136 CHAPTER V. 



If p =j= 2, we can take a = ft p \ since the condition 



can be satisfied by a mark ft in the GF[p 2s ]. In this case, S 2 0%\ 

 replaces | 2 by the function 



2/J6,+ S, + -"+6, 



and therefore belongs to the group by our assumption on w? 1 

 indices. If p = 2, s > 1, we can choose a and /3 among the sets of 

 solutions in the GrF[p 2s ] of 



127) 

 in such a manner that 



Indeed, the condition is (since p = 2), 



Since ^> 2, we may take for a a mark neither zero nor unity in 

 the 6r.F [p 8 ] and then determine a solution /3 of 127) such that 



ft=^ft p \ Then will a* ft*' =%= aft. To prove that such a choice for /3 

 is possible, we note first that 



CCP* = a, a 2 4= a; hence a* 1 + 1 =4= 1, 4= 0. 



Further, if a', /3 f be one set of solutions of 127), then is also a', rft' y 

 where r is any root of 



Not every root r belongs to the G-F[p*] 9 and therefore not every 

 solution /3 corresponding to a given a belongs to the GrF[p*]. Hence, 

 if p = 2, p s > 2, we may suppose that in the* substitution $ 2 the 

 coefficient a n is such that |^ + 1 =j=l, when the proposition follows 

 as above. 



For p* = 2, an additional generator W, for example, is necessary 

 since the only substitutions of the form OJ^ are the products 



T^r^-i and (g.yZVjT,,,-! (P 3 =l). 



Indeed, there exists in the 6rjF[2 2 ] only six sets of solutions of 



viz., a = Q, ft = and a = 0, /3 = (>, where ^ 3 = 1. Hence the 

 substitutions T,- jr and Of'/ can not combine to give a substitution 

 replacing ^ by ^ -j- | 2 + | 3 , for example. It follows readily that the 

 additional generator W is sufficient, together with the substitutions 

 T and 0, to generate the group 6^2,1- 



