CHAPTER
VIII
Blocks with Cyclic Defect Groups
The earliest attempt to determine and describe the irreducible ordinary and modular characters lying in a block was made by Brauer [l], who successfully analyzed blocks of defect 0 and 1. Although Brauer’s methods did not extend to other cases, Thompson [ 11 later gave a new proof of these results which did generalize in certain special cases. By exploiting Thompson’s technique, Dade [2] extended Brauer’s results to all blocks with cyclic defect groups. We shall not prove here the full theorem of Dade, but using his methods we shall give the analysis in an especially important case, namely, for blocks of defect 1 for a group whose order is divisible by the first power ofponly (Theorem 8.5). A number of theorems dealing more generally with blocks with cyclic defect groups lead up to this result. In the final two sections, we apply these results to prove theorems of Brauer, Feit, and Thompson on the existence of normal Sylow psubgroups in linear groups. 8.1 Extending characters from normal subgroups
In the following sections we shall need on several occasions a method of extending characters from a normal subgroup H of a group G to functions that are characters on G. There are several results of this kind known but here we only deal with the simplest ones. Recall that if 8 is an ordinary (or a modular) character of a normal subgroup H of G, then for each y E G we define B Y as an ordinary (respectively, 197
VIII
198
BLOCKS WITH CYCLIC DEFECT GROUPS
modular) character of H by Oy(x)~=O(ylxy)for all x in H (respectively, H"). Our first result deals with extension of a character 0 from a normal subgroup H of G up to a character of G. Clearly a necessary condition for this to be possible is that OY = 0 for all y E G. In some cases this is also sufficient. Theorem 8.1 Let H be a normal subgroup of the group G such that G / H is a cyclic group of order h.
(i) Let x be an irreducible ordinary character of H such that x Y = x for all y E G. Then there exists an irreducible ordinary character of G such that CH =
x*
(ii) Let i+h be an irreducible modular character of H such that i+hY = i+h for all y E G, and let p be an ordinary character of degree 1 of G with H = Ker p. If p Xh, then there exists an irreducible modular character 4 of G such that = i+h and i+hG = 4 p$ + . . . ph '4, where each pi# is an irreducible modular character of G (pi denotes the ith power of p).
+
+
Proof First let T be any irreducible matrix representation of H over an algebraically closed field K O .For each y E G define the representation Ty of H by T ( x ) := T(y 'xy) and suppose that for ally E G,T yis equivalent to T. Then there exists an invertible matrix ay such that F(x)= a; 'T(x)a,for all x E H.In particular, if we choose y so that Hy generates G / H , and write a for ayr then we get u  ~ T ( x )=u ~ Tyk(x)= T ( ~  ~ x = y ~T)( Y  ~ ) T ( x ) T (since ~~) yh E H. Thus T ( Y ~ ) ucommutes ~ with T ( x ) for all x E H.Since T is abso~ for some scalar a. E K O .Since K Ois algelutely irreducible, T ( Y ~ ) u=aol braically closed we can find a E K O such that ah = a. and then T(yh)= (aa)h. Put a. = aa, and define Toon G by To(xyi):= T(x)ab for all x E H, i = 0,1, . . ., h  1. We claim that Tois a representation of G over K O . Indeed, if x,x' E H and 0 Ii, j < h, then T,(xy')T,(x'Y') = T(x)abT(x')a,'ab+j =
T(x)a'T(x')a'ay
= T(x) T(y'x'y ')a;' = T(xy'x'y')p.
Since ub+j= T(yh)ab+jhif h Ii + j c 2h, we see that To(xy')To(x'9) = T,( xy'x' 9) as required. Note that the restriction of the representation To of G to the subgroup H equals the original representation T of H; in particular, this shows that To is also irreducible. We now consider the situations described in (i) and (ii).
8.2 BLOCKS WITH NORMAL CYCLIC DEFECT GROUPS
199
(i) In this case take K O as a field of characteristic 0 and let T be a representation affording x. Since x Y = x for all y E G by hypothesis, and T y affords x y , we have Tv equivalent to T for all y E G (Corollary 1 of Theorem 2.3). Thus we obtain an irreducible representation To of G extending T and the character afforded by To satisfies cH = x. (ii) In this case take K O as a field of characteristic p, and let T be a representation of H over K Owhich affords the irreducible modular character $. Since = $ for all y E G by hypothesis, the (Frobenius) characters of Ty and T are equal, and so TY is equivalent to T. Hence we again obtain an irreducible representation To of G over K O extending T. The irreducible modular character 4 afforded by To has the property 4" = $. Define I); on G by 6 = $ on H and = 0 on G\H. Then
4
c
h 1 $C(z):=
for all z E G,
$(y'zy')
i=O
where y is chosen such that H y generates G / H . Hence if z i s H otherwise. On the other hand, for i = 0, 1, . . ., h  1, pi4 is the irreducible modular character of G afforded by the representation T defined by T(z) := T0(z)p(z)'. Since Ker p = H, p(z) is a nontrivial hth root of unity for all z $I H. Therefore
This shows that
Hence $' = CFZ; p i 4 as asserted.
8.2 Blocks with normal cyclic defect groups Throughout the remainder of this chapter we shall use the notation first introduced in $3.3. Thus G will be a group of order g, A will be a padic algebra which is an integral domain of characteristic 0 containing a primitive 8th root of unity, and K will be the field of quotients of A. Also K A is the unique maximal ideal of A and k = A/nA is the residue field of characteristic p. Both K and k are splitting fields for G and all its quotient groups as well as subgroups.
VIII
200
BLOCKS WITH CYCLIC DEFECT GROUPS
Let Q be a cyclic normal psubgroup of G. As we know from Lemma 6.2A,
Q is contained in each defect group of a block B of kG. In the present section
we shall consider the case where Q is the (unique) defect group of B. The theorem below describes the set of irreducible ordinary and modular characters that lie in such a block. The proof requires the following lemma. Lemma 8.2 Let Q be a normal psubgroup of the group G and put := QCG(Q).Let b be a block of k H . If x is an irreducible ordinary character of H lying in b, then each irreducible constituent of xG lies in bG.
H
Proof Let Wl, V 2 , . ..,Vsbe the conjugate classes of G with class sums cl, c 2 , ..., c, in kG and put h i = l V i J , V F = V i n H and c : ~ = ~ ~ ~ ~ , . x (i = 1, 2, . .., s). If o is the central character of kH associated with the
block b, then (see $6.3) the central character of kG associated with the block bG is mG given by mG(ci) :=~ ( c f ) (i = 1, 2, . .., s). (1) If %" is a class of H with class sum c' and h' := IV' 1, then by Theorem 4.2B we have
o(c') = h'x(y)/x(l) =
c x(y)/x(l)
(where the bar denotes reduction modulo n) with y E W'.
X€W'
Since H
4
G, we have for each i either Wf = Wi or VF = 0.Therefore
for i = 1, 2, ..., s. Now suppose that [ is an irreducible constituent of xG. Then (see Theorem 4.2A) the central character 1(1 associated with the block of kG in which [ lies is given by I&,.) = h i l i / [ ( l )(i = 1, 2, ..., s), where fli is the value of [ on W i . We have to prove that II/ = oG.First of all, by the Frobenius reciprocity theorem (Theorem 2.5A) x is an irreducible constituent of Moreover by Clifford's theorem (Theorem 2.2A) all irreducible constituents of l Hare of the form xy for some y E G (since H 4 G). Since (xy, [ H ) H = ( x , [kl)H = ( x , C H ) H , it follows that each of the different conjugates of x occurs with the same multiplicity in I H , say m. Let x' = x, x2, . . ., x" be the distinct conju+ x"} and so for all x E H we have = m{X' + gates of x. Then
cH.
rH
8.2
Thus if Vi E H and z E Wi, then Ci = ( ( z ) and
(3)
20 1
BLOCKS WITH NORMAL CYCLIC DEFECT GROUPS
hi5i/5(1) = hi =
C X(Y
5( 1) = mnx( I), so
'ZY)/gX( 1)
YeG
1 x(x)/x(1).
XEYi
Together with (2) this shows that (4)
u G ( c i )= $(ci)
whenever W i E H.
Now suppose W i$ H. Let P = O,(H); then P a G because it is characteristic in H. Moreover Q G P because Q a H. Let z E W i . Since W i $ H = QCG(Q),therefore z $ CG(Q)and so Q (and therefore P ) is not contained in C,(z). This shows that the defect groups of%, do not contain P. Therefore by Lemma 6.2A, ciE rad Z ( k G ) and so ci lies in the kernel of every central character of kG.In particular, (5)
u G ( c i )= 0 = $(ci)
whenever W i$ H.
The equations (4) and ( 5 ) prove the lemma. NOTATION During the remainder of this chapter we shall be examining the following situation. The group G will contain a cyclic psubgroup Q and we put C := CG(Q)2 Q and N := NG(Q)2 C . Let B be a block of kG with Q as a defect group. Then by the First Main Theorem (Theorem 6.3) there is a unique block B , of kN with Q as a defect group such that B = BY. By Theorem 6.4 the blocks of kC that correspond to the block B, of k N are all Nconjugate; fix b as one of these blocks of kC. Then Q is the (unique) defect group of b, bG = B, and the Nconjugates of b are the only blocks of kC that correspond to B under the Brauer correspondence. Write b = e(kC),where e is a central primitive idempotent of kC, and define F := {x E N I xlex = e} as the stabilizer of the block b under N. We shall put 4 I F : C 1. Let ( QI = pd. Then the automorphism group Aut Q has order p d  ' ( p  1). Since N / C is isomorphic to a subgroup of Aut Q, this shows that IN : Cl divides p"'(p  1); and since 41 IN:C ( we conclude that 41$l(p  1). On the other hand, b has IN : F I Nconjugates. Therefore if p' is the highest power of p dividing IN : C I, then Theorem 6.4 shows that p' 1 IN : F I. In particular, this shows that 4 = IN : CI/JN: F ( is relatively prime to p, and so we must have q ( p  1. Note that F/C is a cyclic group because the subgroup of order p  1 in Aut Q is cyclic (see Hall [l], $6.2). We next note that ify E F\C, then C,(y) = 1. Indeed, choose z as a generator of the cyclic group Q. Then y  ' z y = z'" for some integer m with 1 Im < pd. Since I F/C I divides p  1, yP' E C. Therefore z = y w l )Z Y P  l = F  ' and ; so m p  l = 1 (mod pd). On the other hand, if ziE C,(y), then zi = y ' z i y = z M i and ; so mi = i (mod pd). If zi # 1, then we I=
202
VIII
BLOCKS WITH CYCLIC DEFECT GROUPS
would have m = 1 (mod p). Since y # C, m # 1 and so we would have m = 1 + $1 for some integers] and 1 with 1 I j < d and p X 1. But then the binomial theorem shows that m = m m P  ' = ( l + d l ) P = 1 +$+' I + . . . =
1
(modpJ"), which contradicts the condition p XI. Hence we have proved C,(y) = 1 for all y E F\C. This implies that Q = {x ' y  ' x y I x E Q)for each y E F\C since the elements x  ' y  ' x y (x E Q) must all be different. The cyclic group Q has pd  1 irreducible ordinary characters (all of degree 1) different from 1,. The group F acts by conjugation on this set of characters, and no character x # 1, is fixed by any y E F\C; indeed xy = x implies 1 = xy(x)x(x) = x(y ' x y x  '), so from what we have just proved, Ker x = Q. Thus each orbit of F on the set of nontrivial irreducible characters of Q has length 1 F : C 1 = q; and there are (pd l)/q orbits. We shall choose a set A of representative characters, one from each orbit. Then IA I = (pd  l)/q and each nontrivial irreducible character of Q is conjugate under F to exactly one character from A.
'
With this notation we can now state the first theorem on blocks with cyclic defect groups. Theorem 8.2 With the notation above suppose that Q is a normal psubgroup of G (so the block B has Q as its unique defect group). Let b be a block of kC such that bG = B, and using the notation of Theorem 6.2B let 4 and tl, t2,...,<,,be the irreducible modular character and the irreducible ordinary characters, respectively, lying in b (to simplify later notation we use lower indices). Then (i) There are exactly q irreducible modular characterw,, $ 2 , ..., $, lying in B. They all have the same degree and satisfy $1 $2 **. $, = f#JG. (ii) There are exactly q + (pd  l)/q irreducible ordinary characters lying in B. These can be divided into two classes xl, x2 ,...,x, and x A ( A E A). The xi ( i = 1, 2, ..., q) are precisely those characters whose kernels contain Q; their degrees are equal and x1 + x2 + . * * x, = { y . The x A all have the same degree and each x A has the form
+ + +
+
cf=
1
1
Remark Suppose 4 has degree m. Then deg $i = N : F m = deg ( i = 1,2, ...,q), whilstdegx,= I N : C l m = q d e g x i ( A E A ) .
xi
Proof With the notation above, G = N. Let a1 = l,, . .., ad be the
8.2
203
BLOCKS WITH NORMAL CYCLIC DEFECT GROUPS
irreducible ordinary characters of the cyclic group Q. Consider the characters that lie in the block b of kC. Since Q G Z ( C )and b has Q as defect group, Theorem 6.2B shows that exactly one irreducible modular character, say 4, and pd irreducible ordinary characters, say t,, . . ., lie in b where for each i
rpd,
ai(z)4(x) ti(Y) = (0
if y = zx ( z E Q, x otherwise.
E
c")
r:
Now bG = B and C = C,(Q) Q G, so all irreducible constituents of lie in B by Lemma 8.2. Conversely, suppose that x is an irreducible ordinary character of G lying in B. Let be an irreducible constituent of xH ,lying in a block b,, say, of kC. By the Frobenius reciprocity theorem, x is an irreducible constituent of tG, and so by = B by Lemma 8.2. Since G = N,it follows from our remarks above that b , is a Gconjugate of b. Thus by Example 1 of 56.4,t = 5; for some i and some y E G. Then 5' = and x is an irreducible constituent of t:. Hence we have shown that the irreducible ordinary characters of G in bG are precisely the irreducible constituents of t: (i = 1,2, . . ., pd). We now turn to the proof of the particular assertions of the theorem. (i) Since ti = 4 on c" for each i, by (l), the irreducible modular characters lying in B are precisely the irreducible constituents of 4'. By definition of F, 4y = 4 for all y E F. Therefore Theorem 8.1 shows that there is an irreducible modular character $ of F with $c = and a character p of degree 1 of F with Ker p = C such that
+ + +
f#JF = $ p$ * * * p q (2) and hence 4' = $' + (p$)' + . * + (pq l$)'. The characters $', (p$)", .. ., (pq '$)' are distinct since they take different values on p'elements of c".To prove (i) it remains to show that each (pi$)G is irreducible. Let be an irreducible kFmodule which affords pi$. Then & 1 ( by definition of Let W be an indecomposable summand of vl; such that I W, .Since (V& is I W,, dim, W 2 IG : F I dim, because there are irreducible and ( IG : F ( distinct conjugates of But dim, W 5 dim, Vp, so that W = Vp. If U is an irreducible kGsubmodule of Vf,then Uc must contain a conjugate and so contain all conjugates. But is a direct sum of I G : F 1 of conjugates of and therefore U = vf' is irreducible. Thus the irreducible modular characters of G lying in B are t,hi := (pi$)'; i = 0, 1, 2, . . ., q  1. (ii) As we noted in our remarks above, the nontrivial irreducible ordinary characters of Q are permuted under the action of F in orbits of length q whilst the trivial character is left fixed. Since 4 is left fixed by F, (1) shows that 5 , is fixed under the action of F but the other ti ( i = 2, 3, .. ., p d ) are permuted in orbits of length q. First consider tl.
v
v.
v,
v),v
v
v
c.
VIII
204
BLOCKS WITH CYCLIC DEFECT GROUPS
It follows from (1) that if y = zx ( z E Q,x E Go) otherwise.
<:
We can choose an Afree AGmodule W that affords (see Theorem 3.3). Then reduction modulo K gives a kGmodule W of G which affords c#P as its modular character. From the proof of (i), 4' (and hence W )has q irreducible constituents that are all inequivalent. This shows that W (and hence <:)has at most q irreducible constituents that must be inequivalent. Now ~#9 =4 exactly when y E F, and so the multiplicity of tl as an irreducible constituent in ((:)c is 1 F : C 1. Therefore Hence
<:
(<:, G):<
= (51,
1 F : c (= 4.
=
(<:)C>C
is a sum of q distinct irreducible constituents, which we denote by Moreover, these irreducible constituents correspond with those of @, so xi = + i on Go for i = 1, 2, . . ., q ; and Ker xi I> Ker 2 Q. ..., < i + g  l Now suppose i > 1. Then ti has q conjugates, say ti, under F. This means that
xl, x 2 , ..., x 4 .
tf(X)
<:
{
= y x )
+
+
if x E C otherwise.
Now (CP,
ej;
(3)
xA=
@l
+ a * . +
t,hq
on Go
for all A E A. As we noted in the proof of (ii) we also have
(4)
xi=
i,!ti
on Go
for i = 1, 2, .,.,q. Equations (3) and (4) describe the decomposition numbers for B and then (iii) follows from Note 4 in 43.7. (iv) This follows at once from (3) and (4). EXAMPLE Suppose under the hypothesis of the theorem that B = Bo is the principal block of kN ( N = G).Then by the Third Main Theorem (Theorem 6.6), the principal block of b of kC is the only block of k C such
8.3
205
GROUPS WITH CYCLIC SYLOW pSUBGROUPS
that bN = B o . Clearly in this case 4 is the trivial modular character, so C$( 1) = 1 and F = N. Hence there are n := I N : C I characters xl, x 2 , . . .,xn and (pd  l)/n characters xi ( A E A) lying in Bo . In this case, it is clear that ti = C$ = 1 on C". Therefore ty = n on C" for each i and C" G Ker l y . This shows that
x1 =
= 2, =
1 on C"
and
xA = n
on C" for all 1 E A.
EXERCISE
With the hypothesis of Theorem 8.2, calculate the decomposition matrix and Cartan matrix of the block B.
8.3 Groups with cyclic Sylow psubgroups In the present section we shall give a description of all indecomposable kGmodules in the case G is a group with a cyclic Sylow psubgroup. It turns out that (up to isomorphism) there are only a finite set of such modules. In contrast, in the exercises that follow this section, it will be seen that if G does not have a cyclic Sylow psubgroup, there are always indecomposable kGmodules of arbitrarily large dimension.
Lemma 8.3 Let P be a cyclic group of order pd. Then up to isomorphism there are exactly pd indecomposable kPmodules. These have dimensions 1, 2, . . ., pd, respectively, and each is isomorphic to a homomorphic image of the principal indecomposable kPmodule kP. Proof Each kPmodule I/ may be considered as a module over the polynomial ring k[X] if we define OX:= uz for some fixed generator z of the group P. Since k[X] is a principal ideal domain, every k[X]module is a direct sum of cyclic k[X]modules. Thus, if V is an indecomposable k[X]module, then V has the form uo k[X] for some uo E V. Thenf(X)w uo f(X) = uo f ( z ) is a homomorphism of the k[X]module k[X] onto V with kernel I , say. Since k has characteristic p, uo(z  l)pd= uo(zpd 1) = 0 and so (X  l)pdE I. Since k[X] is a principal ideal domain, this means I = (X  l)"k[X] for some m, 1 Irn Ip d ; and I/ N k[X]/(X  l)"k[X]. Conversely, if 1 Irn Ipd, then V, := k[X]/(X  l)"k[X] is an indecomposable k[X]module of dimension m over k on which Xpd 1 = (X  l ) P d acts as 0; hence V, may be considered as a kPmodule with uz := uX.Thus, each indecomposable kPmodule is isomorphic to exactly one of V, (rn = 1, . . .,pd). The last assertion of the lemma follows from the observation that for all m Ipd,
v, = k[X]/(X  l)"k[X]
'y
kP/(z  l ) W .
206
VIII
BLOCKS WITH CYCLIC DEFECT GROUPS
Theorem 8 3 Let G be a group with a cyclic Sylow psubgroup P.Then each indecomposable kGmodule is a homomorphic image of the kGmodule kG (and so a homomorphic image of some principal indecomposable kGmodule). Proof Let V be an indecomposable kGmodule. By 65.2 we know that there exists an indecomposable kPmodule W such that V I WG.By Lemma 8.3 we can find w E W such that W = w(kP). Then W G= w(kP)BkPkG = w,(kG) where w, := w 1. Hence the mapping a woa shows that W c is a homomorphic image of the kGmodule kG. Since V is a direct summand of WG,V is also a homomorphic image of the kGmodule kG.
Corollary Under the hypotheses of the theorem there are only a finite number of nonisomorphic indecomposable kGmodules. Proof By the proof above, each indecomposable kGmodule V is isomorphic to a component of WG for some indecomposable kPmodule W. Then the corollary follows from Lemma 8.3 and the KrullSchmidt theorem. EXERCISES
1. Let G = ( a ) x (b), where a and b have orders p, and let k be any field of characteristic p. Let V be a vector space of dimension 2n 1, with a basis {u,, u l , ..., u,; ul, ..., u,}. Define uia= u,b = ui, 0 I i I n, u,(a  1) = ui, 1 5 i I n ; and yi(a  1) = ui 1. Prove that V is an indecomposable kGmodule. 2. Use Exercise 1 to show that if G is a group with a noncyclic Sylow psubgroup, then there exist indecomposable kGmodules of arbitrarily large dimensions.
+
8.4 Some technical lemmas
In the next section we shall be describing properties of blocks of defect 1 in a group G of order g = pg,, where p %go. To prove these results we need information about characters that are sums of constituents of principal indecomposable modular characters of G (Lemma 8.4B). Lemma 8.4A will only be used for the proof of Lemma 8.4B. Let U o be a principal indecomposable AGmodule and let V be a KGmodule such that V 1 V , 6A K. Then there exists an Afree AGmodule V such that U @ A K = V and O is an indecomposable kGmodule.
Lemma 8.4A
Proof Since II, 6A K is completely reducible, we can write U , 0 A K = V 0 V' for some KGsubmodule V' of U , 0 A K . Let W := V n U , and
207
8.4 SOME TECHNICAL LEMMAS
W' := v' n U , (where U , is assumed to be embedded in U , €3" K). We define U := U,/W' and show that it satisfies the required conditions. First, U is Afree. Indeed, if u E U , and ( u + W')a = 0 for some a # 0 in A, then ua E W' c V', and so u E V'a = V' as well as u E U , and hence u + W' = W'. Hence U is Atorsion free and so by the structure theory of modules over a principal ideal domain (see Example 4 of §1.1), U is Afree as asserted. Next, for any Kbasis df V we can multiply by a suitable element of A to obtain a Kbasis of V lying in U , . Therefore W @ A K = V ; and similarly W' @ A K = v'. Since U is Afree, this shows that U @ A K 2: ( U , @ A K ) / ( W ' @ " K ) . V I/: Also since U is Afree, 0 2: 0 , / W . By Theorem 3.4A, 0,is a principal indecomposable kGmodule (since U o is a principal indecomposable AGmodule) and so 0,has a unique maximal submodule by Theorem 3.4B. Therefore 0 must also have a unique maximal submodule and hence it is indecomposable. This proves the lemma. Now let G be a group of order g = p g , with p $go. Let P be a Sylow psubgroup of G and put C := C,(P) and N := N , ( P ) . Let B be a block of defect 1 in kG, and B , the unique block of k N such that By = B (Theorem 6.3). With the notation of Theorem 8.2 we shall take x , , x 2 , ..., x4 and x A (A E A) as the irreducible ordinary characters that lie in the block B,. The following result serves as a replacement for Theorem 6 of Dade [2]. Under the above hypotheses let V be a KGmodule whose irreducible constituents lie in the block B of kG, and let i be the character afforded by V . Suppose that there exists a principal indecomposable AGmodule U , such that V 1 U o 0 A K. Then for some i (1 I i I q ) and some integers n,, n, (A E A) each equal to 0 or 1, Lemma 8.4B
iN
= nixi
+C
LEA
+8
~ A x A
where 0 is a sum of characters such that 0 = 0 on N\N". Proof By Lemma 8.4A we can find an Afree AGmodule U which affords the character iand for which 0 is indecomposable. In particular, U is indecomposable. Since P c N, U is Nprojective and so U I (U,)', and Note 1 of 45.2 shows that for some indecomposable Afree ANmodule W , W I U , and U I V . Write (1)
w.
u,
=
W O O . . . @w,,
where the are indecomposable ANmodules and W, 2: W. By Mackey's theorem (Theorem 2.1A), ( W G ) ,2: @:= ( V'i)N, where x 1 = 1, . . ., x, is a set of representatives of (N, N)double cosets and V', is an
208
VIII
BLOCKS WITH CYCLIC DEFECT GROUPS
A(x; "xi n N)module. If i 2 2, then xi $r N and so P $ x; 'Nxi n N ; hence in this case x; 'Nxi n N is a p'group since I P 1 = p. Therefore for each i 2 2, the vertex of VXiis 1. This shows that (w"),(and hence V,) is a sum of indecomposable ANmodules such that all but one (namely V', = W) is 1projective. Hence in (l), the modules W,, . .., W, are all 1projective. Now for i 2 1, each W; is isomorphic to a principal indecomposable ANmodule (Example 3 of 55.1); so the character afforded by W; vanishes on N\N" by the Example of 53.7. Thus, if 8 is the character afforded by W, @ * *  @ W,, then 8 = 0 on N\N". T o complete the proof of the lemma it remains to show that the character afforded by W z Wo has the form ni xi + , n, x, (for some i) with each ni, n, ( A E A) equal to 0 or 1. Since W is an Afree indecomposable ANmodule, W is an indecomposable kNmodule by Theorem 3.4B. Hence by Theorem 8.3 there is a surjective kNhomomorphism kN + W.Choose w E W such that iv is the image of 1 under this homomorphism. Then at+ wa is a n ANhomomorphism of AN W; and since w ( A N ) = W,this mapping is surjective by the structure theory of modules over principal ideal domains. Thus W is a homomorphic image of the ANmodule AN. Since W is indecomposable this means that W is a homomorphic image of some principal indecomposable ANmodule, say e(AN). Let t be the character afforded by e(AN). We claim that t = xi + X ,for some i. Since the multiplicity of any irreducible character as a constituent of z must be at least as great as its multiplicity in the character afforded by W, this will prove the lemma. Thus (with the notation of Theorem 8.2) suppose that the principal indecomposable kNmodule e(AN) corresponds to the modular irreducible character i,bi, say. Then by Theorem 8.2(iii) and the example of 43.7 we have
1,
1,,,
xi
0
+ C l e A X,
on Go on G\Go.
Thus, using Theorem 3.7B, we conclude that the multiplicity of xi in t is (for
j = 1, ..., q )
(5, x j ) G = (T,
and the multiplicity of
1,, x,
by (4) Of $8*2;
=
x, in t is (for all 1 E
(T, x,)c = (T, i,b1
Thus t = xi +
i,bj)Go
+ . * .+ &)GO
A)
=1
by (3) of $8.2.
(on G) and the proof of the lemma is complete.
8.5 Groups of order g = pgo with p $go
In this section let G denote a group of order g = pgo where p $ g o . The purpose of this section is to describe the irreducible ordinary and modular
8.5
GROUPS OF ORDER
g = pgo WITH P xgo
209
characters lying in a block with (cyclic)defect groups of order p. The general problem of a block, with a cyclic defect group, of an arbitrary group was solved by Dade [2]; but for our purpose a special case is sufficient. Moreover, a knowledge of this special case enables one to grasp the general situation better. NOTATION Let P be a Sylow psubgroup of G. As before put C and N :=N G ( P ) ;note that c" = O,,,(C)and C = P x C".
:= CG(P)
Let b be a block of kC. Then we know from Theorem 6.2B that there is a single irreducible modular character 4 lying in b and the irreducible ordinary characters 11,t, , . .., t,, (all of the same degree) lying in b have the form (2 E P, Y E C") tj(zY) = aj(z)4(Y) where a I = l,, a,, . . ., a,, are the irreducible ordinary characters of P. With the notation of $8.2 there are q + (p  l)/q irreducible ordinary characters of N lying in bN,which we denote by xl, xz, . . . , x, and x1 (1E A). These are related to the characters in b by
5:
=
x1 + x2 + ... + xq
{ti"I j = 2, ..., p } = (xl
and
+
11 E A}
xq on N o for all 1E A. We shall and moreover we have ,yr = x1 + x, + define T : = X ~+ x, + ... + x,. Let us denote by b , = b, b , , . . ., b, the blocks of kC such that br = b N . Then the description above holds for the blocks bi and we shall denote the corresponding characters by (y), aJ!"and 4(i). Theorem 8 5 Let G be a group of order g = pgo, where p $ g o , and let B be a block of defect 1 of kG. Let b be a block of kC such that bG = B. Then with the notation above we have the following:
+
(i) There are exactly q (p  l)/q irreducible ordinary characters lying in B which we denote by ..., cq and ( A E A). =  xu)" for all (ii) There exists E = 1 or  1 such that (L1 I, p E A. (iii) There exist mi = 1 or  1 ( i = 1, 2, . . ., q ) such that
el, c,,
cr
c,)
where the sum is over all irreducible ordinary characters ( of C with tN = t (namely, 5 = 5f)for i = 1, 2, . . ., n), and Bi E Char(C) vanishes on C \ C . In particular ci is constant on P\{l} because P lies in the kernel of each 5 (see Theorem 8.3). (iv) For each 1E A we have (L)C = E
c 5 + 811
<"XI
VIII
2 10
BLOCKS WITH CYCLIC DEFECT GROUPS
where the sum ranges over all irreducible ordinary characters ( of C such that rN = x A (they all lie in the blocks b , , b , , ..., b, by Lemma 8.2), and tIAE Char(C) vanishes on C \ C . (v) For each A E A we have a
ra
Remark The characters (A E A) will be called the exceptional characc2, ..., the nonexceptional ters of the block B and the characters characters. The exceptional characters all have the same degree by (v).
cl,
rq
Proof For convenience, the proof will be divided into several steps. Step 1
The irreducible characters of G determined b y A.
Let S
N\N". We claim that
I=
ylsy n s =
la
S
if if
Y E N y E G\N.
Thus S is a trivial intersection set with N G ( S ) = N. Indeed it is easily seen that S is just the set of x E N whose ppart x p E P\{l}, because P a N. Let y E G and suppose y  ' S y n S contains some element x . Then x and y x y lie in S, and so both x p and ( y  ' x y ) , = y  l x p y lie in P\{l}. Since I PI = p , this means that P = ( y x p y  = y ( x , ) y  = y P y  and so y E N. Since S is a normal subset of N , we conclude y  ' S y n S # 0 implies y  ' S y n S = S as asserted. Now it follows (with the notation above) that for all A, p E A, x, x A = (x,  T )  (xa  T) is a generalized character of N which vanishes on N\S. So using ( 1 ) it follows from Theorem 2.5B that if 1 A 1 > 1 then there exist distinct irreducible ordinary characters la(A E A) such that for some E = 1 or  1 ,
'
')
(2)
( x A  x,)' =
 c,)
',
for all A, p E A and I # pa
This gives 1 A 1 irreducible ordinary characters cA (A E A) of G when [ A 1 > 1. If I A ( = 1, then we define cA as any irreducible constituent of ( x A  ')T and put E = 1. In particular, the condition (ii) holds. Step 2 Analysis of
( x A  T)'.
r2,
We claim that there exist irreducible ordinary characters c,, . . ., c, different from la(A E A) such that for some integers mo, m,, . ..,m, (independent ofA) with mi # O for i = 1, 2, ..., t we have
In the case ) A ) = 1 this is trivial, so suppose ) A ) > 1. Then for each # 1,we have by Theorem 2.5B that
p E A, p
a x ,  T)", T,  T,>
(4)
+
< ( X , ,')T
 T)",
= &(XI
 79 X A  X,>
X , x,,
since both x1  t and 1 p f v, then
 X,)'
= ((x,
&(XI
=&
vanish on N\S (= N"). Moreover, if
[,> < ( X ,  t)", 5,)
 XV)',
= ((2, = (451
Tp>
 T v h 5,)
= 0.
Thus for 1# p we have that
T,>
<(x, 
= mo
9
+
say, is independent of 1and p ; and so by (4), ((x, ,')T T1) = m, E. This shows that (x,  t)" has the form described in (3). The fact that m , , m 2 , .. ., m, are independent of A follows from the equation
<(x,  T)",
Ti>
Using the definition of the relation
 <(x, ,')T
t=
1 + 4 = (x*
2,
 T p ) , ii>= 0.
ti) = (&((A
cf=lxi and Theorem 2.5A we obtain from (3)
x,  T
h
= <(x,  7)", (2,  d G > G I
Em;.
=(~,+E)~+(IA  l )( m g +
i= 1
Hence t
9 = 2mo&
Step 3
+ 1Almi + i E= m f . 1
The restriction of the characters
Ti to C .
By the Frobenius reciprocity theorem (Theorem 2.5A), for a l l j > 1,
((Ti),
9
5j
 51>c = (Ti
9
(xi
 5)")
=
mi
by (3), where 2, = 5;. Thus if we define c := ((CJc, tl>  m,, then the contribution to (Ti)c from the characters in the block b is
We also note that since the 5, all have the same degree, it follows by
VIII
212
BLOCKS WITH CYCLIC DEFECT GROUPS
Theorem 4.2C that P
(7)
tj(x) =
j= 1
C tj(x)tj(l)= 0
1
p
for all x E C\Co.
)j=1
j(
If b, = b, b,, .. ., b, are the blocks of kC such that br = bN,then we can apply the above analysis to each block b,. Since the constants mi depend only on the characters in bN,they are independent of the blocks 6,. Hence from (6) and (7) we conclude that n
(Ci)c = mi
(8)
j=1
tY' + Pi + yi
9
where pi is a sum of characters from the blocks b,, b l r ..., b, with the property pi = 0 on C\Co,and yi is a Zlinear combination of characters from blocks different from 61, b, , . . ., 6,. Since t:"(zy) = &')(y) for all z E P and y E C",it follows from (8) that the generalized decomposition number of ci at qY') is m i . Since mi # 0 by hypothesis, the Second Main Theorem (Theorem 6.5) shows that Ci lies in the block B. Again Theorem 6.5 shows that y, in (8) is 0 on C\Co.This proves (iii). Step 4
The restriction of the characters
to C .
Let m l = ( C , , 7 " ) . Then if tr = x,, ((C&, m = E dA,, mo + m by (3), and (((,),, t l ) = contribution of b to is
+
P
( m + mo) C t i + m t 1 I=2
+E C
ti) = ( C l , (x,  7 ) " )
(el,
7')
+
= m.Therefore the
tj
4N=X1
P
= (m
+ mo) C ti  mot1 + E C tj
As in Step 3, we obtain n
ON = x A
i=1
p
n
*
n
(
(
=
#) + 71
where y A is the sum of characters from blocks different from b , , b , , . ..,b,. Then
(el),=
n
n
C tY)+P,+r,, where is a sum of characters from the blocks b,, b, ,. . .,b, with /?A = 0 on C\Co and y, is a sum of characters from other blocks. Consider the generalized decomposition number d;, of c, with respect to at z E P\{1). From (9)
(9). we get
 m o C1 =t1' i " + E CI =
1 (CjlO)N=xA
8.5
GROUPS OF ORDER
Thus we have (10)
c d;,
=
213
= p g o WITH p $go
molA/
P
+ E X aj(z) =  m o ) A I
 E,
j= 2
deA
because I$ aj(z) = 0’ by = the ordinary character relations. Now molAl E#O;thisisobviousfor ( A 1 2 2 , a n d f o r (A1 =litfollows from (3) and the definition of lain this case. Thus at least one 5, lies in the block B. For any 1,p E A, x,  x,, vanishes outside S. Therefore by (2), la l p vanishes outside UxEG xlSx, and in particular it is 0 on Go. If q is any principal indecomposable modular character of G, then ( q , la lJG0= 0. Therefore if one of the [,appears in q, then all laappear in q as constituents. But at least one lalies in B and so all la(A E A) lie in B. Hence by Theorem 6.4, y, = 0 on C\Co.Thus (9) may be written To complete the proof of (iv) it remains to where 8, :=pa+ y, is 0 on C\Co. show that mo = 0. Step 5 If the characters x, are suitably labeled, then mo = 0, mi = 1 or for i = 1, ..., t, and t = q.
1
Choose e as the block idempotent of AG lying in B. Then (see $4.2) each irreducible KGmodule lying in B is isomorphic to a KGcomposition factor of the KGmodule e ( A G )@ A K. This latter module is completely irreducible by Maschke’s theorem; therefore, if V is a direct sum of nonisomorphic irreducible KGmodules lying in B, then V l e ( A G ) @ , K . Let a be the.character that is afforded by such a KGmodule V. Then by Lemma 8.4B there exists an integer j (1 Ij I q) and some n j , n, (A E A) equal to 0 or 1 such that for some 8 E Char(N) vanishing on N\N” we have aN = njxj
+
dsA
+ 8.
n,x,
Now suppose that p E A and choose the character t j of C so that tj”= X, . Since t j = on C” (see the beginning of this section) and ty = T, therefore 0 = (&., t j = (8, (tj t l ) N )= N(8, xp  7). Hence from the above expression for aN we conclude that ( a N , X,  7 ) = n,  nj = 0, 1, or  1. Thus we have shown that for all characters a of the above type (11)
(a,
(x,  T ) ‘ ) ~
= (aN,
x,  T
)
=~ 0,
1, or  1.
We first apply (11) with a = Ti ( i = 1, . . . , t). Then from (3) we see that  1. Since liis a constituent of (x,  7)‘ by the way in which it was chosen, we conclude that m i= 1 or  1 when i 2 1.  m i = 0, 1, or
214
VIII
c,
BLOCKS WITH CYCLIC DEFECT GROUPS
We now apply (1 1) with a = ,, c,. Then from (3) we obtain m, I A I + or  1. Since E = 1 or  1, this shows that mo = 0 when 1 A I > 2. If [ A I = 2, then either mo = 0 or m, = 1, E =  1; in the latter case, if we interchange the labels on the two x A (A E A), then the equations (3) still hold with the new values rn, = 0 and E = 1. Finally, if I A I = 1, then (3) shows that mo + E # 0 (since [,is a constituent of (x,  t)');hence replacing rn, by 0 and E by m, + E gives a valid equation (3) in which rn, = 0. This shows that in all cases we may take m, = 0. The relation t = q then follows immediately from ( 5 ) . This completes the proof of (iv), and (v) now follows from (3) because (x,  T)' = 0 on Go (Theorem 2.5B). E
= 0, 1,
Step 6 Cl, c,, ..., c,, and (, characters lying in B.
(A E A) are the only irreducible ordinary
It is enough to show that any irreducible ordinary character lying in B is a constituent of (x,  T ) for ~ some p E A. Suppose the contrary. Then for all j > 1,
(rc,
where X,= t;. This shows that ( l c , t j )= 5 , ) = c, say, and so the ti, which is 0 contribution to from characters lying in b is given by c on C\cO by (7). Making a similar analysis at the other blocks b = b , , b , , . .., b, we find that
rc
z=
rc = B + Y ,
where fl is a sum of characters from the blocks b,, b , , . .., b, with fl = 0 on C\Co,and y is a sum of characters from other blocks. By the Second Main Theorem (Theorem 6.4), y = 0 on C\Co. But this implies (c,, , I,,) = c( l)/p, and so p I (( 1). Since B is a block of defect 1, this is impossible by Theorem 5.5B.Hence all irreducible ordinary characters of G lying in I3 are constituents of (xl  T)' ( A E A). On the other hand, Steps 3 and 4 show that these constituents lie in B. Hence ( i ) is proved, and the proof of the theorem is complete. EXAMPLE 1 Let G be a group of order g = p g , , where p $go, and let Bo be the principal block of kG. Then Bo has defect 1 (see the example of 84.9, and by the Third Main Theorem the principal block b, of kC is the only block of kC with bg = B. In the notation of @8.2this shows that F = N and q = n, where n := I N : C I. Moreover, +e unique irreducible modular character of 6 , is lc, so t, = lc and C"c Ker t i for each i. Then the theorem shows that ([JC
= mi lc = ei
( i = 1, 2,
..., n).
8.6
GROUPS WITH REPRESENTATION OF DEGREE d
< +(p  1)
215
I n particular, ii mi lc = 0 outside of Gosince if x # Go,then x is conjugate to an element of C\Co.Now (see $8.2) N ( = F) acts on the set {t2,. . . , t,} of characters in bo permuting them in orbits of length n (= 4).Any two characters in the same orbit induce to the same character of N. If {til,. . ., ti,) is one orbit, then t: =
1 t + 8,
= 4X,)C
+ 0, .
<"Xu
EXAMPLE
2 In the case where I A 1 = I, it follows from the theorem that
where 5 ranges over all characters in the block b,, . . .,b,. It follows from (7) and the conditions of (iii) and (iv) that this implies that eli
In particular,
+ mili = O
iAis constant
on C\Co.
on P\{1). EXERCISES
[In the following exercises suppose that the hypotheses of Theorem 8.5 hold.] 1. Show that G has exactly q irreducible modular characters which lie in B. 2. Prove that the exceptional characters Cl in B form a single pconjugate set of characters, while each nonexceptional character li is pconjugate to itself (see 94.6). 3. Determine the decomposition matrix of B and the Cartan matrix of B. 4. Prove that there are at least two irreducible ordinary characters whose restrictions to Go are irreducible modular characters lying in block B. 5. Let a = Lie,,i AThen . prove that ( a c , & y ) c o = 1 for any y in N.
8.6 Groups with a faithful representation of degree d < +(p  1) We shall apply the results of Theorem 8.5 to show that no simple group of order g = pgo with pXgo can have a nontrivial character of degree d < i ( p  1). This result is due to Brauer [3] and will be used in the next section to prove a more general theorem about groups with a faithful representation of degree d < &(p 1).
2 16
VIII
BLOCKS WITH CYCLIC DEFECT GROUPS
Lemma 8.6 Let G be a group of order g = p g o with p $go. Let P be a Sylow pgroup of G and put C I= C , ( P ) and N := N , ( P ) . Let $ be an irreducible ordinary character of G of degree d < f ( p  1) such that P $4 Ker $. Then (i) $ lies in a block B of defect 1, $ is an exceptional character, and at least two exceptional characters lie in B. Let b be a block of kC such that bG = B. (ii) For each exceptional character lying in B, there is an ordinary irreducible character 5 lying in b such that
r
I
where 5y1, .. ., 5yt are the distinct conjugates of t in N. (iii) We can find an exceptional character lying in the block B such that # $, ( r p , $,) = 0, and $co) = I N : C 14, where 4 is defined as in Theorem 8.5. (iv) 4 Id.
r
r
(rco,
Proof (i) Since p $ d = $(l), I(/ lies in a block B of defect 1 by Theorem 4.5B. We next claim that $ is not constant on P\{l}. Indeed if it were, then we should have the integer ($,., lp) = (l/p){$(l) + (p  l)$(z)} for each z E P\{l}. But then $ ( z ) is an integer and $ ( z ) = $(1) (mod p). Since I J / ( z )I I$(l) < f(p  l), this implies that $ ( z ) = $(1) for all z E P ; that contradicts the hypothesis P $ Ker $. Thus $ is not constant in P\{1). This shows that $ is an exceptional character (Theorem 8S(iii)), and that 1 A I > 1 (Example 2 of 58.5). (ii) In the notation of Theorem 8.5 has the form
rC
for some irreducible ordinary character x of N and some BEChar(C) with 0 = 0 on C\Co. The t are irreducible ordinary characters of C and by Theorem 6.2B we know that Tc0 is equal to the unique modular irreducible character of the block b in which 5 lies. Since is an exceptional character, there are exactly 4 of the t in the sum (1) that lie in a given block b (see Theorem 6.2B). Thus we have
(rC., tcu)= f 4 + (ecotrco)= + 4 + p ( e , t)
because 0 = 0 on C\Co and I C : c" I = p. Since the lefthand side lies between 0 and d, and 4 = (p  l)/[AI < f(p  1) by (i), we conclude that (0, 5 ) = 0 and the positive sign holds in (1). Since 0 is 0 on C\Co, (OP, lp) = 0( l)/p, and so 0( 1) = 0 (mod p). Since the positive sign holds in ( 1 ) and (0, <)= 0 for each 5 in the sum (l), 0 is a sum of irreducible characters of C with nonnegative coefficients. Because I: is an exceptional character,
8.6
GROUPS WITH REPRESENTATION OF DEGREEd
217
<$(p  1)
[( 1) = d < p , and so 6( 1) = 0, which shows that 6 = 0. Hence we have shown that ic = &=r 5 for some irreducible character x of N . By the Frobenius reciprocity theorem the sum is over all irreducible constituents of xc, and by Clifford's theorem these are the conjugates in N of any one of the (. This proves (ii). (iii) Let be an irreducible ordinary character of C . By Theorem 6.2B, tP is a multiple of a single irreducible character of P . Thus it follows from (ii) that the irreducible constituents of t+hc form a single class of irreducible characters of P conjugate under N . Since $( 1) < + ( p  l), there exists an = 0 and irreducible character tolying in h such that ( $ p , l p ) = 0. Let (' be the exceptional character of B associated with the characof N . Then by (ii) ter
<
S
for suitable ui E N . (Actually s = f. Why?) By the choice of tono irreducible constituent of l p equals an irreducible constituent of t,hp, and so ( l p , *P) = 0. Finally, as we noted in the proof of (ii), exactly q of the irreducible constituents tC,of i are equal to the unique modular irreducible character in b. This is true for each of the IN : F I = 1 N : C ( / qblocks b, such that bp = B (see the notation of 48.2). The same is true of the character $. Therefore (ice, $c.) = q 2 * I N : C l / q = q l N : CI.
cN. By
in (ii) are the irreducible constituents of (iv) The t characters Theorem 8.2, q of these constituents lie in b, and so d = [( 1) 2 t 2 q.
Theorem 8.6 Let G be a group of order g = p g o with p $go, and let P be a Sylow psubgroup. If G = G' (the commutator group), then each irreducible ordinary character $ of G with degree d < + ( p  1) has P G Ker $.
Proof We shall suppose that G has such a character $ of degree d < + ( p  1 ) with P $ Ker $ and obtain a contradiction. We shall continue with the notation introduced in $8.2 and $8.5. Note that all the information from Lemma 8.6 is available. Let Bo be the principal block of kG. Then by Example 1 of $8.5 we know that Bo has n := I N : C 1 nonexceptional characters C 2 , .. ., cn and ( p  l)/n exceptional characters (1E A). These satisfy
[,,
(2)
(ii), = milt + 6i
and
(i,)c =
E(xA)C
+
6 2 9
where each mi = & 1, E = 1 and B i , B j E Char(C) vanish on C\Co.We shall not use the corresponding invariants for the block B in which $ lies, but the invariant q will refer to B. The character i will be chosen to satisfy (iii) of Lemma 8.6.
218
VIII
BLOCKS WITH CYCLIC DEFECT GROUPS
ci
Step 1 For each i, is a constituent of $$ ifmi = 1 and a constituent of t+hc fi mi =  1 (the bar denotes complex conjugation).
Each element of G\G"is conjugate in G to an element of C\Co.Indeed if x E G\Gohas ppart x,,, then there exists y E G such that y  ' x p y E P\{ 1) and hence y  'xy E C\Co. Thus it follows from (2) that T i = mi on G\G".On the other hand, since $ and [ are exceptional characters in the same block, t j  [ = 0 on Goby Theorem 8.5 (v). Therefore ($  c)(Ci  rn, lG)= 0 on G. If mi = 1, then $Ci [ = t,h and so ($$, = (*, 2 1 . On the other hand, if mi =  1, then $ = [ [ i i , and so = (*, T l i ) 2 1.
+ [ri
+
$ri +
+
ri)
Step 2 For some 1 E A, [,is a constituent of $$ if& ent of if& = 1 .
$r
*ci)
($r, ri)
=
 1 and a constitu
Since the exceptional characters 'in a given block all have the same degree, Theorem 4.2C shows that for all x E G\G" n
Since Ci(x)= mi by (2), it follows from Theorem 8.5(v) that n
Hence we obtain c,(x) =  E for any x E G\Go. As we saw above, $  ( = 0 on G o ; and so ($  C)(C,.,, r A+ E ) = 0 on G. If E = 1, then C N E A L + =r + C P E A Hence
*
*
and so
C,
($r, c,)
r
r,.
2 1 for some
1 E A. Similarly, if
i,)2 1, and so (@, r,) 2 1 for some A.
Step 3
The degrees of the characters
E
=
 1, then (I,@,
ci ( i = 1, 2, . .., n ) and C, .
By Lemma 8.6(iii), ((l(lc)p,l p ) = ( + p r iP) = 0, and so by Step 1, ( ( [ i ) p , lp) = 0 whenever mi =  1. Thus it follows from (2) that if mi =  1, then 1 = ( ( O i ) p , lp) = O i ( l ) / p and so O , ( l ) = p . Thus ( , ( l ) = p  1 whenever mi =  1. On the other hand it follows from (2) that if mi = 1 , then Ci( 1) = 1 + Oi( 1). Since Oi is 0 on C\Co,((O,), , lp) = Oi( l ) / p so p I Oi( 1 ) . The hypothesis on G implies that only the trivial character has degree 1. When mi = 1, then i i (1 ) 2 p + 1 except for the trivial character rl. From (2) we have i,( 1 ) = q i (1) + O,( 1). Again, since O1 = 0 on C\Co, p ) O A ( l ) .On the other hand, ~ ~ ( =1 n) by the example of 48.2. Hence r,(l) 2 p  n if E =  1 and i,(l)2 n if E = 1 . In fact, if E = 1 then ((Jc =
8.6
GROUPS WITH REPRESENTATION OF DEGREE
d
219
(&, and so C(, 1) = x,( 1). Indeed, we showed above that ( ( i j c ) , , 1,) = 0, and so by Step (2), 0 = ((i,),, I,,) = ((x,),, 1), + 8,(l)/p. By Theorem 8.3, P $L Ker xAand so by Clifford's theorem, 1, is not an irreducible constituent of (zA),. Therefore ((x,),, 1,) = 0 and so 8,(l) = 0; in particular, ((x,),, (O,),) = 0. Thus we conclude that x, and 8, have no common irreducible constituent, and so by (2), 8, = 0 and = (x&. Step 4
The case when E
=
1.
Without loss in generality we may suppose mi = 1 for i = 1, 2, ..., I and m i=  1 for i = I + 1, . . ., n. Then by Theorem 8.5(v) we have I
n
(3) For
E =
1 the information from Step 3 gives 1
n + ( n  I)(p  1) = C(i(l)2 ( I

i= 1
l)(p
+ 1 ) + 1.
Hence n(p + 1) 2 2pI. On the other hand, Lemma 8.6 shows that nq = ( ( i ) c o
7
( i j ) c J = < ( i j C ) C ~ 1c.) 9
and so
(4) by Steps 1 and 2. Now by Step 3 we have for E = 1 that lcD)= ( ( z ~ ) ~l,,,) ~ ,= n
by the example of 58.2.
On the other hand, for mi =  1, 0 I((ii)c 1,) =  1 3
+
(ei
1,) =  1
+ (1/p)((8i)cc
7
1, )
because Oi = 0 on C\Co. Thus ((8JC, l,, ) 2 p and hence (5)
(((Jc0, lco)2 p  1
Thus we conclude from (4)that nq 2 n
whenever mi =  1.
+ (n  l ) ( p  1)
and so I(p  1 ) 2 n(p  9). This combined with the inequality above gives n(p2  1) 2 2pI(p  1) 2 2 p ( p  q ) n
and so p  q < j p . But q Id by Lemma 8.6, and so we have a contradiction. Thus the case E = 1 is impossible.
220
VIII
Step 5
BLOCKS WITH CYCLIC DEFECT GROUPS
The case E =  1.
In this case the equation (3) and the information from Step 3 gives
( n  I)(p  1) = [,(I)
I
+ C i i ( 1 ) 2 (P  n ) + ( 1 i= 1
I)(P
+ 1) + 1
and so n 2 21. On the other hand, we obtain analogously to (4)the relation
by ( 5 ) . Therefore
nq 2 (n  +n)(p  1) = +n(p  I), which implies q 2 j ( p  1). But again this is impossible because q 5 tl by Lemma 8.6. Thus in either case we have a contradiction. This proves the theorem. Remark It has been shown in Feit [3] that if G is a group of order g = pgo with p Xg,, and G has a faithful ordinary representation of degree J(p  l), then either G has a normal Sylow psubgroup or G / Z ( G )s PSL(2, p). EXERCISE
Let G be a group of order g = pg, with p$g0 and let Bo be the principal pblock of G. Suppose that G has an irreducible ordinary character of degree i ( p  1) and that i':= 1, is the only irreducible ordinary character of degree 1 lying in B,. Then prove the following: (i) Ifp = 1 (mod 4), the set of irreducible ordinary characters lying in B, consists of c', $(p  5 ) characters of degree p + 1, two pconjugate characters of degree f(p I), and i ( p  1) characters of degree p  1. (ii) If p = 3 (mod 4),the set of irreducible ordinary characters lying in Bo consists of $(p  1) characters of degree p + 1, two pconjugate characters of degree j(p  I), and i ( p  3) characters of degree p  1.
+
[',
8.7
Criteria for normal Sylow pgroups
In the present section we give a proof (based on Brauer's result in $8.6) of the theorem that a group with a faithful ordinary representation of degree d < +(p  1) has a normal abelian Sylow psubgroup.
d
Nore If G is a group with a faithful KGmodule V of dimension p  1, then its Sylow psubgroups are always abelian. Indeed, if P is a
_<
8.7
CRITERIA FOR NORMAL SYLOW
GROUPS
22 1
Sylow psubgroup, then the restriction V, is completely reducible and the dimensions of the irreducible constituents of V ‘ divide I P 1 (Theorem 2.4B). Since rl < p. this means that the irreducible constituents of V, all have dimension 1. Since V is faithful, this shows that P is abelian.
In proving the general result we shall need to use the following (simpler) result for solvable groups. This result (due to Itb) shows that when G is solvable we can prove a slightly stronger result. Theorem 8.7A Let G be a solvable group which has a faithful ordinary representation of degree ri < p  1. Then G has a normal abelian Sylow psubgroup. Proof Suppose the theorem is not true and let G be a minimal counterexample. Let [ be a character of a faithful ordinary representation of G with smallest possible degree tl. Then d < p  1 and clearly d > 1. Let P be a Sylow psubgroup of G ; P is abelian by the note above. Step 1
iis irreducible.
+
Otherwise we could write i= i l 1, as a sum of characters. By the are not fzithful, and so by the choice of G, O,(G/Ker ii) choice of (, i l and i2 is the Sylow psubgroup of G/Ker ii (i = 1, 2). Then O,(G/Ker i l x G/Ker i,) is the Sylow psubgroup of G/Ker i l x G/Ker i,. Since Ker n Ker 1, = Ker i= 1, there is an injective homomorphism of G into this x Ker c2). But that implies that O,(G) latter group given by x w ( x Ker is the Sylow psubgroup of G contrary to the hypothesis.
rl,
Step 2 For each normal subgroup H of G with H # G, H n P
G
Z(G).
Since G is a minimal counterexample, O,(H) is the Sylow psubgroup of H . Put L = C,(O,(H)) a G . Since P is abelian, P c L. If L # G, then P = O,(L) by the minimality of G, and so P a G since L a G. This is contrary to hypothesis, so L = G ; hence P n H E O , ( H ) G Z ( G ) . Step 3
G = PG‘, P n G’ = 1, and G” E Z ( G ) .
Since PG’ is a normal subgroup of G, and P $ Z ( G ) , PG’ = G by Step 2. Next let T be a representation of G which affords C. Let z E P n G’. Then z E Z ( G ) by Step 2 and so T ( z )is a scalar a l , say. However, since z E G‘ it is a product of commutators and so ad = det T ( z )= 1. On the other hand, since z is a pelement, a is a ppower root of 1. Since d < p , this shows that a = 1, and so z = 1. Thus P n G’ = 1. Finally, since P n G‘ = 1 and G” # G’, therefore G“P # G. By the minimality of G, P 4 G ” P . Since G” 4 G”P, this shows that G“P = G” x P . Hence P E C,(G”) a G. Since P $L Z ( G ) ,Step 2 now shows that C,(G”) = G and so G” E Z ( G ) .
VIII
222
BLOCKS WITH CYCLIC DEFECT GROUPS
Step 4 Z(G’) = G’n Z ( G ) and I G’: Z ( G ’ )1 = d2. Let 1 be an irreducible constituent of (,,. Since G = PG’ and G’Q G, the group P acts transitively on the set of conjugates of 1. Since p$((l), it follows that 1” = 1 for all x E G. By Theorem 8.1, 1 is the restriction of an irreducible character of G to G’,and so cG, = 1, which shows that (, is irreducible. Let T be a representation which affords (. Then using Schur’s lemma, we have T ( x ) a scalar for each x E Z ( G ‘ )and hence x E Z(G).Thus Z ( G ’ )E Z ( G ) n G and the reverse inequality is trivial. We next show that ( = 0 on G‘\Z(G’). Indeed if x E G\Z(G’) then there exists y E G’\C,.(x). Since G” E Z ( G ) ,Schur’s lemma shows that T ( x  ‘ y  ‘ x y ) is a scalar, a1 say, and a # 1 because x y # y x . Thus a T ( x ) = T ( y  ‘ x y ) , which yields a ( ( x ) = ( ( x ) . Since a # 1, ( ( x ) = 0 as asserted. Thus ( = 0 on G’\Z(G’). On the other hand, I ( ( x ) I = d for x E Z(G’) because there T ( x ) is a scalar a1 with I a 1 = 1. Therefore
and I G’: Z(G’)I = d 2 as asserted. Step 5
Contradiction.
Let S := C,,(P). Then S 2 Z ( G ) n G’z G”by Step 3, and so S G . Then S 4 G’P = G. Since P c C,(S) Q G, and P $ Z(G), Step 2 shows that C,(S) = G, so S E Z ( G ) . Hence by Step 4 Z(G’) = S = C,,(P). Now P acts by conjugation on G’and S is the union of its orbits of length 1. All other orbits under P have lengths that are powers of p greater than 1, so p divides Q
IG’I  ( S I = l S l { / G ’ : Z ( G ’ ) l  I } =
I S l ( d 2  1)
(using Step 4). Since p X I G’I by Step 3, this implies p I (d  l)(d + 1). Hence d + 1 2 p contrary to the choice of d. Lemma 8.7 Let H be a normal subgroup of the group G, let ( be an irreducible ordinary character of G such that H $ Ker (, and suppose some x E G satisfies C , ( x ) = 1. Then ( ( x ) = 0.
Enumerate the irreducible ordinary characters of G so that l l , . . ., tr+‘, ..., (, do not. The condition C , ( x ) = 1 shows that I C , ( x ) I = 1 C C I H ( H x )I ; indeed each H y centralizing H x in G / H contains exactly one element of C,(x). The characters ( ’ , . .., (, correspond to a complete set of irreducible characters for G / H , so the usual orthogonality relations give Proof
(, contain H in their kernels and
r
S
i= 1
i= 1
1 I ( i ( x ) l2 = I C G / , ( H x ) I = I C d x ) I = 1 I i i ( x ) 12*
,
Hence i,+ (x) = * . = (,(x)
= 0.
8.7
223
CRITERIA FOR NORMAL SYLOW PGROUPS
Theorem 8.78 Let G be a group with a faithful ordinary representation of degree d < f(p  1). Then G has a normal abelian Sylow psubgroup.
Proof We shall suppose that the theorem is false and obtain a contradiction by reducing to the situation of Theorem 8.6. Let G be a minimal counterexample. Then G has a faithful ordinary character of degree d c +(p  l), and clearly G is not abelian so d > 1. Let P be Sylow psubgroup of G ; by the note above, P is abelian. Step I
For each proper normal subgroup H of G, P n H E Z ( G ) .
The proof of Step 2 of Theorem 8.7A applies. Step 2 G
= G’.
Since P $ Z ( G ) , it follows from Step 1 that PG‘ = G. Suppose G # G’. Then the argument of Step 3 of Theorem 8.7A applies to show that G’ n P = 1. Since P is not normal in G we can choose a prime q so that q G : N G ( P )1. Let Q be a Sylow qsubgroup of G’. For each x E G, x  ‘ Q x is also a Sylow qsubgroup of G’, so x  ’ Q x = y  ’ Q y for some y E G’. This shows that x E N , ( Q ) y s N,(Q)G’. This holds for all x E G, so G = NG(Q)G’.Hence NG(Q) contains a Sylow psubgroup P I , say, of G . Consider the subgroup Q P , . Then Q + Q P , , so Q P , is solvable. On the other hand, since P , is conjugate to P, the choice of q shows that Q $ N G ( P l ) ,so P I is not normal in Q P , . This contradicts Theorem 8.7A. Thus G = G’ as asserted.
1I
Step 3 ( is irreducible. The proof of Step 1 of Theorem 8.7A applies. Step 4 Z ( G ) is the unique maximal normal subgroup of G. It is a p’group and G / Z ( G ) is simple and noncyclic.
Let H be a maximal normal subgroup of G. Then the subgroup P H # G ; for otherwise G / H z PIP n H is abelian and so H = G by Step 2. By Step 1, P n H G Z ( G ) = Z ( G ) n G’; so we can apply the first part of the argument of Step 3 of Theorem 8.7A to show that P n H = 1. Since PH is a proper subgroup of G, the minimality of G shows that P a P H . But H 4 P H , so P H = P x H . This shows that P E C G ( H )a G.Since P $ Z(G), Step 1 now shows that C G ( H )= G, so H c_ Z ( G ) . Since G is nonabelian and H is a maximal normal subgroup, H = Z ( G ) .Since H n P = 1, Z ( P ) is a p’group. Since H is a maximal normal subgroup, G / Z ( G )is simple. Finally, G / Z ( G ) is noncyclic because otherwise G would be abelian. Step 5 Let N := N , ( P ) and N o := {x E N I C,(x) # 1). Then (1)
x‘Px n P = 1
for all x
E
G\N
VIII
224
x'Nox n No =
BLOCKS WITH CYCLIC DEFECT GROUPS
if if
(;:G)
X E N X E G\N.
I n particular, both P\{ l} and N,\Z(G) are trivial intersection sets. First suppose x ' P x n P # 1 for some x E G . Then x  ' P x n P $ Z ( G ) by Step 4, and so H I= C,(X 'Px n P) is a proper subgroup of G. By the minimality of G, H has a (unique) normal Sylow psubgroup. Both P and x ' P x are Sylow psubgroups of H because P is abelian; and so P = x ' P x and x E N . This proves (1). It is easily verified that N o is a normal subset of N , so to prove (2) we must show that x ' N o x n N o $L Z ( G ) implies x E N.By hypothesis there exists y such that x y x y both lie in N o \Z(G). Put P , := C , ( x y x  ' ) and P , := C,(y); these are nontrivial by the definition of N o . Now x ' P , x and P , are psubgroups of C,(y) and C,(y) # G by the choice of y. By the minimality of G, C,(y) has a unique Sylow psubgroup, which will contain both x ' P , x and P , . Since the psubgroups of G are all abelian, this means that x ' P , x E C,(P,). But P E C,(P,) and C,(P,) # G by Step 5 . Therefore P is the unique Sylow psubgroup of cG(P2) and x  ' P , x C P . Hence 1 # PI G x P x  n P, and so x E N by (1). This proves (2). We now introduce the following notation. Put p" := I P 1 ,n := I Z ( G ) I and p"nh := IN I. Let = 1, , , . .., c, be the irreducible ordinary characters of G and write
',
'
el
rz
(i=1,...,~),
(ci)N=ai+bi
where ai is the sum of the irreducible constituents of ([r)N, none of which contain P in their kernels, and pi is the sum of the remaining irreducible constituents. Since the representation affording ci is scalar on Z(G), Ifli(x)\ = &(l) for all x E PZ(G). Step 6 For each i > 1, fli(l)2pm< ci(1)'.
It follows from (2) that N o \Z(G) has I G : N I distinct conjugates that are mutually disjoint. Therefore for i > 1
2
(G:NI
For each irreducible character Therefore
C I
xeNo\Z(C)
el,
I2
+ Ci(l)'/B*
Ic,(x)I = C,(l) whenever x E Z(G).
8.7
CRITERIA FOR NORMAL SYLOW
+ 2(ai,
225
pGROUPS
+ fii(1)2/h Ci(lI2/hprn since ai = 0 on N\No by Lemma 8.7 and I pi(.) I = pi( 1) on P Z ( G ) by the definition of pi. However, the definition of tli and pi shows that (ai, p i ) = 0. 2 ( a i , ai>
pi>
Moreover since Ci # Cl = l,, Step 4 shows that P $ Ker Therefore we conclude
C,(q2
ci, so ai # 0.
+
= prn{p,(1)2 h(a,, a,>  h) 2 PmS,(l)Z.
Step 7 (1,)' = C;=1
pi( 1 ) C i .
ci)
By Frobenius reciprocity, (l:, = ( I p , ( C i ) p ) = ( I p r (ai)p+ ( p i ) p ) . By the definition of a, and pi and Clifford's theorem we have (l,, (ai),) = 0 and ( I p r (pi)p) = p i ( l ) ,and so the result follows. Step 8 The conclusion.
We first show that d I 1 +
i
i= 1
(ci
3
( C i , Cr)fli(l).Indeed by Step 7,
CT>
Cr>fii(l)= ( ( l p ) ~ , = (1P
9
m P ) = (CP
9
CP)
2 c(l) = d
since
C p is a sum of
((1) characters of degree 1. Since
(C, c ) = 1, the assertion follows.
(Cl, [g)&(l) =
Therefore, from Step 6 it follows that d 5 1+
I;=,
=1
i
i=2
9
cz)ci(l)/p*m
+ ((lG, rr>  I } / P
because 1' = Ci(l)Ci (see the Example of 52.5). Since (lG, Cr) = [(l)' = ti2, this shows that p*" I(d2  l)/(d  1) = d + 1. But d + 1 < f(p + 1) by hypothesis, and so m = 1. However by Step 2 this means that the hypotheses of Theorem 8.6 are satisfied; and that shows that our counterexample is impossible. This completes the proof of the theorem. EXERCISES
1. Let G be a group with a normal pcomplement for some prime factor p of I G I. If G has a faithful irreducible ordinary character of degree d < p  1, show that G has a normal Sylow psubgroup.
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VIII BLOCKS WITH
CYCLIC DEFECT GROUPS
2. Let G be a group. If every irreducible ordinary character c of G has degree d Ip  1, prove that G has a normal Sylow psubgroup. 3. Let G be a group and c a faithfulordinary character of degree d. If R is the set of all prime factors p of IG 1 such that d < p  1, then show that G has an abelian Hall Rsubgroup. 4. Let P be a Sylow psubgroup of a group G such that C,(u) = P for every u # 1 in P and let p" be the order of P. If G has a faithful ordinary representation of degree d < f(p"  l), prove that either P 4G or no proper subgroup of P is normal in N,(P), N,(P)/V is nonabelian and d2  d 2 2(p" 1). 8.8 Notes and comments
The earliest attempt to analyze the characters in a pblock of a group goes back to a paper of Brauer [2], who successfully determined the structure of blocks of defect 0 and 1. In [3] Brauer investigated the pblocks ofa group G whose order is divisible by p only to the first power and then applied this information to prove that such a group with a faithful representation of degree d < f(p  1) is not simple. In fact, in [3] Brauer proves Theorem 8.6 without the hypothesis that G = G . There are further applications of the results of Brauer [3] on the blocks of defect 1. For example, the following result of Brauer [13] leads to some important theorems on simple groups whose orders are divisible by a prime to the first power only. Theorem 8 8 A Let G be a group with G = G' and let z be an element of order p such that C&) = ( 2 ) . Then the following are true.
I p  1 such that g = ( P  l)P(l + nP)/rl
(i) There are integers n and t with
t
where g IGl,1 + np is the number of subgroups of order p, and t is the number of conjugate classes of elements of order p. (ii) Assume that there are no positive integers u and h such that n(u + 1) = u(u 1) h(pu 1). Then one of the following holds I=
+ +
+
(a) n = 1, t = 2, and G Y P S Y 2 , p ) , p > 3; (b) n = (p  3)/2, t = ( p  1)/2, p = 2" 1 > 3, and G z PSY2, p  1);
+
(iii) if n < (p + 3)/2, then G 2: PSY2, p ) , p > 3, or G z PSY2, p where p = 2" + 1 > 3 is a Fermat prime.
 l),
Using this result, Brauer and Tuan [l] proved the following result. Theorem 8 b B If G is a noncyclic simple group of order g = pq'g,, where p and q are distinct primes and go < p  1, then G Y PSY2, p) where p = 2" + 1 or G Y PSY2, 2") where p = 2" + 1 > 3.
8.8 NOTES A N D COMMENTS
227
As a consequence of this result, the only simple groups of order prqb, where p , q, r are distinct primes, are PSY2, 5) and PSL(2, 7). As another application of the above ideas we have a result of Brauer and Reynolds [l]. Theorem 8.8C If the order g of a simple noncyclic group G has a prime factor p > g1'3, then G is isomorphic to PSL(2, p) where p > 3 or to PSL(2, p  1 ) where p > 3 is a Fermat prime.
After the two papers of Brauer [3] on the structure of blocks with defect groups of order p , the theory remained stagnant until Thompson [l] generalized the result of Brauer in a special case. Using this idea of Thompson, Dade [2] generalized Brauer's results [3] to blocks with cyclic defect groups. Theorem 8.80 Let p be a fixed prime, B a pblock of G with a cyclic defect group D of order pd, b the unique pblock of N := N,(D) such that bG = B, c a fixed pblock of C := C,(D) such that cN = b, and E the set of all elements x of N such that cx = c. Then
(i) q : = I E : CI is a factor o f p  1. (ii) E/C acts on the set of nontrivial irreducible ordinary characters of D such that each orbit has (pd  l)/q elements. (iii) If A is a set of representatives of the orbits in (ii), then B has q irreducible modular characters $l, $', ...,$q and q (pd  l)/q irreducible ordinary characters r l , c2, . . ., [ q and cA(A E A). (iv) The decomposition numbers associated with B are 0 and 1. The decomposition numbers d,,associated with (l,,&') are all equal for a fixedj. If d, = 1, then there is exactly one ('such that d, = 1. If d,, = 0, then there are exactly two ['such that d, = 1.
+
Moreover the generalized decomposition numbers are also determined. The situation bears a close resemblance to that of blocks of defect 1 (see Theorem 4.6B). The proof of Theorem 8.8D given in Dade [2] is based on a complicated induction argument, but the paper should be accessible to someone who has followed the arguments of the present chapter. There are two other proofs of this result, one due to Peacock [l] and the other to Feit [5]. These results of Brauer and Dade do not seem to extend to other types of blocks. The theorems of $8.3 are due to Green [1,2]. Theorem 8.7B was proved by Feit and Thompson [l]. For further results in the latter direction see Blau [2], Lindsey [l], and Winter [l]. In Feit [3] Theorem 8.7B is extended as follows. Theorem 8.8E Let G be a group of order g which has a faithful representation of degree n over the complex field. Let g = g1h where (gl,h) = 1 and p > n + 1 for every prime p dividing h. Then G contains a normal abelian subgroup of order h or h/p for some prime p.
228
VIII
BLOCKS WITH CYCLIC DEFECT GROUPS
There are other applications of Theorem 8.6 and Theorem 8.7B. For example see Brauer [9,11,12], Brauer and Fowler [l], Hall [2], Herzog [l], and Wales [1,2,3,4,5]. An interesting problem in finite group theory has to d o with the determination of all simple groups with orders p"qbr' divisible by exactly three distinct primes p, q, r. By Thompson [2] the primes involved in such groups are 2,3,5,7,13 and 17. No example of simple group is known in which all three exponents a, b, c are greater than 1. The first step is the determination and classification of all simple groups of order 2"3'r, r prime. In this case Brauer's method is powerful enough to give a complete solution. Leon and Wales [ 13 have extended the result to the case 2"3'r' where the Sylow rsubgroups are cyclic. The results of Brauer, Leon, and Wales can be stated in the following combined form. Theorem 8.8F The only nonabelian simple groups of order 2"3'rC, with cyclic Sylow rsubgroups, are A,, Sp(4, 3), U3(3), PSL(3, 3), and PSL(2, r ) for r = 5, 7, 13, and 17.
The cases r = 5, c = 1, 2 were settled by Brauer; the cases r = 7, 13, 17 with c = 1 were treated by Wales; and the general case was given by Leon and Wales.