36 CALIFORNIA ACADEMY OF SCIENCES. [Proc. 3D Ser. 



The final condition X 2 -J- fx -J- u 2 = 1, becomes, on applying 

 7 2 52 = — 7 2 62 and setting a> = 1 + 2JL — __ , 



7 31 7"62 



762 



or 



L. „+ S Y - 2 ^ 2 W* + 7 2 4l) - 3 2 7 2 31 

 V 762 / 7 2 31 7 2 62 



Here co-^Z.0 and 2 (7^1 -f- 7 2 4 i) is a square not zero. 

 Thus 8 = gives an immediate solution for ft. To prove 

 that there is always a solution, we note that (§ 7) an arbi- 

 trary value except zero may be assigned to y &2 by a previous 

 transformation by O5G. Further, 1 in a Galois Field of 

 order fl n for which — 1 is a square, (fl a — 5)/4 of the 

 squares are followed by squares (not zero). Hence, if 

 ft n >5> there exists values of p" 2 making p l — 1 a square. 

 Thus the right member of our condition may be made a 

 square. 



Consider the case that our substitution is commutative 2 

 with 756. Now the conditions that S, given by formula (1), 

 shall be commutative with every T {] are seen to be 



«ii = «11 , «ij = «12 (*\J— 1 - - '#)• 



The conditions (2) that S be orthogonal then give 



« 2 n + (w — i) a 2 12 =1, 2fln a 12 + (m — 2) a 2 12 = o. 



Thus an = «i2±i, »a 2 i2= + 2a 12 . 



Hence /S is the identity, or iV= C\ CV  ' C m , or, if m is not 

 a multiple of/, the substitution 2 of determinant — (±i) m : 



Thus our substitution of .determinant -f 1 affecting only six 

 indices can not be 2, and, by our first hypothesis, is neither 



1 By a generalization of Jordan, Traits des Substitutions, § 198, or Gauss, 

 Commentatio prima, 16. 



2 Of course the same treatment applies if it be commutative with Tx. 



