248 



the eight triads has 1, none of them has both 7 and 5. 

 Hence none of them has a transposition which fractures 

 either of the circles 2 4 6 8, 7 5. Consequently the effect of 

 any one of the eight is to unite the circles of (BC)4 into one ; 

 that is A(BC)4 irreducible is always a substitution of the 

 seventh order. It is easily demonstrated that if A^A.^ be any 

 two of the eight values of A^ A^BC can be no power of 

 AaBC. Hence there are not less than eight substitutions of 

 the seventh order, no one of which is a power of another : 

 that is, there are 8-6r substitutions of the seventh order. 



The number of irreducible triplets is eight times that of 

 the substitutions of the fourth order, written each under four 

 forms, BC=CD=DE=EB in terms of the didymous radicals 

 BCDE. That is, there are 8-2-21'4 irreducible triplets. 

 This must be divisible by 4*8"6r, because each of the sub- 

 stitutions ABC of the seventh order is formed with four 

 values of BC ; consequently r=l or r—7, the latter of which 

 is easily seen to be inadmissible. 



It is thus proved that there are in reality 48 different irre- 

 ducible triplets, and so many substitutions of the seventh 

 order. We demonstrate easily that every quadruplet ABCD 

 is reducible to a triplet ; hence every quintuplet ABCDE, &c., 

 is reducible to a triplet. And we thus prove that the 21 similar 

 radicals form with their products a group of 



28-23-|-21-24-f2M2+8-67+li=7'6-4 

 substitutions. 



The equivalent groups will have each 48 substitutions of 

 the seventh order, The number of these substitutions is 

 6'5'4:'S'2'. Hence if each be r times found in the equivalents, 



r6-5-4-3'2 



1 have shown, ill the iMeinoir above quoted, that two groups 

 of 7*6*4 can hs formed to contain the powers of ^3345671 ; and 

 the same thing is deducible from the fact that we might have 



