THE REV. T. P. KIRKMAN ON NON-MODULAR GROUPS. 311 



stitntions (^^g, ')(^^ permutables with the first two 6-plets. 

 Hence all the triplets (R) are permutable sets, 



AB«,= i=ArZ>,, &c. 



The letters AB, &c. are those which we dropped in forming 

 the sextuplets. 



The triplets possible with the twenty-five elements are 

 next to be examined. We have to consider the forms ABC, 

 abc, Abe, ABc, bAc, be A, AcB, cAB. Let the triplet be 

 ABC; AB = «i or 'BC=d^j giving two ways of reduction; 

 and thus every triplet of three capitals is a duad. The 

 triplet flti^^c, has its first duad in (T), and we have the 

 reduction a^b^ . Ci = b^e^ . e^=.b^c^. The triplet a^^c^ has its 

 second duad in (T), and we have ay.b^Ci^a^.a^^^^aJ:)^, 

 The triplet a^^e^ in (S'} is FIB, reduced before. 



ABf?, = «,c, by (U), Bc,A=F«,A=FB, by (S)(a), 



CiAB = c,«i. 

 «,5,C = FIC, «AC = «,D, a^bf,^b^cS^=.b^. 



Thus every triplet may be proved reducible. It will be 

 useful to lay down the conditions of reduction of any triplet 

 of radicals xyz, having any elements capital or small. Let 

 {xy\, \xz), or {yz\ be the multiplet containing the duad 

 xy, x%, or yz. Then xyz is reducible, 



(1) if {xy\ contains z permutable with j, for we have 



xyz=.tz z=.tz" ; 



(2) if {yz\ contains x permutable with x, for we have 



(3) if {xz} contains u permutable with zyz=Uj for 



xyz=xzzyz=xzu = tuu = tu" . 



Here uz=zy, or u is in {zy} equidistant from z with y. 



It is easy to see that, if none of these conditions is ful- 

 filled, the triplet xyz is irreducible. 



If now we add to the 25 square roots of unity a^a^a^, &.c-> 



p2 



