’},.n≧2,with a cyclic factor group G/X=〈Xy>of order pt, t≧LThen we
may write
(2.1) G=〈x,ylxpnニ1, Ypt=xpm, yxy−1ニx「〉,
(2.2) n≧2,n≧m≧1, t≧1, p↑rand r≠1(mod pn).
1・fa・t, if y・“−x・m・, P↑・, th・n w・ha・・th・f・・m(2.1)・もy・epl。,i。9’・t with xα. Furthermore, we have (2.3) rpt≡1 (2.4) r≡1 (mod pn), (mod pη一m). Since(Z/ρZ)*have no element of orderρ, n≠1. Conversely, it is wel1− known that if n, m, t and r satisfy(2.2),(2.3)and(2.4), then(2.1)defines ametacyclic group G・=〈x, y>of orderρη+¢. Lemma 2.1. The integers r satisfyi皿g(2.2)and(2.3)are givenりy the 五)110wf119: ア≡1十pn−tk (mOdρη),
励ere 1≦1≦min(ちn−1)andρ仕,1≦k<pl.
Pm(’f, Cleat.□ We suppose the integer r of(2.1)is a primitiveρε一th root of 1(modρn),1≦1≦min(ちn−1). By Lemma 2.1 we have r≡1十pn−tK・(mod
ρη)fbr some.k such that p†kand 1≦k<pt. There exists an integer〃★ (p↑.uk)such that 1十p刀一1≡(1十pη一lk)レた(modρη). Then we have G=b,y>=〈〆た, yUk>, (〆りPπ=1,(y”り〆=(x・り・m, (ピり(〆り(y・り一1=(x・り1+P ‘. Hence we may assume that the group G defined by(2.1)and(2.2)is of the fbrm: (2・5).C−〈x, ylx・” 一 1, ypt−x・m, Y・y−1−x’+P”“t>. We also note that m≧1, because by(2.4),1十pn−1≡1(modρπ一m).(C…1)1≦1〈・t.W・hav・Z(G)=〈x・t, yρt>and・IZ(G)1−pn+・−2t.
Since Z(G)i・cy・li・, w・hav・y・t−x・t.1・deed, ifm>1.・th・・Z(G)w。。ld have no element of order p九+t−21, because fOr i,ゴsuch that O≦i<ρn−1‘ and O≦ゴ<pt−t, (・p∨ゴ)・叶?”2t“1−。pn+£一’−1ゾ+t+1」 =(x・m)P ‥1ゴ =1.82
A METACYCLIC p−GROUP・
On the other hand, if m=1, then Z(G)is clearly cyclic. So th6 group G in(2.5)is of the fbrm: Cl = 〈x, !ノ l xpn =1, Ypt=xpt, 9/XZノー1 :=xl+pn−1>. (Case II)1≦1=t. Z(G)is necessarily cyclic because Z(G)=〈xp¢〉.・ Then the group G in(2.5)is isomorphic to the fbllowing(]2: G、=〈x, y・1・xpn=1, Yρt=1, y・y−1−x’+・π一z>. Indeed, put r=1十ρn−¢. Because pn l l rP£−1andρη一£ll r−1, we have t P£1ド≡1.S・th・・e exi・t・ゴ・u・h th・t t ゴ「≡1+・m≡・(m・d・P・・), t . ¢ ・7・ρ 一1 t because m≧1=t. Fbr thisゴ,(xJy)P =x3 r−1 Yρ =1. ConsequentlyG=〈x,y>=〈x,内〉竺G2.
Proposition 2.2.. Let p be an odd pr∫me. Let(D be a faithful meta− cyclic p_group de五ned by(2.1)一(2.4). Then G js isomorl)hjc to one of the fbllowlillg groups: Gl=〈x, y l xpn=1, Ypt=xpi, Yxy−1=xl+ρπ一1>, 1<1<n−1and l<£. G、=〈x,y1・・” =1, y・‘−1, yxy−1−・c’+・n”t>, 1 <t<n−1.3.Determination of the faithfu1 representations.
We quote the useful Lemmas. Lemma 3.1(Iida−Ya血ada[3, Lelnma 4.2D. Let G be a毎∫th血1 meta− cyclic group of order nm. Let G=〈α, b l an=1, bm=α8, bab−1=.α「),where(n, r)=1, rs≡8(mod n)and〆π≡1(rpod n). Let t≧2be the
least integer such that r£≡1(mod n). Then eve】ry faith血1 jrreducible representation pf G is induced from. a one−dimensio皿al representation of Ht=〈α, bt>.Lemma 3.2([1, Corollary(45.4)D. Let G be a finite group alld H a subgroup of G. Let・T be a one−djmensjonal rePrese皿tatio刀of H. Then the・induced・m・皿・m」al r()presentati・n TG・fG・is・irreducible・if・and・nly if, f・・ eaCh X¢H, there・eXi・t・y∈X−iHX∩H・SUCh・that・T(y)≠T(XyX’1). We will determine the faithful representations of the groups GI and G2. Fbr a natural number 8, a primitive ps−th root of unity is denoted by 〈P・・ (1)The groups Gl. P・tG=G、一〈x, y>・nd・H=〈・, Y・’〉. lt・f・ll・w・fr・m Lemm・3.1 that every)(∈FIrr(G)is induced from a linear character of H such that
θ・,。・・一(麺pZH(㌫.t.21,
where O≦u<pn,0≦μ<pn+t−2t andガ≡μ(mod pn一り.
C−∪±51Hy・. Th・n w・hav, th。 i・duced,ep,e,ent。ti。n。f G,Uv,μ:XH
yト〉脇
0
ζジ1+pn−∼) ζ㌫・t−21 1 . . . る . −r. 1 0 ζ㌧・+・n”t)アに1 ◆ ,Let
By Lemma 3.2,σ“,μis irreducible, if and only ifレ≠レ(1十pn一りz(mod pn),1≦i<pl. This is clearly equivalent to the condition p tレ. Befbre we show that every irreducible representationσレ,μ(p tレ)is faithful, we determine the character field Q(θGv,μ), whereθ緩μis the char− acter ofσレ,μ. We have the fbllowing lemma. Lemma 3.3. Let p be an odd prjme. Letα, b, m and K be integers such that m≧2,1≦b<m, a≧m−band P tκ. Then the f()110wing」holds:0
= .ア κ・ bp㌃
φ
ーゴΣ⊇
84
AME.TACYCLICρ一GROUP
ProOf. Fbr eachαandβsuch that O≦α<pa一πt†b and O≦β〈−pm←b,
we have
くll{+,b・)ハ+’一ζ舞・b・)β. So we have pa−1 pM−b−1 Σζ雷bκ)’−P・−m+bΣζ㌫pbκ)∵ i=O i=OFbr eachλand 7 such that 1≦λ≦p−1and O≦ty<pm−b−1,
斑pb・)ρ ト1λ+ツー斑・m−1κ・λ・・+・‘・)t「 一窮pm−1κλ)(1十pbκ)t「 −cλ(・+ρb・)寧ρbκ)ty 一ピλ群pbκ)守.SinceΣ《ゴご一Σ《ゴζ;一一1, w・hav・
pm−b−1 pm一ト1−1Σ雷pbκ)’一(1+Cp+…+〈3−1)Σ、ζ舞pbκ)z−0・・
i=O i=O Si・・e Uv,。 i・i・duced加m th・n・・m・1・ubg・・up Hニ〈x, y・’〉,it・f・ll・w・』th・tθ£・μvani・h…nG−H. A・d・1…ly w・hav・, f・・鋤・tj(0≦i〈
P?,o≦ゴ<Pトり,
θ緩。( ・ ↓・xtyP’2)一く㌫,−21θ三。(め・ By the way, fbr any integers 8 andゴsuch that O≦8〈landρ†ゴ, pl−1 θ三。(・psゴ)一Σθ“,。(躍8ゴy一り ㌶ 〒Z) e。,。(x)pS」(1+P ‘)’ 冨 一Σζ鐸3ゴ(1+Pヲ 芦 一亙(くbu{一・)9+ρ九一1)’一゜by L・mm・3・3・1・i・ea・ily・een・h・tθ5。(xpsゴ)−P’,1;・{一・ f・r・・y i・t・gers 8andゴsuch that l≦8≦namd p fゴ. Thus,、 (3.1)
嚥・)一
o1:ζ蹴・ll隠1あ
It follows that Q(θGUIμ)=Q(〈pn−t, kpn+t−・t)=Q(Cpn+・−21). Now we show that every irreducible representation U〃,μis faithful. Let x・”・・’J’ クk・・θ緩。(・≦i・〈・pn−‘,・≦ゴ<pt一り. Th・nθ6。(x・’i・ptゴ)− P’(芸一・(㍑ト・1 ・pt・C・n・eq・・n・1・,(嬬;邊・」−1and・・ pt−lyi十μゴ≡0 (mod pn+t−21).Then we have p£一’1ゴbecauseρ†μand n十¢−21>舌一1. So we have
ゴ=0・and i−Ob・・au・e G−1−1 and・P t v・lt・f・ll・w・th・t k・・θ£。=1・ (II)The groups Gl2. Put G=G2ニ〈x, y>and H=〈x>. It follows from Lemma 3.1 that every X∈FIrr(G)is induced from a linear character of H such thatθ・・XH(脇,
wh・・e O≦〃<P・. L・t G一噛1W・. Th・n w。 h。。。・he i。duced
representation of C:UV:Xト〉
yト〉0
1 1 ・ φ1+P”’t) . ふ . ’・D 1
. 0 ・ ● . . 1 一 t 、 P ︶ t 一 n P 十 1 ︵n己
, By Lemma 3.2, Uu is irreducible, if and only if u≠v(1十pn−1)i.(mod pπ), 1≦i<pt. This is clearly equivalent to the condition p f v. Thenθ〃is a faithful character of五r, and soθ9∈FIrr(G).86
A METACYCLIC p・GROUP
we now consider the character field Q(θ9), Whereθξis the character ofσ“. Sinceσレis induced from the normal subgroup H=@〉,it fbllows thatθg vanishes on G−H. Fbr any integers 8 and j such that O≦8<t and p†ゴ, pt−1 θ9@ρ8ゴ)一Σθ.(Y’xpsゴy一り ㌶ 一Σθ。(x)・3ゴ(1+・ り’ 昌 一Σ(㍑8元(1+P”’t){ ㌶ 一Σ(ζち.,)(1+pn一り㌧o i=o by・L・mm・3・3・lt i・easily・een th・tθ9(xpsゴ)−ptζ㌫、 f・・ally irlt・gers・andゴsuch that t≦5≦nand p{ゴ. Thus,
(3・2)θ9(…1)一
o1:雛,llぽ!=°,
It fbllows that Q(θ夕)=Q((;pn−・).Summarizing, we have
Proposition 3.4. Notation js the 5ame a8 jn Proposition 2.2. Th en,(1)FIrr(G・)一{θ£1。11≦・<pn,1≦μ〈pn+t}21, P↑・,・≡
μ(mod pπ一り}, Q(θξ1μ)=Q(ζpn+t−21), whereθ51μjs given by(3・1)・ (II)FIrr(G2)={θ9・11≦u<pn, P↑〃}, Q(θ9・)=Q(ζP・・一・), whereθ9・ is given乏〕y(3.2).4.Determination of the rational representations.
Now we will give a required pair丘)r each X∈FIrr(Gi),(i=1,2). (1)The groups Gl.P・tG == G、=〈x, y>and’・H=〈x, y・t>. lt i・easy.t・・ee th・t f・・ ・v・・yθ緩。(1≦〃<pn,1≦μ<pn+t’21, P t・, u≡μ(m・d・pn一り)th・・e ・xi・t・an i・t・g・・μ’(μ’≡1(m・d P”’1))・u・h th・tθξ。’i・alg・b・ai・ally
conjugate toθ£μ・Hence we may assume u=1and simply writeθμin
place ofθ1,μ. Note that 1≡μ(mod pη一り. We will distinguish two cases. (i)t−21≧0. We have Q(θμ)= required pair fbrθS・ Q(〈pn+t−21)−Q(eS). S・{H−〈x, Y・t>,θ。}i・a (ii) t − 21 < 0. We have Q(θμ)=Q(Cpn)anq Q(efi)=Q(〈pn+・一・・),(n>n+t−2の. Put Fニ〈x, yPt−1>・First we■ill show that Q(θξ)=Q(θS). Let.F= U碧ご一IHy・t−’i. Since・H・i・a−・1・ubg・・up・f F,θ‘(,)一・f・・ 9∈F−H.・lt・i・e・sy t・・ee th・t f・・掬・tゴ∈H(0≦i〈・pn,0≦ゴ<pt一り, θS(∂・・l」)一(㌫、.、、θs(・‘)・ It follows from Lemma 3.3 that fbr integers 8 and元such that O≦8<21−t and p↑ゴ, eS(xps・)rΣ碧ご一1θ。(y・t一㌔・SjY−・t’‘i) 一Σ馨ご一1θ。ωρ8・(・+・n−’)pt”,” 一Σ≧♂−1〈謬ρ1+t−21z)’=・, t一ε 2t−t t−t 2t−t where z is some integer such that p↑z. Since YP xp y一ρ =xp ; we:have・ ’ eS( 21−t・¢P J)−P2‘“t(b”:’−tゴーP21−‘くln+,.、、,・≦ゴ<P・+‘−2t’ So we have鋼)一{ll↓一螂認, ll三↓1,ご
188
A METACYCLIC p−GROUP
It follows that Q(θ‘)=Q(Cpn+t−・・)=Q(θξ)=Q((θ5)G)・ There exist・integers c andκsatisfying (4.1) cκ≡二1(mod pn−1), (4.2) P’・一{(1+P・一’)・¢−1}/{(1+P・一‘)pt”‘ 一 1},(P†・). We take an integer c satisfying(4.1)and put ’ 21−t t−t >, K6=〈xρ・.,Wρ Lemma 4.1. Let Kc be as above. Then, 刎一£ ¢−t(i) κc = 〈コ;P >×〈xCyP !・ (’i)The f°11°痂g卿.ζe6ne/s a 1’nra「gha「hcte「°fK・: t−l 21−t ψc :xp ..・←→〈;pn+・二2・, xCyρ ・→L P・・Of lt foll・wS easily伽m(4.1)th・t th…d…fx・y・t−‘i・pl. N・t・ th・t・xp2t−t c・mm・t・・With・W・t”.・lf g∈〈¢・2‘一¢〉∩〈x・yPt”t>, th・n th・・e ・xi・t i・t・gers〃,λ・u・h th・t O≦〃<pn+£−21,0≦λ<pl a・d 9−(xp2t7t)・一(X・y・t’t)λ. So YPt−1λ∈〈x>. This means that ptいand soλ=0. It fbllows that 9−1.C・n・eq・・ntly・K。 i・a・di・ect p・・d・・t・f〈・・2‘−t>・nd〈x・y・£一‘〉. Th・statement(ii)now fbllows immediately.口
we observe thatQ(ψ。)ニQ(〈P・+・−2t)=Q(θξ)=Q(θξ)・we will show that there exists an integer c satisfying(4.1)such thatθξ=ψぎ. Lemma 4.2. Leげ, H and K, be a8: ≠b盾魔?D Then‘ the f()110fUings hold.、 (i) F = H」Kc. (ii)H∩K。=〈xp2‘’t, .ypt>. (iii)[F・珂=P2t’tニ[F・’K。】.Proof. Clearly(i)holds since F>H∋xandκc∋xcyPt一ε.Le七山be ’the
integer SatiSfying (4.3)P・・一・ω一{(・+pn−1)pL・}/{(1+P・一り・t∵−1},(plω).Because yρ1・=(xp2t−t)一ω(x・yPt−1)P2t−t∈K。, H∩Kc⊃〈xp2t−t, Yp−t>. Oh the other hand, if H∩Kc∋h, then there exist integers v,λ, i alld j such
that O≦v<pn+t−2110≦λ<pl,0≦i<pn,0≦」〈pt−l and
九一(xρ2t’t)〃(w・←’)λ一掬〆ゴ. Th・n y・’ゴーハ∈〈x>,・・Pε」−P・−1λ≡0(m・d・pt), a・d h。nce・P2t−t lλ. It・f・ll・w・easily・fr・m thi・th・t・P2t−t l i. C・n・eq・・ntlyん∈〈xρ2’”t, y・t>. The proof of(ii)is complete.(iii)is clear, because I FI=pn+l and I Kc l= pn+t−1. □ Lemma 4.3. Letωandμbe aS above. There exists an in’ teger c satisfying (4.1)such that (4.4) 一ω≡μ(mod pη+t−2り. Proqf All the integers c(modρπ)satisfying(4.1)are ・=一κ’ {pn−1λ, o≦λ<pl, whereκ’is an integer such thatκ’κ≡、1(mod p九一り. So it is enough to show that there exists an integerλsuch that O≦λ<pl and (4.5) 一(一κ’十pn−1λ)ω…≡μ(mod pn+t−2り.We have
xpt │〆一(y・’)・目一((Xρ2t−t)一ω(w・t”t)ρ2‘−t)pt−1 −(〆一£)一ωρt−1(wρt−’)ρ1−。一・‘ω. S・P’ E−P↓ω(m・dpn), and h・nce一ω≡1(m・d pn一り. fU・th・rm・・e, by (4.1)we have cκ≡−1(modρπ一り. We also recallμ≡1(modρπ一り. Let ω’be an integer such thatω’ω≡1(mod pη+t−2り. Using these relations,we have
Cκω(κ’一μW’)≡Cκκ’ω一Cκμωω’ :=ω一Cκ≡−1十1
≡0(mod pn一り.’90
A METACYCLIC p−GROUP
Because p t crccv, pn−l lκ’一μω’. Letλbe the integer such that λ≡(’・1’−Pω’)/pn−1(m・d pt−1)・nd O≦λ<pt−t. Then it is easy to see thatλsatisfies(4.5)and that O≦λ<pt(pt−1<Pり.ロ
Take an integer c satis」lying.(4.1)and(4.4). Let Kc andψc be as in Lemma 4.1 .. We are now ready to show thatθξ=ψ『..By Lemma 4.2 andthe Mackey subgrOup theorem,
(θS,ψζ)F=(θ司κ。,ψc)K。=((θμIH∩K。)κ・,ψc)、κ. =(θμIH∩κ。,ψciH∩K。)H∩k。. W・wi11・h・wθ。一ψ。・n H∩K。一〈xp2t’t, y・’〉. Cl・a・ly, 2t−t θμ(〆 ) 21−t 2t−t=¢n =〈;pn+t−2t=ψc(xp ). Using the decomposition’@of yρ!in the proof of Lemma 4.2, we have ψ・(y”’)一ψ6((x・21”t)一ω(w・t”)・21’t)一(㌶.、、 一箏.ト』、一θ。(y・t),by(4・4)・Henceψc=θμon H∩Kc, qnd so(θS,ψξ)F=1・Since
θξ(1)=p2t−t=ψ三(1), it fbllows thatθξ三ψ『, as desired.We haveψξ=(ψ三)G=(eS)G=θfi and Q(ψ。)=Q(ζp・tt二・り〒
Q(θS)・=Q(θff)=Q(ψ三)・Thus{Kc,ψc}is a required pair forθfi・ (II)The groups G2. The groups G2 have been dealt with by Ford[2, p.599】. Put G=G2= 〈x,,,y>.、and., H.r〈x>. It・is . easy to、5ee that・.everyθ9・(1−・≦u〈pn, P†〃) is algqbraittqlly. ponj ugate to each. other. Hence we.・May assume u=、1 and. simPly Writeθ:in lplace Ofθ1・Sgt ・・ 二 … 、:‘ . ...・ 「 ξ κ=¢・t, y>, ψ・:.塗ρもト・㍍π一;ニダ亡→.1.‘\『 Then we have Q(ψ)・=9(〈P竺二ζ.i=. Q(θ『)・..We will show thatθG=ψG・Lemma 4.4.‘Let G, H and K be a5 abbve. Then・ the followings hold. (i)G:=HK. (ii)H∩K=〈xpt>. (iii) [(] : H] = ρt = 【G : 」K’】・