6

CHAPTER 1.GENERAL LEMMAS

we are done. So we assume that t 1 and derive a contradiction. By hypothesis,

KK = PK, the regular character of the abelian group K, and thus TTK is multiplicity-

free. Thus

{^IK^JK)

~® a n d (fiiKi

^K)

= 0 for all distinct i and j . As t 1, each

(fiK is thus the sum of some but not all the nonprincipal irreducible characters of

K.

Suppose that for some i, fa is constant on K#. Then faK is a linear combination

of the form faK — apK

+ M K »

and so (f)iK has either all or none of the nonprincipal

irreducible characters of K as constituents. This contradicts what we just showed,

so no fa is constant on K#.

In particular N is not transitive on K# and fa G £ for all i. Thus for any

y G X not conjugate to an element of K#,

/op\ (1) 0i(2/)

is a

non-negative integer independent of i, and

^ (2) ^ ( j / ) 0

4

( l ) i f j / ^ l .

Indeed by (25), /;(?/) = c is independent of i and then n(y) = 1 + £c, whence

c is a rational number, hence an integer by [1^; 32.9]. Moreover as 7r(y) 0 and

t 1, c is non-negative. In particular if the equation fa(y) = fa(l) holds for one i

then it holds for all i, whence 7r(y) = 7r(l) and y = 1. This establishes (2C).

At this point we may imitate the proof of Burnside's Theorem [Hul; V.17.3],

setting s = fa(l) and letting rii be the number of y G X not conjugate to an

element of K* but with fa(y) = i. By (2C(2)), ns = 1. In view of (2C(1)), the

relations

(0i, lx ) = 0 = (faK,lK) and {fa, fa) = 0 = {faK^2K)

yield X^=i

^n* =

1-^

:

N\s and Ylt=i ^

n

*

=

\X '

N\s2,

respectively, whence

s

] T i{s - i)rti = \X : N\s2 - \X : N\s2 = 0.

i=i

Therefore n^ = 0 for 1 i s. But then the preceding equation implies that

\X : N\ = 1 and so X = N, a final contradiction.

We digress to note Burnside's classical corollary.

COROLLARY

2.7. Let X be a transitive permutation group of prime degree p.

Then either X is 2-transitive or X has a normal Sylow p-subgroup of order p.

PROOF.

Clearly a Sylow p-subgroup K of X is a regular subgroup of X of

order p generated by a p-cycle and K = Cx(x) for every non-identity element x of

K. The result now follows from the previous proposition.

We conclude with three further small results.

LEMMA

2.8. Let X act on 17 with rri2(X) = 1 and \Q\ even. Suppose that for

every a,/3 G ft with a =^ /3, there is an involution s G X with as = /3. Then \il\ = 2.

PROOF .

Let s be any involution of X. Since rri2(X) = 1, all involutions of X

are conjugate to s, so the hypotheses apply to (sx) and we may assume that X =

(sx). By the Brauer-Suzuki Theorem [IG; 15.2,15.3], s G Z*(X), so X = Z*(X) =

02'(X){s). The hypotheses imply that X is transitive on £2, so \Xa\ = \X\/\Q\