LINEAR GROUP WITH QUADRATIC INVARIANT. 185 



Given, inversely, a Q"$, where a, /3 are marks of the GF[p n ] 

 satisfying 169), we can determine a square co in the GF[p 2n ~\ which 

 satisfies 170). In fact, 170) may be written in the solved form 



of which the second follows from the first in virtue of 169). That 

 the first can be satisfied by a square o in the GF[p 2n ~\ follows 

 from the relation 



For a a not-square in the 6rF[jp 2n ], Q" is the product of 

 an 16 , not a Q 16 in the CrF[p n '], by the substitution A, neither 

 factor belonging separately to G". 



Under the transformation of indices 165), F becomes 

 \i/ _ fcs _i 722 y y i TT v 



1 "T w ?6 J -13 j: 24"t" J- 14^23^ 



where, by 166), J 2 belongs to the GF\j) n ~] but is a not-square in it. 

 We introduce in place of the Yy new indices such that 



r 15 r M -r u r w = sj + sj + ! + 



5 



Then V becomes JPgwvjg^fc Therefore, by 189, ^L" wiU be 



i = l 



transformed into Oi(5,p w ). 



For 1 the square of a mark i in the 6rF[j9"], we may take 



r 13 = i 2 + ;| 3 , r M =8,-|,, -r 14 =! 4 +*| 6 , r 23 =| 4 -^ 5 . 



As in 187, ^4. becomes an 6)4,5, which is a 4,5 if and only if co 

 be a square in the 6rjP[jp 2n ]. Hence G-" is isomorphic with a sub- 

 group of O r (6, p 7 *). The subgroup contains every Q^Q and every 

 16 45 , neither factor a $, but does not contain the separate factors. 

 For 1 a not- square, so that p n = 41 -f- 3, we may take o so that 



Then A multiplies Y u and I^ by 1. The required transformation 

 of indices, transforming G" into a subgroup of 1 (6,j9 n ), is the 

 following: 



r 24 = {, + ? 4 - /jfc, r M = | 3 + /jg 4 + a $,. 



As in 189, ^L becomes in the new indices C^C^O^T^^^, the last 

 factor being not of the form 4>5 , while C 5 C 4 = Ql',l belongs to 

 Oi(5,j) n ). Hence G" is isomorphic with a subgroup of Oi(6,p w ). 

 The subgroup contains every 16 an( i every OjgO^, neither factor 

 being a Q, but does not contain the factors separately. 



