LINEAR GROUP WITH QUADRATIC INVARIANT. 187 



If j = m, we transform S by T^Tu (k not 1, I or m) and obtain a 

 substitution in G which replaces i by 



.?! H ----- h a<mim- 



From the resulting substitution S in which a 12 , a 13 , . . ., a lTO are 

 not all zero, we derive a substitution ^ belonging to 6r and having 

 ii + "?2 4 s L We g et #1 immediately if a^ + f, 4= 1 for j = 2, 3, . . ., 

 or m 1. In the contrary case, we have 



172) oj, + &-!, ! 8 =4, = - = ?-!. 



If 12 = 0, then 0^=1 and therefore i w =0 by 147), contrary to 

 the assumption that a 12 , # 13 , . . ., i m are not all zero. Hence a 12 =|=0. 

 Transforming $ by a suitable product of the d, we can take 



<*12 = a !3 = ' ' * = airo-l 4= 0. 



Transforming 1 ) the resulting substitution by OJ'S, we obtain a sub- 

 stitution which replaces f^ by 



If p w > 5, we can determine a and /3 in the 6rJP|j) n ] such that 

 a+/J-l, .L+^i + ^^ + l. 



Indeed, since ai 2 = i3=H^? an( i "u+^L^l? ^n e second condition 

 becomes 2a/3 4= 0. But, of the p n s sets of solutions in the G-F[p n ] 

 of the first condition, where s = 1 according as 1 is a square 

 or a not -square in the field, only four sets of solutions have either 

 cc or ft equal zero. Hence, if^) n >5, there exist other solutions. 

 For p n = 3, we transform S, in which 



"11 = > = a is = i4= a i5 = 1> 

 by TF" 8 | 45 and obtain a substitution in 6r which replaces ^ by 



for which therefore a^! -f ^ 2 = 0. 



For p n =b, S has a u = 0, J, = J 8 ^ 4 = 1 in virtue of 172). 

 Transforming S by a product of the C iy we may take 



The resulting substitution is transformed by .R^ into a substitution 

 of G which replaces | t by 2| 2 -f 2 3 + cf 15 | 5 + -., for which 



1) If the transformer does not belong to 0^(w, p), we afterwards trans- 

 form by OJ'|. Since the product OJ'JPJ^ belongs to the main group, the 

 transformed' substitution will belong to G. A like remark is to be understood 

 throughout this section. 



