M.-P.— Vol. I.] DICKSON— THE ORTHOGONAL GROUP. 39 



These combine into the single condition 



a 2 (S 2 3 + S 2 4 ) — 2 a Si o 3 + 3 2 4 7 2 := 8 2 4 — a 2 x . 



For S 2 3 -}- S 2 4 = o , and therefore S 3 ^ , a solution is given 

 by 7 = when o\^e> , and by 7 = 1 when Si = o. For 

 S 2 3 -\- S 2 4 ^ , there exist solutions of the equivalent equa- 

 tion of condition, 



-{ a (8 2 3+8 2 4)-Sl 83 ^ 2 + 8 2 4(S 2 3 +a 2 4) 7 2 = ^4(8 2 3+8 2 4-8 2 l). 



Similarly by transforming the result by U5 we can (since 

 8' 5 =zB 5 = o) make S" 5 = S" 6 . 



Hence /contains a substitution (not the identity) Say T\% 

 commutative with both Tu and 7 56 . Thus / contains the 

 substitution 



( So) T12) 7\a Z5G ( So) Tvi)~ l 7^56 T26 = Sq) 7l6 So)' 1 Tl<o = 7l6 726. 



Since the alternating group on w>4 letters is simple, the 

 group / containing 71 6 7 2 e contains the whole alternating 

 group. 



Now C\ C 2 transforms T n T\& into Tn 7i 3 C\ C 3 , so that 

 /contains C\ C 3 and thus every C { C i . Hence, by the 

 lemma of § 5, the invariant subgroup /, under II, coincides 

 with H, whose maximal invariant subgroup is thus of order 

 1 or 2 according as m is odd or even. 



16. Consider the orthogonal group H on 5 indices. 

 Suppose an invariant subgroup / contains a substitution 



S: fi^So^, (*=i.. 5) 



not the identity and not a product of the C { (see § 5). By 

 a previous transformation of S we can suppose that a\ x ^ 1. 

 Transforming S by 



f £ ' 



05432 : \ , 



f 5 = \ f= 5 -f fl f 4 + V £3 + p I a 



we have a' n = «n , a'n = \ a )5 -f /* a H + v «i3 + £ a i2 • By 

 §§ 8-9 we can make a\ 5 = o . Transform the resulting sub- 

 stitution S' by 6>432 . We reach the substitution 



