288 CHAPTER xni. 



relations not derivable therefrom. The order 0(ti) of 6r(&) is 

 therefore 5> & 



Denote by G the subgroup G(k 1) generated by B 19 B 2) . . ., 

 142 and consider the following sets of operators 1 ) of 



It will be shown that these sets of operators are merely permuted 

 amongst themselves upon applying as right-hand multipliers the 

 generators B r (r = 1, . . ., & 1). Since B? = I, we have 



O r+1 B r = GB k -i . . . Br+lBr = O r , 

 O r S r EE GrjBjei . . . S r S r EE O r _j_i. 



If * > r + 1, we find, on applying 262) to move B r to the left of 



^ . . . fi= 0,-. 

 If ^ < r, we find, on moving l? r to the left of B if -B/+i, . . ., B r 2> 



Of JB r = GB k i . . . BiJB r = 



By 263), we may replace B f B r -iB r by B r ^S r B r -^ We then 

 move the first ^ r _i to the left of JB/--{-i> . . ., -B^- i and merge it 

 into G and get 



QJ&r = GBki 5/-+i5 r J5 r _iJ5 r _2 . . . jB^ EE O/. 



Hence the right-hand multiplier 5 r gives rise to the transposition 

 (O r O r +i) on the k sets O x , . . ., O k . It follows that the product of 

 any operator of these ~k sets by an arbitrary operator of 6r(&) is an 

 operator belonging to these sets. Taking for the former operator 

 the identity, we see that these & sets include all the operators of the 

 group G(Jc). The number of operators in 6r(&) is therefore at most 

 ~k times the number in G(k 1). Hence 



Combining this result with the earlier one, we have 0(k) = kl 

 The proof of the holoedric isomorphism of GQt) and 6 is there- 

 fore complete. 



The relations 261), 262), 263) may be combined into the formulae 



264) J=Bf=( J B, ;+1 ) 3 



1) It turns out that these sets form a rectangular table for Gr(K) with the 

 operators of Cr in the first line. 



