[Update: I've now asked the same question on mathoverflow.]

For a semigroup $G$ with a left action on itself, the axiom for compatibility becomes:

$$ \forall f,g,h\in G:hg(f)=h(g(f)) $$

Now suppose there is additional axiom, or constraint if you prefer, called consistency:

$$ \forall f,g\in G: f(g)f=g(f)g $$

This can be represented by a commutative diagram:

enter image description here

If I chain two of these diagrams together I get the following:

enter image description here

The consistency of $f$ and $hg$ can be represented by the following:

enter image description here

Comparing these last two commutative diagrams suggests the following two identities:

$$ \left. \begin{array}{l} hg(f)=h(g(f))\\ (hg)=g(f)(h)f(g) \end{array} \right\} $$

The first is compatibility of course but now there is a second identity which suggests that compatibility has a dual if we require consistency.

These identities have applications in rewriting theory, however it has been suggested to me that a semigroup or monoid with a consistent left action on itself may have interesting mathematical properties in its own right. Is this true? Have semigroups or monoids such as this ever been studied?

  • 1
    $\begingroup$ I'm also wondering whether this question wouldn't be better on mathoverflow but don't want to get marked negatively yet again. $\endgroup$ – James Smith Jun 22 '16 at 20:42
  • $\begingroup$ It does sound like mathoverflow material if your interesting characteristics are indeed interesting, but we can't guess now can we? $\endgroup$ – Patrick Da Silva Jun 22 '16 at 20:46
  • $\begingroup$ Okay, I've rewritten the question pretty extensively. $\endgroup$ – James Smith Jun 23 '16 at 19:20
  • $\begingroup$ What side is the group action from? I can't make the condition make sense. $\endgroup$ – Tobias Kildetoft Jun 23 '16 at 19:55
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    $\begingroup$ I think the condition might be more clear if you added the parenthesis to it (they are not needed for specifying the condition, but I think they would make it easier to read). $\endgroup$ – Tobias Kildetoft Jun 23 '16 at 19:59

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