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 3, 7 o. Consequently the effect of 
any one of the eight is to unite the circles of (BC)i into one ; 
that is A(BC) 4 irreducible is always a substitution of the 
seventh order. It is easily demonstrated that if AiA, be any 
two of the eight values of A, A 4 BC can be no power of 
A 2 BC. 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 
BCHE. That is, there are 8 - 2'2T4 irreducible triplets. 
This must be divisible by d’S'Gr, because each of the sub- 
stitutions ABC of the seventh order is formed with four 
values of BC ; consequently f=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 
28-2 3 +21-2 1 +2M 2 +8-6 7 + l 1 =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‘3'2‘. Hence if each be r times found in the equivalents, 
?- 6 - 5 - 4 - 3-2 
- — —r‘5'3. 
48 
1 have shown, in the Memoir above quoted, that two groups 
of 7'6'4 can be formed to contain the powers of 2345G71 ; and 
the sarile thing is deduciblc from the fact that we might have 
