Functions give^i by Groups. 321 



We read (5,z, "G^ under S" ; 0(3^, "that value which 

 6^Gd the derivate of G^ becomes under S" ; 0a(BA~\ "that 

 function which the equivalent ^Qia^aT^ of G^ becomes under 

 S." In 0(5^ is no operation of on (3^ denoted: such 

 tactical operation on a sum of algebraic products is for- 

 bidden, both in the order 0(5^ and Q>a^. 



In solving our problem, then, if we can prove of two 

 groups G^ and G^ in (A„), that 



(Bcz-0„(3., (5) 



we thus prove that the distinct groups G^ and G^ do not 

 give under our selected S distinct functions, because one 

 function is what a derivate of the other group algebraically 

 becomes under 2, merely by a changed order of elements 

 under the same index. 



If, then, G^ is our standard of comparison, we are bound 

 ^y (5) to mark Gg out of our table (A„), as useless to us 

 under our selected 2. 



We come to handle G^ thus. Having our Ij+i, which 

 is determined by our selected S, written in a column as in 

 art. 6, we seek a group in (A„) which has m-\-\ substitutions 

 in common with our index-group under the barred 2, so that 

 no group has more than in-\-\ of them. Many G's may 

 have such a greatest subgroup },«+! in common with I^+i. We 

 take one, G^^. Let the subgroup be 



J^+i = I + 61 + 62 + ... + e„, ; (6) 



We find these G's in I^+i, and name them so where they 

 stand in it. Its remaining t-in substitutions we call its 0's, 

 none of which is in G„:, which has all the 9's. I^+i can be 

 written thus : 



i,+i=i + ei + e2 + . . . + a,, + 0i + 02. . • +0«-m; (7) 



9. The next proposition in M.M. p. 342 is double, 

 affirming what had been proved in preceding pages. 



Oa = Q>o^i ; and O, = O^Oj ', (8) 



whichever among the above-written be G> or 0,. 



