THE HYPERORTHOGONAL AND RELATED LINEAR GROUPS. 143 



For the case p* = 3, we first prove that I contains the sub- 

 stitution QCg. We have shown that I contains an Ol'J not the 

 identity and therefore 0'/ given by 132). If /3 =|= 0, we can make 

 '= 0; indeed, if a be not itself zero, we have in the 6r.F[3 2 ] 



and we need only take r = 1. But the square of Oj',2 gives C^C^ 

 since 0' 4 = 1 when ' = 0. If, however, ft = 0, then a =f= 1. If 

 a = 1, we have at once Oj'a = C^C^. If a =}= 1, then the square 

 of 0?;J gives Of '} = &. 



Having C^Cg, I contains (as above) the substitution 



= ccp (mod 3). 

 Taking for a and /? an arbitrary set of solutions of 



ft 4 = - 1, /3 4 = - 1, whence a 4 + /3 4 = 1, 



we have 0?', 3 where ft = a/3 is an arbitrary solution of ft 4 = 1. 

 Hence I contains 



Transforming the latter by Oi,'f, we obtain by 132), 



(f'f'T T 



Hence I contains every such O"/ . For a = 0, /3 4 ==1, we have 

 a' = ^, /3 r = 0; for a 4 = - 1, /3 4 = - 1, we have ' = - 1 ^. We 

 have therefore reached in the group I every \ in which x = ji, 

 0, 1 + ft, where ft is an arbitrary one of the four roots of ft 4 = 1. 

 Defining the GF\$*\ by the irreducible quadratic congruence, 



* 2 = -l (mod 3), 



we have x ='0, 1, +i, 1 i. Hence % takes every value in 

 the GF\Z*]. We thus reach aU 24 substitutions O\. It foUows 

 that I coincides with H m , 3,1. 



For the case p'= 2, we have in I a substitution 0"^ =f= 1. By 

 the result at the end of 146, it must be one of the six substitutions 



The transformed of the latter by T^ t T^ t i gives 



(i^V^-i"^^,*- 1 - 



Hence, in every case, I contains a substitution of the form 



