AUXILIARY THEOREMS ON ABSTRACT GROUPS, etc. 295 



Finally, 



We have now derived from 276) all of the relations 275). It remains 

 to derive 273). Since B 2 , B s , B 4 are conjugate with B t by 277), 

 they are of period 2. By 275), B^B Z is conjugate with B 19 B 2 B 3 

 with jB 3 , B 3 B with B s . Hence they are of period 2 and therefore 

 B 3 is commutative with B 2 and J5 4 , B^ with B z . Since E 

 is its own reciprocal, we have 



so that B 1 B 3 = B S B 1 . Since B^B^B^ was shown to be the trans- 

 formed of B 3 by E*E%, we have 



Hence J5 2 is commutative with B. Since B is commutative with 

 B 3 , E z and E z , it is commutative with B by 277). 



272. Theorem. Every substitution of FO (5, 3) is given once 

 and 'but once l)y the following 27 sets, in which A denotes the sub- 

 group J, 60 : 



Since w; is not in J., a substitution of E t belongs to Rg if and only 

 if t = r. If a substitution of E t belong to E 9itJ the product 



must belong to A, whereas it replaces | 5 by a linear function of 

 li, | a , | 8 , | 4 , every coefficient being 1. 



If a substitution of JS, f< belong to R j<t, the product 



8 = w^WCfeWw (feWftWw- 



must belong to J.. Supposing first that ^ r =)= 0, we show that S 

 replaces | 5 by a function involving more than one index and there- 

 fore does not belong to A. In fact, w~~ s S replaces | 5 by a function 

 of the form 



where a, Z>, c are three of the integers 1, 2, 3, 4. Then w* replaces / 

 by /i | 5 , where ^ is a linear function of ! Ij, | s , 4 with coefficients 

 not all = (mod 3). Hence S replaces 5 by /i | 5 , involving two 

 or more indices. Suppose, however, that = r. If then i=%=j, 

 S replaces 5 by a linear function of $ lf | 2 , | 3 , ^ 4 with coefficients 1. 

 If ^=j, S = w- a , which belongs to J. only if s = 0. But in the 

 latter case, the two sets P 3it and E jt are themselves identical. 



