An element $a$ of a monoid $M$ is invertible iff there exists $x\in M$ such that $axa=1$

Question

An element $a$ of a monoid $M$ is invertible iff there exists $x\in M$ such that $axa=1$

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I can't do this one. How do I get started? It looks like it is saying there is only an inverse if $x=a^{-1}a^{-1}$ is in $M$, e.g. it is only invertible if there is an $x$ that is a left and right inverse of $a$, which makes sense, but then isn't the answer 'true by definition'?

Answer 1

$(\Longrightarrow)$If $a$ is invertible, then $$az=za=1\implies \exists x,axa=1$$ Take $x=zz$, then $az=za=1 \implies (az)(za)=(1)(1)=1$

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$(\Longleftarrow)$ There is some $x$ such that $axa=1 \implies az=za=1$

Proof: $$axa=1\implies ax=a^{-1}\implies (ax)a=a^{-1}a=1$$ So $a$ has a left inverse, similarly $$axa=1\implies xa=a^{-1}\implies a(xa)=aa^{-1}=1$$ so $a$ has a right inverse. It is know that the left and right inverse are equal(if both exist), so $ax=xa=a^{-1}=z$

Therefore $az=aa^{-1}=za=a^{-1}a=1$

$\blacksquare$

Answer 2

The only hard part is proving that the existence of such an $x$ implies invertibility of $a$.

You have that $ax = ax(axa) = (axa)xa = xa$, so $a$ and $x$ commute.

Now you just need to conclude that $ax$ (which is equal to $xa$) is the unique inverse of $a$.