256 CHAPTER XL 



Of the Jc 2 sets of two integers each < k, k 2 /qj have their integers 

 chosen from the &/9> multiples of gi and are to be excluded. We 

 thereby exclude, in particular, the sets of integers each of which is 

 one of the k/%iC[j multiples of qiq_j. Hence, in afterwards excluding 

 the sets of integers each of which is a multiple of #y, we subtract 

 the number k 2 /qj k 2 /qjq 2 . After the required exclusions have all 

 been made, there evidently remains the number of sets indicated 

 Among the latter sets, the couple 0, does not occur since 





1-2 1-2 



235. A cyclic group generated by a substitution C of period 

 p n 1 will be called special if two of its substitutions C a } C b of 

 period p n 1 are conjugate within 6r, i. e., have the same set of 

 multipliers. Since a and b must be prime to p n 1 , the condition 

 requires that C and C bcti shall have the same set of multipliers, 

 where % is determined from afl^^l (mod p n 1). It thus suffices 

 to investigate when C and C m have the same multipliers, m being 

 prime to p n 1 and 1< m < p n 1. The three distinct ways in 

 which the two sets 



a r a s a*- a mr a ms a mt 



may be identical in some order will be considered in turn. 



i) If a mr = a r , a ms = a s , a mt = a', then r (m 1), s(m 1), and 

 therefore also </(w 1), are divisible by p n 1. Since g is prime 

 to p n 1, m 1 must be divisible by p n 1, contrary to hypothesis. 



ii) If K mr =a s , a ms =a r , a^^a*, then must 



w 1). 



Then r must be prime to p n 1 ; for a common factor would divide s 

 in virtue of the first congruence, whereas the greatest common divisor 

 of r and s is prime to p n 1. Hence, by the last congruence, 



246) w 2 EEl (mody-1). 



Inversely, if m be any solution of 246) and if r be any integer 

 less than and prime to p n 1 and if s be determined by 



s = mr (mod p n 1), 



then (7 and C m have the same multipliers. Moreover, C is the r ih 

 power of a substitution with the multipliers a, a, a~ m ~~ l , which 

 may therefore be taken in place of C as generator of the special 

 cyclic group. 



If 2 k be the highest power of 2 contained in p n 1 and if % = 

 when & = or 1, H =1 when & = 2, ?c = 2 when k ^> 3, and if ^ be 



