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


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'?

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    $\begingroup$ The problem isn't very hard, but you're tricking yourself with the notation $a^{-1}$, which you are not allowed to use until you have shown that $a$ does in fact have a two-sided inverse. But certainly, if $a$ is invertible and you put $x=a^{-1}a^{-1}$, then the given formula holds. So now you need only prove the converse. $\endgroup$ – Harald Hanche-Olsen Jan 16 '15 at 15:16
  • $\begingroup$ @HaraldHanche-Olsen I just did $axa=1\implies ax=a^{-1}\implies x=a^{-1}a^{-1}$. You are saying I just need to say now $x=a^{-1}a^{-1}\implies ax=a^{-1}\implies axa=1$? $\endgroup$ – Makoto K. Jan 16 '15 at 15:18
  • $\begingroup$ For the opposite take $x=(a^{-1})^2$. $\endgroup$ – Janko Bracic Jan 16 '15 at 15:20
  • $\begingroup$ Don't forget that $a$ can have a left inverse and no right inverse, or vice versa. Do you know a result about the case when both a left inverse and a right inverse exist? $\endgroup$ – Harald Hanche-Olsen Jan 16 '15 at 15:20
  • $\begingroup$ @HaraldHanche-Olsen I don't think so, actually I haven't read anything other than a passing comment on my previous question about left and right inverses $\endgroup$ – Makoto K. Jan 16 '15 at 15:21
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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 $a$.

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  • $\begingroup$ Oh wow that is a novel approach! I wouldn't forget that! Did you think of that?? $\endgroup$ – Makoto K. Jan 16 '15 at 15:51
  • $\begingroup$ Novel? It's a completely standard "trick" to write an expression and use the associative law to put parentheses two different ways to show two things are equal. $\endgroup$ – kahen Jan 16 '15 at 15:52
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    $\begingroup$ Novel having not seen it before. Everything is novel the second before it becomes trivial(it was certainly novel when it was made aswell!) $\endgroup$ – Makoto K. Jan 16 '15 at 15:54
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    $\begingroup$ See also this question. $\endgroup$ – Harald Hanche-Olsen Jan 16 '15 at 16:37
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$(\Longrightarrow)$If $a$ is invertible, then $$az=za=1\implies \exists x,axa=1$$ Take $x=zz$, then $az=za=1 \implies (az)(za)=(1)(1)=1$


$(\Longleftarrow)$ 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 $az=aa^{-1}=za=a^{-1}a=1$

$\blacksquare$

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    $\begingroup$ "$x = a^{-1}a^{-1}$" has no meaning when you haven't proved that $a$ is invertible yet. $\endgroup$ – kahen Jan 16 '15 at 15:44
  • $\begingroup$ Your $\Leftarrow$ calculation actually belongs in the $\Rightarrow$ section. What you have done is assume that $a$ is invertible and then verified by calculation that $axa = 1$ (making use of the fact that $a$ is left- and right-invertible). So far there is no justification for $\Leftarrow$ at all here :). $\endgroup$ – Erick Wong Jan 16 '15 at 15:51
  • $\begingroup$ @ErickWong Can you fix my arrows haha, I can't understand the last part of your post because of the first part $\endgroup$ – Makoto K. Jan 16 '15 at 15:53
  • $\begingroup$ @MakotoK. It might be better not fix this for you because it is an error of understanding and not a simple typo :). Can you see why the first chain of implications doesn't prove by itself that $axa = 1$? $\endgroup$ – Erick Wong Jan 16 '15 at 15:57
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    $\begingroup$ Looks pretty good now, thanks! $\endgroup$ – Erick Wong Jan 16 '15 at 17:57

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