LINEAR GROUP WITH QUADRATIC INVARIANT. 173 



For r = t 2 , 155) may be expressed in the form 



For r a not- square, 155) is not of the form 157), since that would 

 require y*= (* + l) 2 /4r. It follows that to the subgroup G^z of 6r 6 

 of index two there corresponds a subgroup of O^G,^") of index 

 two, where is extended to 1 (6,p n ) by any substitution Of; 4 not 

 of the form 3,4. We proceed to prove that is identical with the 

 group 0((,p n ) defined in 181. We first show that contains all 

 even substitutions on the six letters f^, . . ., f^. Expressing the sub- 

 stitution 



in terms of the indices 



2 = %s> 3 = 1? 

 - it takes the form 



' = 4,5, 6) 



12 



13 



14 



23 



34 



By inspection this substitution is the second compound of 















-~ 



o 



having determinant unity. Hence contains the substitution 

 In the transformation of indices 154), the pairs ^ and 2 , 3 and ^ 

 J 5 and 6 enter symmetrically. Hence contains the substitution 

 (&&*)? ^ wo ^ ^ e distinct integers i, j, ^, each < 6, being chosen 

 from one of the above pairs. But the linear substitution denoted 

 by (ijg 6s) transforms (Jj^fe) i^ (iMs)- Hence contains every 

 cyclic substitution (% r t> s t>i) on the six indices and therefore every even 

 permutation of the six indices. 1 ) 



1) Netto-Cole, The theory of substitutions, p. 35. 



