THE HYPERABELIAN GROUP. 117 



For proof, we reflect on its main diagonal the determinant of 

 S~ , then change the signs of the 21 1 st row and column for 

 I = 1, . . ., m, and finally interchange the 21 1 st row with the 2 I th 

 row for I = 1, . . ., m, and likewise interchange the corresponding 

 columns. We obtain the determinant 



Hence AA^ = 1, being the determinant of the product S3 1 . 



131. Theorem. The maximal subgroup M of the hyperabelian 

 group H(2m, p 2n ) which transforms into itself the Abelian group 

 SA(2m, p n ) is given by the extension of the latter by the substitution 



wJiere Q is a primitive root in the GF[p 2n ]. The index of S A(2m, p n ) 

 under M is p n -\-l. 



We determine all hyperabelian substitutions 



2m 



fc (* = !,..., 2m) 



which transform the Abelian group into itself. Now S transforms 

 the Abelian substitution, affecting a single index, 



| 2 r 1= ?2r 1+ ?2r 



into the substitution 



(* - 1, . . ., 2m), 

 whose coefficients must therefore belong to the 6rJPf_p n ], viz., 



f -2r-i ^L-i ft J ^ ? 2m; r = 1, . . ., m). 

 Likewise, S must transform the Abelian substitution 



?2r == ?2r H- ?2r 1 



into a substitution belonging to the GF[p*]. Hence the products 



r ft j = 1, - -, 2w; r = 1, . . ., m) 



must belong to the GF[p n ]. The reciprocal 8~ L must transform 

 the Abelian group into itself. From the above results, it follows 

 therefore that the products 



2r-i* _!, , ,.. .,.f; r = , . . ., m 

 must belong to the GF[p"]. Combining our results, every product 



