OPERATORS AND CYCLIC SUBGROUPS etc. 249 



of p n 1. Similarly, if a =j= 7, then must a = /3, ' =j= 0. Finally, 

 if a = p = 'y, each may be taken equal to unity. Then, by induction, 

 j^r. ft* __ x if 1 = I/ s' = z -I- TK'IJ I ra fl x 



so that 12 would have the period j?. Hence either (i) or (n) must 

 be satisfied. 



Suppose, inversely, that relations (i) are satisfied. Setting 



y__ _</_ y _ , tt " X 



a "' 



E takes the form 



f 2 ~\7~f ~V f7l 



x == a x, JL === a JL j == ex, 



and is thus of period -jp (p n 1) if, and only if, a be a primitive 



root of the GF[p n ~\. Interchanging x with y, the proof follows for 

 case (H). 



Using the theorem just proved, we proceed to determine the 



number and conjugacy of the cyclic subgroups of order -^p (p n 1) 

 which leave the symbols {x} and {y} fixed. For case (*), 



E: x'=a~ 2 x, y'^ccy, tt' = az + a'y + a" x (a' 4= 0, a 3 =f= 1), 

 where a is a primitive root of the GF[p n ~\. By induction we find 



a'e 



_ 

 I 



In order that Q r E t shall be identical with the substitution 

 x' = a~*x, y' = ay, z' = az + qty + () ff ^, 



it is necessary and sufficient that 



Let J^- denote any one of the (p n l)/(p 1) distinct marks M^ 

 Jf 2 , . . . such that no two have as their ratio an integral mark 1 ). 

 If a be a fixed mark =j=0 and Jfan arbitrary mark, ihep n (p n l)/(p 1) 

 substitutions 



238) x'=a-*x, y'-xxy, e' = as + My + MX 



have the property that no power of any one of them reduces to one 

 of the set. We therefore obtain that number of cyclic subgroups of 



order ^p(p n -l). 



Furthermore, every substitution V of the subgroup leaving {x} 

 and {y} fixed, and having a = /3, and of order a divisor of -^p(p n 1) 



1) The marks M lt M S1 . . . are evidently the multipliers in a rectangular 

 array of the marks =J= of the GrF[p*], the first row being formed by the 

 integral marks 1, 2, . . ., p 1. 



