410 MR T. B. SPRAGUE ON A NEW ALGEBRA. 



or each constituent in P is got by adding g to the preceding one. For instance, if n = 7, 

 g = 3, and we take 4 as the first constituent in P, we have 



.s 3 £(473G2d1) = s 3 (7362514) = (4736251) . 



I have treated of the case where n is a composite number, in a paper which has been 

 printed in the Proceedings of the Edinburgh Mathematical Society, vol. ix. 



As another example, I will enquire whether the inverse of a permutation can be 

 contained in the same set as the permutation itself. If possible, let iP = s h t k P ; then 



P = i s ,l t k P = isH'^isYV) 



= *Wr¥P 



= s 2 *P. 



This equation cannot subsist unless 2/?,=0 ; that is, unless 2h = or n. The former 

 of these values gives us *P = t k P, or 



{a n a n ., . . . a s a x ) = (a k+1 . . . a n a x . . . a k ) ; 



whence a n = a i+u &c, which are impossible. 



Taking 1h = n, we have h = n/2, so that n must be even. If n = 2N, we have iP = s K t k P, or 

 («„«„- 1 . . . o: 2 « 1 )eee(%. +1 - N . . . a„-N, %-N" . . . f^-N). 



Hence a„ ^=a, k+1 — N 



a n _ 1 ^=a k+2 — IS 



a i+1 ==a n - N 



a,. ^^a l — N 



«! z=a k — N 



If £ is odd, these conditions give us a {k+x)ft =• « ( /, +1)/ . 2 — N, which is impossible. Hence h 

 must be even. 



Suppose, for example, that n = 6, k= 4 ; then 



a 6 =« 6 — 3 a 4 =«! - 3 #2 — a 3 — ^ 



a 5 =a c — 3 «3=« 2 — 3 «!=a 4 — 3 



We may give the constituents any values that are not inconsistent with these 

 conditions. 



