Have semigroups with actions on themselves that have a dual to the compatibility axiom ever been studied?

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 standard commutative diagram. If I chain two of these diagrams together I get the following:

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

Comparing these two commutative diagrams suggests the following two identities:

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

The first is compatibility of course but now there is a second identity which indicates that compatibility can have a dual, which I'm going to call co-compatibility.

These identities have applications in rewriting theory, however it has been put 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?

I do not have much intuition for the identities presented in the question, the LD-monoids are the monoids with self actions that satisfy identities that look similar to consistency, compatibility, and co-compatibility. An LD-monoid is an algebra $(X,*,\circ,1)$ such that $(X,\circ,1)$ is a monoid and where

1. $x\circ y=(x*y)\circ x$ (The braid identity)

2. $x*(y\circ z)=(x*y)\circ(x*z)$ (distributivity of $*$ over $\circ$)

3. $(x\circ y)*z=x*(y*z)$ ($*$ acts on $(X,\circ,1)$)

4. $x*1=1,1*x=x$ (Identity)

LD-monoids satisfy the self-distributivity identity $x*(y*z)=(x*y)*(x*z)$.

Notice how the distributivity of $*$ over $\circ$ resembles co-compatibility and the braid identity resembles consistency.

LD-monoids originally arose in set-theory since the algebra of all elementary embeddings $j:V_{\lambda}\rightarrow V_{\lambda}$ is an LD-monoids. LD-monoids also appear in other contexts since every Heyting algebra is an LD-monoid (where $x\circ y=x\wedge y,x*y=x\rightarrow y$), every group becomes an LD-monoid where $x*y=xyx^{-1}$, and every self-distributive algebra can be extended to a larger LD-monoid in a canonical way. More information on LD-monoids can be found in Chapter 11 of Dehornoy's book Braids and Self-Distributivity.

• Many thanks for the input. I'd given up on this question, to be honest. I've marked your answer up. Even though I haven't taken the time to fully digest it yet, I'm sure it'll be helpful. – James Smith Jan 20 '17 at 16:14

I cannot answer the question about whether or not semigroups or monoids with consistent (what I call) self-actions have ever been studied. However, I can shed some light on consistency and its relation to co-compatibility, as well as the relation between co-compatibility and compatibility. These turn out to be equivalent for such an action on groups but not it seems on semigroups or monoids.

I can start with the observation that consistency can only hold for self-actions. We have both $f(g)$ and $g(f)$ therefore both $f$ and $g$ must be in both the $G$-set and the underlying set. Since this holds for all $f$ and $g$, these two sets must be the same by extensionality.

Next if we assume compatibility then equating the two commutative diagrams above we get the following identity:

$$f(hg)f=g(f)(h)f(g)f$$

For groups we have an inverse $f^{-1}$ for $f$ and the result follows. It is also easy to see that if we start with this identity we can recover the identity $hg(f)=h(g(f))$ with the same kind of reasoning. Hence compatibility and co-compatibility are equivalent for groups with a consistent self-action.