SUBGROUPS OF THE LINEAR FRACTIONAL GROUP LF(2, p). 271 



with those of G p m in the first row has the property that the sub- 

 stitutions in each row are all found in a corresponding row of the 

 rectangular array for G. In fact, two operators A, B of G' lie in 

 the same or in different rows of the array for G' according as 

 AB~ l is or is not in G p m . But AB~* belongs to G' and hence 

 belongs to G p m if, and only if, it occurs among the substitutions in 

 the first row of the array for G. Hence each row for G' lies wholly 

 in a row for G. The quotient -group G' / G p m is therefore a subgroup 



6r<j_ of the quotient -group G/Gg\ the latter being a cyclic 6r,_i. 



"all 



Indeed, these quotient -groups may be obtained concretely as groups 

 of the permutations of the rows of G induced by applying as right- 

 hand multipliers the substitutions of G or G'. But all the substitu- 

 tions in the same row of G (and, a fortiori, all in the same row 

 of G f ) give rise to the same permutation. Hence G d -. is an abstract 



cyclic group. Now G contains s cyclic 6?J i , where x runs through 



Tjl 



the series of marks of the GrF\s\ 9 all conjugate under the trans- 

 formers Sp. Leaving different elements t fixed, they have no sub- 

 stitution other than the identity in common. Counting also the s 1 

 substitutions of period p, we have accounted for all the substitutions 

 of G. Besides the cyclic subgroups of G^\ G therefore contains no 

 cyclic subgroups other than the G ( a'*\ for the various divisors d 



n -t 



of -zr' Among these cyclic groups occurs one whose substitutions 



p-i - 1 



may be chosen as the right-hand multipliers in forming the above 

 array for G'. In fact, within the row of G 1 corresponding to the 

 generator of the quotient cyclic G d there must exist a substitution A 

 such that A d , and no lower power, belongs to the group G p m whose 

 substitutions form the first row. The right-hand multipliers for the 

 array may thus be chosen to be J, A, A 2 , . . ., A d ~ 1 . Hence G' is 

 given by the extension of the G p m by a certain G ( '*\ within which 

 G p m is self - conjugate. But the largest subgroup of GM(S) within 

 which G p m is self -conjugate is ( 249) the group H of order sK, 



K^ ^, given by the extension of G by a cyclic G?' 0) . In 

 particular, d must be a divisor of K, so that d depends upon the Gyn. 

 The cyclic 6rj?' 0) contains a single cyclic 6rrf' 0) . Hence, by trans- 

 forming G' by a suitably chosen /S^, we obtain a group G p m d (con- 

 jugate with G' under 6r) given by the extension of G p m by the sub- 

 group #ST' 0) of #' 0) . The substitutions & of G p m transform that 

 subgroup into p m conjugate cyclic 6$' 0) , since Si replaces the fixed 

 elements oo, by elements oo, I. These p m groups together with 

 G p m contain all the substitutions of G p m d , as shown by simple 



