LINEAR GROUP WITH QUADRATIC INVARIANT. 165 



The following substitution of determinant unity, 



y. -i rr "Y &$ 



a" p a 



2yd y 2 d 2 



leaves %% rj* absolutely invariant. Written in terms of the indices 

 li> 2* %s> ^ takes the form 



cud + 3y i(cty + /3d) ccy Su 



x= ~ i( 



afi yd 



It follows that X has determinant unity ( 101) and leaves If + 1| + 

 absolutely invariant. Giving to the substitution Y the notation 



we readily verify the formula of composition 



ra' 0nr 01 [-' + /?/. /j' + /sdn 



L' d'JL dJ La' d f S' + dd'J 



The group of the substitutions X, being isomorphic with the group 

 of the substitutions Y, is isomorphic with the group of the linear 

 fractional substitutions 151). But Y and therefore X is the identity 

 if and only ifa = d = +l, = y = 0. Hence the isomorphism is 

 holoedric. 



If 1 be a square in the 6r.F[jp n ], so that the coefficients of 

 X belong to that field, the substitutions X form a group 0{(3,jp n ), 

 a subgroup of Oj(3, p n ), which is holoedrically isomorphic with 



If 1 be a not- square in the (S^PyrJ, the coefficients of X 

 will belong to the 6r.F[p n ] if we choose or, /3> y, d in the GrF[p* n ~] 

 such that a is conjugate ( 73) with d, /3 with y, with respect to 

 the G-F[p n ~\. By 144, the resulting substitutions 151) of determinant 

 unity form a group holoedrically isomorphic with LF(2jp n \ The 

 corresponding substitutions X form a subgroup 0{(3,# n ) of 0^3,^"). 



In each case, the subgroup 0{(3,^ n ) has the order -crp n (p* n 1), 



since it is holoedrically isomorphic with LF(2,p"). We proceed to 

 prove that this subgroup does not coincide with (^(S,^ 71 ). In order 

 that 0^3 shall be of the form X, it is necessary and sufficient that 



