SLOCUM. — FINITE CONTINUOUS GROUPS. 



243 



(3) 



9x1 

 h 



of- = <Au («) in &) + fu («) £h 0*0 

 (*- = l,2; 4=1,2). 

 The determinant of the if/s, namely, 



"Ash ^22 



— 1, «i 

 0, -1 



not being identically zero, equations (9) may be solved for the |'s, giving, 

 (10) 



(* = 1, 2 ; y = 1, 2), 



a n (a u a 2 ) = — 1, a 12 (a 1; a 2 ) = 0, 



9x- 9x- 



iji (*/, * 2 ') = «,-,- (a) 7^7- + a,. 2 (a) ^-i- 



where 



a 22 (a l5 a 2 ) — — 1. 



a 21 V&1} ^2) — #1} 



It is to be observed that 



*i £11 0*0 + e 2 £21 (a/) = — «2J 



e l £l2 0*0 + g 2 ^22 (#') = — ^i — e 2 X 2 . 



Therefore, these expressions linear in the £'s cannot both be simulta- 

 neously zero (for all values of the x"s) if e x and e 2 are constants other 

 than zero. That is to say, no two constants e 1 and e 2 , not zero, can be 

 found for which 



«i6i(a0 + *&( a = 0, 



for i=l and t = 2, simultaneously. 



We come next to the demonstration of the second part of Lie's first 

 fundamental theorem. Starting with a system of equations which define 

 a family with r essential parameters (r = 2 in the case considered), and 

 satisfying differential equations of the form (9), we proceed to follow the 

 steps in Lie's demonstration that this family (provided certain other 

 conditions are also satisfied) constitutes a group. 



As shown above, the family of go' 2 of transformations T a defined by 



(1) 



x 1 , = x 1 + a 2 = f x (x,a), 

 x 2 — e a 2X + «! =/ 2 (x, a), 



satisfy differential equations of the form (9) in which the determinant of 

 the if/ Jk is not identically zero ; also the differential equations 



