THE HYPERORTHOGONAL AND RELATED LINEAR GROUPS. 141 



Consider the substitution belonging to H, 



T^T^T^T^-t 

 where i > 2. The group I will contain the product 



since T and 0*\2 are commutative. Since S' leaves j^ fixed, our 

 theorem is proven unless S' reduces to the identity. In the latter 

 case, we find by comparing the values by which S^T und TS t 

 replace 2 that 



2,-=0 (j 3,. . ., w; j=H)> raj/ *-*,. 

 If m > 3, i has at least two values and therefore 



ij=0 (j = 3, . -,*). 



If w = 3, the same result holds if p* > 2. For then a value of r 

 exists satisfying T^ <! + 1 = 1 but not r 3 = 1. Hence must 2j - = 0. 

 Excluding the case m = 3, _p* = 2, it follows that 8 l (which was seen 

 to leave ^ fixed) alters | g at most by a constant factor L Hence 



8 -OR TV*., 



where leaves ^ and | 3 fixed. Hence I contains 



S' = S' 1 (Ti,T tt -*)- l 8(Ti t r,,-i) [.+' = 1] 



which leaves g s , . . ., | m fixed. If $' =(= 1, the theorem is proven. If 

 $' = 1, we find by comparing the values by which STi t T 2r -i and 

 TulW-iS replace ^ that 



X 6 T (5. 



Hence, taking for r a value for which t 2 =f=l, we have 6 = 0. The 

 only case left for consideration is therefore that in which 



S=2'i,r 2 ,x-i2'Ml. 



If S be not commutative with every 0' we obtain at once a sub- 

 stitution =|= 1 i n I which leaves 3 , . . ., | m fixed. In the contrary 

 case, i-=*Vand therefore 



S^T^T^Z. 



If m = 3, Z = T^ % 2, the determinant of S being unity. Trans 

 forming S by (^g)^, we obtain the substitution 



belonging to I. Then I contains 



leaving | g fixed and not reducing to the identity. For that requires 

 jt 3 = 1, when we should have 



S = Ti, x Z\ x l\ x 

 contrary to the hypothesis made in 148. 



