120 CHAPTER IV. 



Va'Vr 1 : Sii-i-a-'o'Sii-i, 62i= p V-" B | 2 , (Z = l,..., w) 



belongs to SA(2m, p n ~) if and only if a~ 1 a' belongs to the GF[p n ]. 

 It follows that the substitutions V a . (i = 1, 2, . . ., > w -f 1) give the 

 totality of substitutions V a such that V a 'V~^ 1 does not belong to 

 SA(2m,p n ). Hence an identity of the form 



U V { = U' V a . (i and j ^ p- + 1 ; * + j) 



is impossible when U and V' both belong to SA(2m, p' l \ Every 

 hyperabelian substitution 103) is therefore of the form UV a ., i being 

 chosen from the series 1, 2, . . .,p n + 1, while an identity C7F a . = 27' F. 

 requires i = j, 27= U f . Hence the number of distinct substitutions 103) 

 is (p n + l)SA\2m, p*~\. The second part of our theorem is there- 

 fore proven. 



132. Those substitutions of the hyperabelian group H(2m,p 2n ') 

 which have determinant unity form a self - conjugate subgroup H 1 of 

 index p*+l. In fact, for c any mark 4= of the GrF[p 2n ], the 

 substitution 



li-fffe, li-a-'-fe, 65-1, (-3, ...,2m) 



belongs to T(2m, jp 2 "). Its determinant <? (pn ~ 1) can ; by choice 

 of 6, be made equal to any one of the p n -f 1 roots of A pW + 1 =l. 

 Hence there exist hyperabelian substitutions whose determinant A is 

 any root of this equation. By 130, there are no other values of A. 

 The group H r contains a self - conjugate subgroup formed by the 

 substitutions 



106) T.,: |! = v^ (i - 1, . . ;, 2m) [**- 1, x? n +i - 1]. 



The quotient -group will be denoted by the symbol HA(2m, p 2 "). 

 It will be proven simple except in the special cases m = 1, p n = 2 or 3 

 ( 138, 145, 148). By the same references its order HA[2m, p 2n ] is 



where ^ denotes the greatest common divisor of 2m and p n -\- 1. The 

 order of H(2m,p 2n ) is 



m-2) ^ ^ ^ (^,2 w 



The Abelian group SA(2m, p n ) has an invariant subgroup formed 

 by the identity and T_I. The quotient -group A. (2m, p n ) is simple 

 except in the three cases m = 1, p n = 2; m = 1, p n == 3; m = 2, p n = 2 

 ( 119). But H(2m,p 2n ) contains SA(2m, p n ) as a subgroup. In 

 order that T y , shall belong to the latter, the coefficient K must belong 

 to the GF[p n ]. But jc^=5c and p n + 1 =l require K 2 =l. Hence 



