1. The problem statement, all variables and given/known data Let [itex](S,\cdot )[/itex] be an algebraic structure where the operation [itex]\cdot [/itex] is associative and commutative and also the following axiom is satisfied: [tex] \forall x,y\in S,\exists z\in S: zx = y\ \ (1) [/tex] Prove that if for every [itex]a,b,c\in S, ac = bc[/itex], then [itex]a=b[/itex]. 2. Relevant equations 3. The attempt at a solution Assume by contradiction that [itex]a\neq b[/itex]. Then for every [itex]c\in S[/itex] either [itex]ac = bc[/itex] or [itex]ac\neq bc[/itex]: 1. if [itex]ac\neq bc[/itex] then the proof is concluded. 2. provided [itex]a\neq b[/itex] and for every [itex]c\in S, ac=bc[/itex]. By axiom [itex](1)[/itex] there exist [itex]x,y\in S[/itex] such that [itex]ax=c[/itex] and [itex]by=c[/itex]. It follows that [itex]a(ax)=b(by)[/itex], which is the same as [itex]a(by) = (ab)y = b(ax) = (ba)x = (ab)x[/itex]. We wind up with the equality [itex](ab)y = (ab)x\ \ (2)[/itex]. Stuck, I feel like I have reached nowhere with the last statement, wandering in circles. If [itex]x(ab) = y(ab)[/itex], then [itex]x\neq y[/itex] as I have assumed at the beginning of 2. but this is a circle. The function [itex]\cdot : S\times S\to S[/itex] need not be injective, in other words if [itex]x\neq y[/itex] it doesn't necessarely mean that [itex]kx\neq ky[/itex] for some [itex]k\in S[/itex], in fact I have assumed if [itex]x\neq y[/itex], then [itex]kx=ky[/itex], somehow this has to yield a contradiction. Idea: Maybe I can show that the function [itex]\cdot [/itex] is, in fact, injective.