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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'?

Makoto K.
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    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. – Harald Hanche-Olsen Jan 16 '15 at 15:16
  • @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 $x=a^{-1}a^{-1}\implies ax=a^{-1}\implies axa=1$? – Makoto K. Jan 16 '15 at 15:18
  • For the opposite take $x=(a^{-1})^2$. – Janko Bracic Jan 16 '15 at 15:20
  • 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? – Harald Hanche-Olsen Jan 16 '15 at 15:20
  • @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 – Makoto K. Jan 16 '15 at 15:21
  • Well, it is a classical result that if both a left inverse and a right inverse exist, they are equal (and hence a two-sided inverse). You might try to prove that first. – Harald Hanche-Olsen Jan 16 '15 at 15:22
  • The sad thing is, I don't know which direction of the iff I have proven :. – Makoto K. Jan 16 '15 at 15:26
  • @Harald I am fairly sure I have proven to the right, and now I have to prove to the left, e.g. There is some $x\in M$ such that $axa=1$ hence $a$ is invertible, is this correct – Makoto K. Jan 16 '15 at 15:27
  • Yes, that is the direction that needs proving now. – Harald Hanche-Olsen Jan 16 '15 at 15:35
  • @HaraldHanche-Olsen Can you review my answer please? – Makoto K. Jan 16 '15 at 15:38

2 Answers2

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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 $a$.

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

$\blacksquare$

Makoto K.
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    "$x = a^{-1}a^{-1}$" has no meaning when you haven't proved that $a$ is invertible yet. – kahen Jan 16 '15 at 15:44
  • 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 :). – Erick Wong Jan 16 '15 at 15:51
  • @ErickWong Can you fix my arrows haha, I can't understand the last part of your post because of the first part – Makoto K. Jan 16 '15 at 15:53
  • @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 $axa = 1$? – Erick Wong Jan 16 '15 at 15:57
  • @ErickWong Yes I can whoops I am silly! Will edit – Makoto K. Jan 16 '15 at 15:58
  • @ErickWong If you are still there(and won't judge me for taking 40 minutes lol), can you tell me if this is right? – Makoto K. Jan 16 '15 at 16:43
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    Looks pretty good now, thanks! – Erick Wong Jan 16 '15 at 17:57
  • @ErickWong Why are you thanking me?? Thank you! – Makoto K. Jan 17 '15 at 12:37
  • @MakotoK. One comment I'd add is you have a tendency to abuse and overuse $\Rightarrow$ and $=$. You should strive to express your reasoning more clearly, using words. E.g.: $axa=1 \Rightarrow ax=a^{-1} \Rightarrow (ax)a = 1$ should really read more like "$axa=1$ implies $(ax)a = 1$, so $a$ has a left-inverse, namely $ax$". As kahen points out, it's circular to refer to $a^{-1}$ at all until you've proven that it exists. – Erick Wong Jan 17 '15 at 16:20