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Let $M$ be a finite monoid with identity $e$. If $x\in M$ has a left inverse $b$, I want to show that $b$ is also a right inverse.

There’s already a proof here In a non-commutative monoid, is the left inverse of an element also the right inverse?.

However, in my book the hint to this problem goes as follows. Since the set $\{x^n: n\in \mathbb N\}$ is finite, there are two distinct positive integers $l,k$ with $x^l=x^k$. But because $M$ is not a group there’s no cancellation law. How should I proceed?

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"But because $M$ is not a group, there's no cancellation law"...

But $x$ has a left inverse, so you can certainly cancel $x$ on the left: if $xr=xs$, then multiplying by $b$ on the left you get $bxr = bxs$, hence $er=es$, so $r=s$.

Remember, the cancellation law holds in groups because you have inverses. In a monoid, the existence of a (one-sided) inverse for $x$ means that you can cancel $x$ (on one side).

Assuming that $l\lt k$, from $x^l=x^k$ you get, multipying on the left by $b^l$, that $e=x^{k-l}$.

Now consider $xb$. Multiplying on the right by $x^{k-l}$ we get $$xb = xbe = xbx^{k-l} = x(bx)x^{k-l-1} = xex^{k-l-1} = xx^{k-l-1}=x^{k-l}=e.$$ Now that we know that $b$ is a two-sided inverse, we may deduce that $b=x^{k-l-1}$.

Arturo Magidin
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  • Thanks for your detailed answer –  Jul 27 '22 at 20:48
  • For an alternative way of finishing the proof: $y:=x^{k-l-1}$ is clearly an inverse of $x$, in particular it's a right inverse, thus $$y=ey=bxy=be=b,.$$ – Berci Jul 27 '22 at 21:36