THE HYPERABELJAN GROUP. 121 



would T x be the identity or T_I. It follows that A (2m, p n ) is a 

 subgroup of HA(2m, p n ). We proceed to determine the number of 

 conjugates to the former group within the latter group, using the 

 result of 131. 



133. Theorem. -- The largest subgroup M' ofHA(2m,p 2n ) which 

 transforms A(2m, p n ) into itself is identical with A(2m, p n ) ifp = 2 

 or if p > 2 and p n -f 1 contains a higher power of 2 than m contains; 

 in tJie remaining case, the order of M' is double the order of A (2m, p n ). 



The determinant of S=UV a being supposed to be unity and 

 that of U being unity, it follows that V a has determinant 



107) a -m( P -i) =1> 



Now V a and T y .V a correspond in the quotient -group HA(2m, p 2ti ) 

 to the same operator. We investigate the conditions under which 

 T y V a has its coefficients in the GF[p n ]. The necessary and sufficient 

 condition is seen to be 



(xtf)"*- 1 -!. 



Hence must x 2 =a pn ~ 1 and therefore 



or a must be a square in the GF[p 2n ]. The remaining condition 

 v? m = 1 becomes an identity in virtue of 107). Hence, if the solutions 

 of 107) are all squares in the GF[p 2n ], the substitution 8 UV a 

 will correspond in the quotient -group to an operator belonging to 

 A(2m, p n ). But, if there occur not-squares as solutions of 107), the 

 resulting substitutions F may be expressed as products V v Vp*, v being 

 a particular not -square. Then F^ corresponds in the quotient -group 

 to an operator of A(2m, p n ), while F does not. In this case the 

 group A (2m, p n ) is transformed into itself by a subgroup of 

 HA(2m,p* n ') of double the order of A(2m,p n ). 



For p = 2, the theorem follows at once since every mark of the 

 GF[2 2n ] is a square. For p > 2, we are to determine in what 

 cases 107) has as its solutions in the GF[p* n ] only squares. A 

 common solution of the pair of equations 



i 



is required to be a solution of a 2 (p * 1} = 1. A common solution 

 of 108) satisfies a d ^ n -^=l, where d is the greatest common divisor 

 of m and p n + 1. The condition is therefore that d shall divide 



y (P n + !) It is satisfied if, and only if, p n + 1 contains 2 to a 

 higher power than m does. 



