THE REV. T. P. KIRKMAN ON NON-MODULAR GROUPS. 207 



If we begin with the first system,, a{a^){a^), we shall 

 merely reproduce the non- modular group of 60 above men- 

 tioned made with 12345^ and the same group made with 

 67890, which will teach us nothing. 



We therefore begin with aa^a^. above written, which have 

 not all the same undisturbed elements. The first thing 

 required is bj^ such that «Z>, and ajjj^ shall be of the fifth, 

 and «i6, shall be of the third order. We write 



« =1325468709, «, = 63245 10987, 

 «a = 62354i9078, b^=pqrstuvwxy. 



As a^bj is to have circular factors of three elements, we 

 shall have either ^ = 5, or 5 = 4. Put ^ = 5, and try the ver- 

 tical circles 412, 413, &c., till our problem is either solved 

 or proved impossible. The circle 412 gives b^ =^0541 52 . . . , 

 which gives, when written under a, a circle of 5 only on con- 

 dition that j9 = 6, which is inadmissible, for by (G) a 2, and i, 

 can have no element in the same place. The same objec- 

 tion lies against the circle 41m, whatever m may be. After 

 a few trials we find that 407 is a proper circle oi a^b^; and 

 we readily complete the systems 



«i = 63245io987 

 ^, = 9860537214 



€ =2197584630, 



©=1325468709 «i = 62354i9078 



^>, = 986o5372i4 ^>, = 986o5372i4 



^1 = 4731082695 6^ = 7492061835 



0=5289176340 c?, = 1578293460 



<? =0674923851^ ^, = 3014876592; 



where ee^e^ are mutually permutables, as are cc, and dd^. 

 We have next 



4^C?^,=: 0938657421, 



c^ = cc, = 8o59367i42. 



To find b, b^, we write c, under «, which will give either a 

 triplet or a quintuplet containing a. It turns out to be a 



