> (O) 



THE REV. T. P. KIRKMAN ON NON-MODULAR GROUPS. 205 



a h^c^d^e 

 ajb^c d^e, 

 a J) c^de^ 

 ajb^cd^e^ 

 aj) c^d^e, ] 



in whicli we disregard for the present the order of the ele- 

 ments_, we exhaust once, and once only, the dnads of the 

 fifteen elements. If next we define that the triplets (E) are 

 triplets of mutually permutables, giving i^=aa^a^=^bhj)j^, 

 &c._, that the triplets (F) are didymous factors often substi- 

 tutions of the third order ajSyBe^rjdXfju, and that the 

 quintuplets (G) are also didymous factors of six substitu- 

 tions V p(TT (f>co of the fifth order, we can easily prove that 

 every triplet not exhibited, and possible with with the fif- 

 teen elements, reduces to a duad. We assume that the 

 fifteen elements are all of one form ; then it is sufiicient to 

 prove this of the triplets beginning with «, as all the fifteen 

 are alike involved. 



It is evident that any triplet containing the same letter 

 in its first or final duad reduces by (E). The triplets abaj 

 aba^y aba^ are 



aba =^aac =c, aba^z^caa^z^caj^, 

 aba^:=caa^=ca^, all by (F) and (E). 



The triplet abcj^-=bcc^ — bc-^i by (F) and (E), 

 The triplet ab^Cj,=^b^c^c^=b^Cy by (G) and (E); 



and thus every triplet can be shown to be reducible. 



We have, then, by adding to the fifteen elements the 

 twenty substitutions aa% /S/3^, &c., and the twenty-four 

 substitutions v vV^v"^, p p^p^p"^, &c., a group with unity, of 



Such a group, non-modular, is obtained by selecting from 



