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Show that a semigroup $S$ is a rectangular band if and only if $ab=ba \Rightarrow a=b$. (For all $a,b\in S$)

I have the definition of a rectangular band as $aba=a$.

When I try to prove this I keep getting stuck. This is my best effort.

$aba=a$

$ab=ba \Rightarrow aba=baa \\ \Rightarrow a=baa \\ \Rightarrow ab=baab=b \\ \Rightarrow ab=b $

But I can't get from here to the final result. Any hellp is much appreciated

user2973447
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2 Answers2

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The opposite direction can be proved as follows. Since $a$ and $a^2$ commute, $a = a^2$. Now, since $(aba)a = aba^2 = aba = a^2ba = a(aba)$, $a$ and $aba$ commute and thus $aba = a$.

J.-E. Pin
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You're almost there. If we replace the variables to the opposite of what you have above, that is $$bab=a$$ $$ba=ab \Rightarrow bab=abb \\ \Rightarrow b=abb \\ \Rightarrow ba=abba=a \\ \Rightarrow ba=a$$ This is valid because $a$ and $b$ can be any elements of $S$. Therefore,$$b=ab=ba=a \\ \Rightarrow a=b$$

This proves the forward direction.

Nambiar M.
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    Thanks! I think I have the reverse direction too but I'm not sure. Is this correct?

    $ab=ba \Rightarrow a=b $.

    Let $b=a^2$ $\Rightarrow a=a^2$ (as $a=b$) $\Rightarrow aba=baa=ba=a^2=a$ (Again using that $a=b$ and that $a=a^2$)

     – user2973447 Oct 26 '16 at 01:44
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    Remember, to prove the backwards direction, what you have to do is assume $ab = ba \Rightarrow a = b$. You don't know for a fact that $ab=ba$, only that if it is true, then $a=b$ I haven't fully done the proof, but a valid method would be to consider $aba \neq a$ and $ab=ba$ then show that $a \neq b$. Again, I haven't actually done the proof, but this would be a valid approach if you could get it to work. – Nambiar M. Oct 26 '16 at 02:53
  • I couldn't get it to work =/ I'm sure it would work but I seem to be going in circles when I try to do it – user2973447 Oct 26 '16 at 09:39