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Define $\mathcal L^*$ by $a \mathcal L^* b$ if and only if

$∀x,y \in S^1$, $ax=ay \iff bx=by$

Show that $\mathcal L ⊆\mathcal L^*$ on a semigroup $S$

This seems like it should be a very simple question but it isn't working out as I would have hoped. Any hints would be much appreciated!

EDIT. The Green's relation $\mathcal L$ on a semigroup $S$ is defined by $a \mathrel{\mathcal L} b$ if $S^1a = S^1b$. Here $S^1$ is $S$ with an identity adjoined if $S$ is not already a monoid.

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  • $\begingroup$ @zachary-selk Both $S^1$ and $\mathcal L$ are standard notation in semigroup theory. I have edited the question to include these definitions. Why do you think that the definition of $\mathcal L^*$ does not make sense? $\endgroup$
    – J.-E. Pin
    Nov 25, 2016 at 14:50
  • $\begingroup$ In my opinion, this question should stay open. $\endgroup$
    – J.-E. Pin
    Nov 25, 2016 at 14:57

1 Answer 1

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Suppose that $a \mathrel{\mathcal L} b$. Then there exist $s, t \in S^1$ such that $sa = b$ and $tb =a$. Suppose now that $ax = ay$. Then $sax = say$, that is $bx = by$. Similarly, if $bx = by$, then $tbx = tby$, that is $ax = ay$. Thus $a \mathrel{{\mathcal L}^*} b$.

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