PRIZELL: 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. ao{aoh) = aoaoh = uoh = }), and 

 {boa)oa=^ho{aoa) = hou = h, that is, § 45 is sat- 

 isfied by p = a o /> and q = boa. 



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, g) = (n,p) when and only when mop = noq and 

 set up a rule defined by the relation (m, g)®(n, r) = 

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

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

 the sign o. 



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

 by hypothesis and §1. Hence either {m,q) = {n,p) 

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

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

 meaning, since this is so for nog' and mop. It re- 

 mains to verify the euclidean postulate. 



Suppose that {m,q) = {l,r) and {l,r) = {n,p). 



Then mor = loq and lop = nor. 



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

 But since G is an abelian semigroup {mor)o{lop) = 

 morolop = {mop)o{lor) and likewise (nor)o 

 (I o q) = (n o q) o (I o r) . Therefore (mop) o (lor) = 

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

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

 of equality. 



2) Let (m\ q') = (m, q) and (n',r') = (n, r). There- 

 fore m' oq = moq' and n' or = nor'. Whence as 

 under 1) (m' on') o (qor) = (mon) o (q' or') dJidi 

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



