THE ABELIAN LINEAR GROUP. 109 



Suppose first that a u = 0, so that ft 1 y n = l. Transforming 

 Z 2 by jRi,fB,a, where 1 + ^ fti = 0, we reach a substitution equal to 

 a product $!$', where S 1 affects 1? % only and S' affects 



only . a sj ,, * ^ (i-2,...,m) 



Suppose, however, that n =4=0. Transforming Z 2 by L m ^^^, 

 ym-iro-i-f- 2pcf n == 0, 



we obtain a substitution Z 2 f the form Z 2 and having ^ TO _ lw _i= 0. 

 Transforming Z 2 by .L',,, ^, where (l mm 2pGr n = 0, we obtain a sub- 

 stitution Z 2 of the form Z 2 , but having ft mm = y m _i m _i = 0. 



If ft 1 =^ 11 =0 ? we transform Z 2 by Q m 1,1,*, where 1 2Aa n =0, 

 and afterwards by 1, m , ? , where 1 2^ n =0 ; and obtain a product 

 $!$', where S affects only | 1? % and 5' affects only | f , ^,- (i > 1). 

 /J n H=0, we transform Z 2 ' by 



which has the form of Z 2 with y 1]L =)= 0. We therefore treat the 

 latter case only. Transforming Z 2 by L' m ^ f where 



m-lm+ 9^11= 0, 



we obtain a substitution U of the form Z 2 , but having cc m i m and 

 y m _i w _i both zero. Transforming C7 by E m ^ m ^ Hi, i, where 

 a n + ^7n = 0, we get a substitution of the form 



Transforming this by Ri, m , JL, where 1 -j- ^7n = 0, we get a similar 

 substitution with the elements 1 replaced by zeros, and therefore 

 the product of a substitution on | 1? ^ by a substitution on the 

 indices ,-, i? t - (* = 2, . . ., m). It is therefore conjugate with 



M, M, . . M m . 



