M.-P.-Vol. I.] DICKSON— THE ORTHOGONAL GROUP. 41 



By § 14 we may suppose that 83 is not zero. Since 2 is 

 a not-square, we can not have S\ -f- S 2 2 = o. 



If 8 2 x -\- S 2 2 is a not-square, we may make S t = S 2 by § 7. 

 Transforming by 6)345 , we can make 8' 3 = \ S 3 = 81 ; for the 

 condition \j? -\- v 2 = 1 — 8\/S% can be satisfied. Since 

 S 1 = S 2 = S 3 , <5g is commutative with Tn and T 23 . Hence 

 / contains the substitution 



If, however, S 2 i -f- S 2 2 is a square, we can make 8 2 = . 

 Then 6*g will be commutative with 7^ and Z45 . Thus 



(og 7ia) Ti*> Z45 ( Z12 ■•S'g) 2^5 Tu = T 2 *> Z15 



belongs to group /. 



In either case it follows (see end of § 15) that //is sim- 

 ple if m be odd, but has a maximal invariant sub-group of 

 index two if m be even. 



18. Suppose next that 8 2 2i -f S 2 25 is a not-square. By 

 § 7 we may thus suppose that 7?g C x C 2 is commutative with 

 T i5 . If it be commutative with every T {i (i,j— 1, . . 5) it 

 has the form (see end of § 12), 



fi--&+Y<& + &+&'+& + M (* = i,..5). 



But the transformed of this substitution by Ci C 2 is not 

 of the same form. We may thus assume that 7? = 7?g d C% 



is commutative with Z45 but not with Tn , for example. 

 Thus / contains 



R~ 7i2 745 7? Z45 7l 2 = "Jp T11 , 



not the identity. As at the end of § 12, we have 

 P 



K-fe--ftip,6 (* = i--5) 



If p 5 = o we are led to the case solved in § 17. By § 8 

 we may make p 5 = o unless p 2 3 -\- p 2 t -j- p 2 5 = o with — 1 a 

 not-square. In the latter case, p 2 i -f p\ = 2, when we may 

 make pi = p 2 — -\- 1. Further, we can not have p 3 = Pi = P5', 



