144 
We cannot always form a positive and transitive group of n~ 1, 
corresponding to our group over A ; but the preceding reasoning 
applies with little change to the solution of the general equa- 
tion of the ?ith degree. When n is a power of a prime number, 
the process will be somewhat abridged. Nothing prevents us from 
breaking up the function above handled into a number of 
separate functions, which will each contribute a quartic root to the 
solution, and thus most probably simplify the computation. 
To solve the general sextic we proceed thus : — Let 
x Q x x x i x z x i x b be the roots of the sextic, G=0. We can write 
x o= ^ { x< r -x z +a(x l —x?)+ a(x 2 —x b ] 
+ { *o— * 4 + <t(x 2 —x b ) -f a\x tg—xi } 6 )* 
+ *0 #5+ a (^3 X l) 4“ 0. 2 (x 4 X 2 ) 1 
4" g ^ { x o — x \ 4" a(x 5 — x a) 4” ®?( X 5 # 3 ) } 
+ l ^{ x 6— x 2+^(x6—xz(+a 2 (x--x 4 ) Y J 
=^+ 1 g (s+ Hi ) i +i(s+H 2 )‘ + J( S + H 3 ) i 
4~^(s4“H 4 ) 4“^S4-H 5 ^ j 
where a is an imaginary cube root of unity, and S is symmetrical 
in the roots and rational in the coefficients of G=0, and where 
Hj H 2 H 3 H 4 H 5 are five values of a 120-valued function of 
those roots. Our object is to transform H x , &c., into irrational 
functions of the coefficients of G=0. 
Since Hq H 2 , &c., after involution, are invariable by the cyclical 
permutations of x Q x l x 2 x 3 x i x 5 , they receive all their possible values 
by the permutations of x x x 2 x z x 4 x 5 . 
The positive substitutions composed of five elements are as 
follows ; — - 
