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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?

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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.

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    $\begingroup$ I have a strong feeling of déjà vu. I think I have seen this strategy (namely, an idempotent element with a left or right inverse must be the identity element) used before in some other questions. Does anybody know any examples? $\endgroup$
    – user1551
    Jul 9, 2020 at 19:32
  • $\begingroup$ There is a similar proof of the statement that if a monoid is finite and $ab=1$ then $ba=1$. The point is that if $ba\ne 1$ then all elements $b^ka^k, k=1,2,...,$ are idempotents and if $b^ka^k=b^{k+m}a^{k+m}$ then $b^ma^m=1$, and so $a$ has a left inverse which then must be $b$. $\endgroup$
    – markvs
    Jan 2 at 4:42

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