Suppose $G$ is a set with a binary operation such that:
- (Associativity) For all $a, b, c \in G$, $(ab)c = a(bc)$.
- (Identity) There is $e \in G$ such that, for all $a \in G$, $ae = ea = a$.
- (Left inverse or right inverse) For all $a \in G$, $ba = e$ for some $b \in G$ or (note the difference with and) $ac = e$ for some $c \in G$.
Does this imply that every element $a \in G$ has an inverse, i.e. an element that is both a left and right inverse? That is, for all $a \in G$, is there $a’$ such that $aa’ = a’a = e$? In other words, is $G$ a group?
Best Answer
The answer is yes. Suppose $a$ has right inverse but not left inverse: $ab=e$. Then let $f=ba$. We have $ f^2= baba=ba=f$ and $f\ne e$. The element $f$ has a left or right inverse $c$. Suppose $fc=e$ , Then $f=fe=ffc=fc=e$, so $f=e$, a contradiction. If $cf =e$ then $f=ef=cff=cf=e$, again a contradiction.