THE ABELIAN LINEAR GROUP. 97 



The product of two such substitutions gives L*, i ^ + 1 ^). But, by 

 64, marks r v and r 2 can be found in the G~F[p n ], p > 2, such 

 that t\ 4- 1\ has an arbitrary value ^ in the field. Hence J contains 

 -L 2)/u . Then I contains the product 



Hence 7 contains Z/ >/t and .Mi, the transformed of Z 2 ,/u and Jf 2 

 respectively by P 2? . Finally, 7 contains 1 ) 



^3) -2VJ-, y, ^ = Qi, j, i L^ p Lj, p Qi, j, i Li, p Lj, n 



b) Suppose, however, that $ 3 = I. Then S 2 is commutative 

 with NI, 2,^? s that 



n = !; fti = 0, d n -= 1, d\ 2 = cf 21 . 



Applying the Abelian conditions E 13 = R> 5 = 0, we find that d l2 = 0, 

 y 12 ='y 21 , so that 5 2 becomes 



= la 4- 



/S 2 is not the identity since $ t is not. If y n = 0, $ 2 is of the form 81) 

 considered under case a). If y n =f= 0, J contains $' 2 , the transformed 

 of 2 by 2 ,i,^ where A = y u /y u , viz., 



For d = 0, /Sa = ^i, yil - For d 4= 0; ^ contains the transformed of /S'g 

 by Ti t il2 ifl , K and ft being arbitrary marks 4= 0,- giving the sub- 

 stitution 



Forming the product of two such substitutions and noting that, 

 for p > 2, the equation AJ -f ;| == ^ has solutions in the 6r-F [p n ] for ^ 

 an arbitrary mark 4= f ^ ne field, we find that J contains 



where a and /3 are arbitrary marks =(=0. A suitable product of two 

 such substitutions gives 



Ll, a L% } ^ - LI, a L%, p = LI, 2 a 



In every case we reach in J a substitution LI, 2, where A =j= 0, and 

 therefore also L 2 , *. It follows as in case a) that J = G. 



117. Theorem. For p = 2, SA(2m, p n ) is simple except when 

 m = 2, p n = 2, and when m = 1, p* = 2. 



1) We might reach J^i, 2, a by 82) and then obtain JV/, y, /* in the group J. 



DlCKSON, Linear Groups. 7 



