LINEAR HOMOGENEOUS GROUP IN THE F[2] etc. 211 



These substitutions must therefore belong to the group C, the second 

 compound of HA(^, 2 2n ) when expressed in the indices |/, r\i. Also 

 C contains L and therefore also MM m by formula 196). Hence C 

 contains all the generators of J^ (m = 3). But the order of C, being 

 equal to that of HA (4, 2 2 "), is . 



which equals the order of J% (m = 3). Hence J^ = C. 



208. Theorem. If m = 3, /&e number of substitutions of 

 which leave m fixed is 



If a substitution $ of Jx does not alter m and replaces y m by 



.7=1 



we must have, in virtue of the relations 78) and 194), 



m 



212) d mm = 1,5? p m jd mj + Aftm = 0. 



We proceed to prove, inversely, that if ft mj , d m j be any set of solu- 

 tions in the 6rF[2 w ] of 212) there exists a substitution X in J^ 

 which leaves TO fixed and replaces rj m by f m . 



If ft m j- d m j= (j = 1, . . ., m - 1), then fi mm = or A- 1 . Hence 

 we may take as I the identity or M 2 M m L respectively. 



In the contrary case, let /3 m 2=(= 0, for example. Then Jv contains 

 a first hypoabelian substitution T leaving | m and r] m fixed and re- 

 placing 7^ 2 by 



since TO id jn i+ /3 m2 d = in virtue of 212). Then we may take 



For w = 3, the number of sets of solutions in the (rjF[2 B ] of 212): 



Pmldml. + /3 m 2^m2 + /3 mw + A/3^ m = 



is (2 2M + 1) 2 2 ". Indeed, there are 2 n ~ 1 distinct values in the GF[2 n ] of 



f == Pm il i " Pm m 



By 204, /3 ml d m i+/3 m2 (? m2 =T has 2 Sw 4- 2 2n - 2" sets of solutions 

 if % = 0; while, if T have any one of the 2 n1 1 possible values 

 =)= , it has 



U* 



