FRIZELL: FOUNDATIONS OF ARITHMETIC. 393 



49. Proposition XVIII. Every group contains the 

 inverse of every one of its members. 



50. Proposition XIX. A semigroup with modulus 

 is a group if it also contains the inverse of every one 

 of its members. 



Proof, a o (a o h) — a o a o h = u o h = J)^ and 

 (!) oa) o a = h o (a o a) = h o u = h , that is, § 45 is sat- 

 isfied by p = a o /' and q = hoa. 



51. Proposition XX. Given an abelian semigroup 

 G with reference to a rule denoted by o , if in the class 

 [ ( m, n ) ] = C of pairs of elements of G we declare 

 [m,q) = (n,p) when and only when mop = noq and 

 set up a rule defined by the relation {m,q)Q{n,r) = 

 {mon,qor), then 1 ) C will be a K-class, 2 ) o a C-rule, 

 3 ) C an abelian group as regards the rule denoted by 

 the sign o. 



Proof. 1) Either mop = nog' or mop is not =noq 

 by hypothesis and §1. Hence either im,q) = (n,p) 

 or(m, g) is not =(n,p). Obviously the assertions 

 {n,p) = {m,q) and {m,q) = {n,p) are identical in 

 meaning, since this is so for Tto ^ and mop. It re- 

 mains to verify the euclidean postulate. 



Suppose that (m,q) = (l,r) and (^, r) = (?i,p). 



Then mor = lo q and lop = nor. 



Hence (mor) o (lop) = (nor) o (loq). 

 But since G is an abelian semigroup (mor)o(^op) = 

 mor o lop = (mop) o (lor) and likewise {nor)o 

 (loq) = (noq)o(lor). Therefore (mop)o(^or) = 

 {noq)o{lor). Whence mop = noq by definition 

 of semigroup and finally i'm,q) = (n,p) by definition 

 of equality. 



2) Let (m', q') = (m, q) and (?t', r') = (n, r). There- 

 fore m' oq — moq' and n' or = no r'. Whence as 

 under 1) {m' on') o {qor) = {mon) o {q' or') ^Xidi 

 hence (m' on', q' or' = {mon,qor). That is {m',q')Q 



