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
a n d (fiiKi
= 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
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
( 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
(0i, lx ) = 0 = (faK,lK) and {fa, fa) = 0 = {faK^2K)
yield X^=i
^n* =
N\s and Ylt=i ^
\X '
respectively, whence
] T i{s - i)rti = \X : N\s2 - \X : N\s2 = 0.
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.
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.
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.
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.
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\
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