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



ABCj &c.^ the 30 substitutions of the fourth order, 

 A«i = ^j Aa^ = 6^j kc, given by (Q), the 20 of the sixth 

 order Bci = ^, B6, = 0^, &c., and the 20 of the third order 

 BF=^% BI=^^ &c., given by (S), and the 24 of the fifth 

 order aj}^=pj a^Cj,:=p^, &c., given by (T), we shall com- 

 plete a group of 



20.l6 + 24.l5 + 30.l4.+ 20.l3 + 25.l3,+ 1=120. 



To construct such a group let us take six symbols, 

 123456. We see by the first quadruplet and sextuplet that 

 we have to take B«jC, such that afi shall be of the fourth, 

 c,B of the sixth, and c^^a^ (in the third quintuplet) of the 

 fifth order, and that «, permutable with d^ shall be the fourth 

 in the6-plet Be,. 



All the didymous radicals of p of the fifth order made with 

 six elements wiU have a common element undisturbed, and 

 each a difierent second element undisturbed. We take at 

 random 



c,= 165432, giving c,aj of the fifth order. 

 «3= 154326. 



It follows from the theorem (art. 77 corrected^ vide art. 76), 

 Whenever the order K of the group i + <^ + ^* + . . , made 



on the partition 



N=N.i, 



or 

 or 



N=M. I + 1.2, 

 N=(N— «). i + i .«, 



is even, the didymous factors of <p^"* are two of the same form, 

 but those q/'<^^"*+^ are two of different forms ; that B is not of 

 the form of c, : it will have no letter undisturbed. We 

 cannot make a^B of the fourth order, if 61 is one of the three 

 transpositions of B ; for by the above theorem applied to 

 a circle of N=4, we must have i and 6 in the circle of 4. 

 Try for a^B the vertical circles 1265 and 1364. We get 



