LINEAR GROUP WITH QUADRATIC INVARIANT. 



171 



183. Theorem. The group 0{(3, p n ) just defined is identical 

 with the subgroup of 0^ y p n ) of index two defined in 178. 



It is only necessary to show that every QfJ and every 

 are of the form X or, if we prefer, 151). We have 



H-l-H 

 2 



-f- 1 i G 



2 



T T 



-*-t9 J-13. ^^ 



1213 



In particular, we reach Z 12 T IZ and J 12 T 23 . These suffice to transform 



0*tfO} a into Ol\lOl\l and 0? O?;?. Transforming these products 

 by C^Oj, CtCj and C 2 (7 3 , we obtain every OJ/ Ogf, since 0< trans- 

 forms Of/ into 0%?. Transforming Ql\l (which is of the form X 

 by 178) by T 12 T 13 and T ]2 T 28 , we obtain every $/. 



184. Theorem. T^e group Oi(3,p w ) is o/" 

 second orthogonal group O r (3, p n ). 



Consider the substitutions, in which a 2 -f-/3 2 



two under the 



0: 







1 



Since transforms J? + g + vij into v ( 2 + 6J 4- g), it transforms 

 the group O^S,^? 7 *) of the latter into the group O v (3,p n ) of the 

 former. But is commutative with O^'J. Hence if OJJ serves -to 

 extend the subgroup 0^(3, jp w ) to Oj(3, jp n ), there exists a subgroup 6r 

 of O w (3,j) n ) which 0J extends to the latter. We readily prove that 

 ^ is identical with the Oi(3,p ra ) defined in 181. For example, 

 transform OjOJCC into OilC where 



Here Of;? is not a $i, 2 since (1 + p)/2 = a?/v is a not -square. But 

 CiC 3 is a 1>3 in O i (3,p n '), but not in O y (3,p"). It follows that 6^ 

 contains the product Ojja^Cg, neither factor being a . 



185. It will be shown in the following sections that 

 is not identical with On(m,p*) in the cases m = 4, 5, 6 and there- 

 fore, by 182, that its index under Op(m, p n ) is exactly two. By 

 181184, the same result is true for m = 2 and m = 3. For 



