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If $G$ is a group, we see that the bijection $y\mapsto y_r$, where $(x)a_r = xa$, functions written from the right, is a group isomorphism.

In this way, we have an interesting interpretation of groups as translations.

Inspired by this, let $S$ be any magma. Let's consider the same operation of a right translation on $S$. We can ask a question, when is this map $y\mapsto y_r$ a homomorphism.

Clearly, this is iff $(x)a_rb_r = (x)(ab)_r$ for any $x, a, b$, that is, $(xa)b = x(ab)$.

In this way, from a simple question about translations forming a homomorphism we naturally arrived at the definition of a semigroup.

We can ask ourselves another question, when is this map an isomorphism? This is an isomorphism iff $y\mapsto y_r$ is injective, and that is iff $(x)a_r = (x)b_r$, that is, $xa = xb$ implies $a = b$ for all $x, a, b$.

Such semigroups are called left weakly reductive, lwr semigroups for short.

However, there is a serious issue with this, lwr semigroups don't form a variety among structures with binary operation. Clearly, if $S$ is a semigroup which isn't a lwr semigroup, then $S^1$, where $S\mapsto S^1$ is the operation of adding an identity element, is a lwr semigroup. But then $S^1$ contains $S$ as its subsemigroup, so that a subsemigroup of a lwr semigroup doesn't have to be an lwr semigroup.

But there is a class of semigroups which does form a variety, maybe not as structures with binary operation, but with addition of a nullary element. Those are left monoids $(S, \cdot, e)$ where $\cdot$ is a binary operation, $e$ is nullary element, and there are two identities: $x(yz)\approx (xy)z$, $ex \approx x$.

As a class of semigroups which comes from forgetting the operation $e$ i. e. $(S, \cdot, e)\mapsto (S, \cdot)$, all left monoids are lwr semigroups.

My question is, if we could find some structure $(S, \cdot, \mathscr{F})$ where $\mathscr{F}$ denotes the set of all $n$-ary functions for $n\in\mathbb{N}_0$ other than $\cdot$ , the binary function, and identities, so that a set of algebras satisfying those identities forms a variety, and the map $(S, \cdot, \mathscr{F})\mapsto (S, \cdot)$ for any algebra from those variety turns it into an lwr semigroup, and so that the class of sets obtained this way is maximal, or if it's impossible.

For starters, if a variety which would cover a class of semigroups between left monoids and lwr semigroups.

Update:

I've been able to find a potential candidate for such structure. Consider $(S, \cdot, e)$ where $e:S\to S$ is an unary operation, with identities $e(x)x \approx x$ and $e(x)e(y)e(x)y \approx y$.

Such structures generalize monoids since if $f$ is the left identity, then $e(x) \equiv f$ gives us a structure of above type. Moreover, any semigroup which satisfies those relations is a lwr semigroup, since $xa = xb$ implies $a = e(a)b$ and $b = e(b)a$ so that $a = e(a)e(b)e(a)b = b$.

However, I don't have an example which isn't a left monoid.

Update 2:

I found that $e(x)x\approx x$, $e(x)e(y)e(x)y \approx y$ are equivalent to $e(x)^2 y \approx y$, $e(y)e(x)y \approx e(x)y$.

In particular, we can fix $x\in S$ and take $ e = e(x)^2$, so that $ey = y$ for all $y\in S$. This means $S$ is a left monoid.

Update 3:

Small lwr semigroups which aren't left monoids:

There is a unique lwr semigroup of order $3$ which isn't a left monoid. It's given by the matrix \begin{bmatrix} 1 & 3 & 3 \\ 3 & 2 & 3 \\ 3 & 3 & 3 \end{bmatrix}

By my calculations, there is $18$ semigroups of order $4$ which are lwr but not left monoids, with Id's $54, 67, 69, 69^t, 70, 70^t, 77^t, 88, 92, 98, 99^t, 100, 101^t, 102, 102^t, 103^t, 110^t, 112$ in GAP package Smallsemi, where $t$ means the multiplication table transposed (anti-isomorphism).

New structure:

Let's consider $(S, \cdot, e)$ where $e$ is a binary operation $e:S\times S\to S$, with identities $e(x, y) \approx e(y, x)$ and $e(x, y)x\approx x$.

You can think the binary operation as a choice of local left identities, which bind two elements $x, y$ together by demanding $e(x, y)$ to be both local left identity of $x$ and $y$.

If $S$ is a left monoid with left identity $f$, we can take $e(x, y) \equiv f$.

Any such structure is a lwr semigroup because $xa = xb$ for all $x$ implies $e(a, b)a = e(a, b)b$, and that means $a = b$.

Such structures are discussed here and here.

Update 4:

There seems to be another class of semigroups which are weakly reductive! Those are semigroups induced by (small) categories. The construction can be found here in the answer of @J.-E.Pin.

If $(S, \cdot)$ is induced by category $\mathcal{C}$, and $x\cdot f = x\cdot g$ for all $x\in S$ and morphisms $f, g$ of $\mathcal{C}$, then taking $x = 1_X$ where $X$ is the domain of $f$, $f = 1_X\cdot g$. In particular, $1_X\cdot g = 1_X\circ g$ is defined so equal to $g$. If $x\cdot f = x\cdot 0 = 0$ for all $x\in S$ and morphism $f$ of $\mathcal{C}$, then as before $f = 0$. Contradiction.

This proves all semigroups induced by (small) categories are weakly reductive.

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You can describe the class of all lwrs this way, using one extra binary and one ternary operation. Denoting the two operations by $w(a,b)$ and $r(a,b,c)$, consider the variety defined by associativity of $\cdot$ and identities $$r(a,b,w(a,b)\cdot a)=a,\\ r(a,b,w(a,b)\cdot b)=b.$$ Note that if those identities hold, then for any $a\neq b$ we must have $w(a,b)\cdot a\neq w(a,b)\cdot b$, which implies the semigroup $(S,\cdot)$ is lwr for any such algebra $(S,\cdot,w,r)$.

Conversely, if the semigroup $(S,\cdot)$ is lwr, then for any $a\neq b$ there is some $w(a,b)$ such that $w(a,b)\cdot a\neq w(a,b)\cdot b$. Use those values to define $w$, setting $w(a,a)$ arbitrarily. We can now define $r(a,b,c)$ in such a way that it equals $a$ if $c=w(a,b)\cdot a$, $b$ if $c=w(a,b)\cdot b$ and takes any value for all other $c$. Then $(S,\cdot,w,r)$ satisfies the identities above.

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  • $\begingroup$ This is awesome! Thank you so much! $\endgroup$ – Jakobian Dec 21 '20 at 14:59

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