# Does existence of a left or right inverse imply existence of inverses?

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?

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.
• 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 $b$. Jan 2 at 4:42