8 MEMOIRS NATIONAL ACADEMY OF SCIENCES. [Vol. xiv. 



Triads of a system may contain elements from one cycle only, or from two, or from three. 

 Further subdivision gives us 10 classes. Indicate the numbers of triads in these several classes 

 as follows : 



5Wi of type abc, Bu^ of type a/37, Bu^ of type 123, 



5v, of ty])e aba, 5^2 of type afil, Bv^ of type 12a, 



5w, of type ahl, 5^2 of type a0a, Bw^ of type 12a, 



5t of type aal. 



As all possible pairs must occiu", we have the diophantine equations to satisfy: 



The solutions are 



Of these the third solution may be dropped, since it is related to the above arrangement of 

 elements in the operator Si exactly as the first solution is to the operator S^, (or S{~^). The 

 first and second solution yield two kinds of systems, as follows: 



First Und.—t = l, Wi = 0, ^1=2, 1171 = (i = l, 2, 3). 



The triads enumerated by 5^ = 5 shall be these: aal, 6/?2, cy3, <?54, ee5. With the pair ab 



must occur 5; for if the triad were ahy, then must follow in cyclic order hc5, , ea/3. There 



remains, therefore, no letter of the second cycle to join with the pair ac, since we have abeady 

 the pairs aa, cy, and ay, a/3, c5, ce. The same reasoning excludes the combination ahe. This 

 leaves only the possibility ab5. Under 5, there follow from ab8 the pairs ce, ay, and there is 

 left for the pair ac only the letter /3, hence ac/3. Similar reasons apply to triads from the second 

 and third cycles and from third and first cycles. The entire system is accordingly determined 

 uniquely, and is given by the following seven triads and the conjugates derived from them by 

 the operator Si : 



System II: aal; abS, ac0; a/34, a72; 12d, lob. 

 Second land. — 1 = \, Ui = 0, 'i'i=w'i = l (i = l, 2, 3). 



After the five triads, aal and its conjugates under Si, the apparent possibilities are a&7 

 and abh. 



Assume fii'st a triad a67. Its conjugates contain the pairs ac, ad, and /3a, /3f, so that there 



remains a possibility of the triad affc. But also there are found the pairs yc, yb; whence the 



pair a7 can join with no letter from the first cycle, but must be completed by either 2, 4, or 5. 



We examine these successively. 



(1) Triad a72. 



Conjugate triads will involve the pairs U, 1/3, 2a, 27, 3/3, 35; and already we had la, 2/3, 37. 

 Therefore the pair 13 can not be completed from the Greek cycle, nor can its four conjugate 

 pairs. But ^3 = 1, and for the pair 12 we have available the letter 5 only. Necessarily one 

 triad is 125. We must complete 



13 by either b, d, or e, 

 ac by either 2, 4, or 5. 



The hypothesis of 136 would result in excluding ac from 2, 5, and 4. The hypothesis 13e leaves 

 open only the alternative ac2, having ebl as a conjugate, which is inconsistent with 13e. Dis- 

 missing, therefore, 136 and 13e, we have 13d, requiring ad. The remainder of the system 

 follows uniquely from these. 



