484 Dr Burnside, Certain Simply-Transitive Permutation- Groups 



Suppose now that T is that irreducible representation to which 

 ^;^ belongs. Then 



Xm^in^J = ^ S e^'^j. 

 IP 3 = 1 



Now choose the set 

 SO that it contains the symbol x^^q. Then 



P3=l 



If m is not a prime k may be chosen so that e^ is a primitive 

 g-th root of unity, where g is a prime factor of m; and when k is 

 so chosen, 



Hence, if m is not a prime, G has a self-conjugate subgroup 

 containing M". 



If m and w are different primes, and V is that irreducible 

 representation to which ^^^ „ belongs, so that 



It P ^ 



then - S e«:? = S e'^J'^i rf'fii. 



Pj-i i=i 



Unless each ^^ is zero, in which case the group has a self- conjugate 

 subgroup containing N, this equation actually contains powers of 

 7) on the right. Hence when the indices of the powers of rj are 

 reduced (mod. n) each power must occur with the same coefficient. 

 This shews that ^ must be a multiple of n, and that the reduced 

 variables for F must be 



baj,i5 bag, 15 •••} ^ar,i (* ^^= "j J-j ^j •••> n I). 



The family of representations to which F belongs accounts there- 

 fore for (m — 1) w of the reduced variables. The remaining reduced 

 variables are t t t 



bO,05 S0,1' •••' &0,n-l3 



and in each irreducible representation whose reduced variables 



belong to this set ^ _ ., 



° AM ~ Xe • 



Hence when m, n are different primes G has a self-conjugate 



subgroup containing either M or N. It is clear that the same 



method of proof will apply, when the transitive Abelian subgroup 



has three or more independent generators. Hence: — 



A simply- transitive permutation-group, which contains a regular 



transitive Abelian subgroup, always has a self-conjugate subgroup, 



except possibly when the operations of the Abehan subgroup are 



all of the same prime order. 



if 



