SUBGROUPS OF THE LINEAR FRACTIONAL GROUP LF(2,p*). 277 



For each of the fp k extenders Vj ( j > 0, % =(= 0)? eac ^ Ya l ue f n 

 gives a single value of A, which may or may not belong to the GF[p k ~\. 

 Hence there are at most fp k (p k l] substitutions V n ^Vj of period 2. 

 The second alternative therefore holds, so that G-Q contains a sub- 

 stitution of the form 



o, - 



Also (1-j- /!>*) is an integer so that /" is odd. 



In case a Fj ( j > 0) gives rise to one or more substitutions 

 T = V^iVj, we replace Vj by one such T, so that the new Vj has 

 ccj -f d ; = 0. Let N denote the number of these Vj for which there 

 exists a product V^iVj distinct from Vj and of period 2. For such 

 a Vj the equation 



will be satisfied by a pair 77, A =)= 1, 0, such that rf and A belong 

 to the GF[jf\. Hence wiU 



belong to that field. Inversely, if ,/% belong to that field, and rf 

 be an arbitrary mark =|= of that field, there exists an unique 

 solution A in the field, so that there will be p k 1 substitutions 

 V^\Vj of period 2. By the lemma at the end of 251, the N sub- 

 stitutions Vj have distinct values for Oj/yj, here shown to belong to 

 the 6r.F[_p*]. Hence JV<^p*. Let M denote the number of the Vj 

 leading to a single V^Vj of period 2. Then M<^fp k -N. The 

 total number of the V n ^Vj ( j > 0) of period 2 is therefore 



^(.pfc- 1) + M ^ JV(jp* - 1) + fp k - N. 



The second member is greatest when N has its maximum value p k . 

 By comparing the minimum and maximum numbers for the 



. 

 of period 2 in GQ , we have 



255) |(/y- l)p k <iP k (p k - 1) + (f- 



Hence must /*= 1 or 3, leading to the two cases: 



(/-=!) _p>2, n[k even, Q = (p* + Dp*(l>* 1) = 2 

 (f-3) p = 3, fc=l, n even, Q = 60. 



Consider first the case /*= 1. 6r^ contains the transformed of 

 Tl by &, 



