THE REV. T. P. KIRKMAN ON NON-MODULAR GROUPS. 223 



fine (I57), which determines the first quadruplet, aspermu- 

 table with each of its four substitutions; i. e. (l57)=A^, 

 (327)=B^, &c. 



We have before us every duad of the 2 1 triads ; we have 

 next to give an account of all the triplets possible. The 

 condition that a triplet RPQ of these triads should be 

 reducible to a duad have been already given. Suppose 

 P=l57, Q=327, and that 



R. P. Q=IM57. 327=11. 327. 467 = ^467.157 



is irreducible. RP cannot have a common capital, nor 

 RQ, otherwise they would be permutable ; nor can R be any 

 permutation of 157, 327, or 467, for the same reason. Then 

 R can only be a permutation of 261, or 635, or 524, which 

 has not the capital 1, 3, or 4. Whatever it may be, it cannot 

 have a small figure in common with both 57 and 27 ; and 

 RP, in one of the above forms, will be a consecutive pair in 

 one of thequadruplets (B) . Let thisbeRPST; then RP = TR, 

 and RPQ=TRQ, where R has neither of the small figures 

 of Q, so that RQ is a consecutive pair in some one of the 

 quadruplets (B) ; and RQ is of the fourth order. We have 

 thus proof that every irreducible triplet RPQ either has PQ 

 of the fourth order, or can be written as TRQ where RQ 

 is of the fourth order. We have then occasion only to con- 

 sider the irreducible triplets of the form D . 617 . I26. How 

 many values can D have, so that this shall be irreducible ? 

 The capital of D must be one of 76432- {5i7 • I26} = (I57) 

 has 751 permutable with 723, therefore D is not 723: 

 {746 .517} = (715) has I57 permutable with I26, therefore 

 D is not 746; and D cannot be 7 15 permutable with 617. 

 Therefore D has not the capital 7. But {635 .517}= (636) 

 has no permutable of I26, nor has {5i7- 126}= (I57) a per- 

 mutable of 635 ; nor has {635 . l26} = (612) a permutable 

 of I26. 517 .126=751. Therefore 635 .617 . I26 is irre- 

 ducible. Next, {647. 517} = (636)7 ^^^ no permutable of 



