Is there a name for this object? (Like a group, but the inverse is not necessarily a member of the set) A group is a set $G$, together with a binary operation $\cdot$ that


*

*is closed - if $f\in G$ and $g \in G$ then $f\cdot g \in G$

*is associative - $(f \cdot g) \cdot h = f \cdot (g \cdot h)$

*has an identity element $e$ such that $ef=f$ for all $f\in G$

*has an inverse function: for every $f\in G$ there exists $f^{-1}\in G$ such that $ff^{-1}=e$


I'm interested in a related concept where $f^{-1}$ exists, but isn't necessarily a member of $G$. This means that the identity function doesn't need to be in $G$ either. 
As a simple example, consider the set $T$ of transformations $x\to x+a$, where $a>0$. Each element of this set has an inverse ($x\to x-a$), but neither the inverses nor the identity transformation are members of $T$.
So I'm looking for something very roughly like this:
A <<insert name here>> is a tuple $\langle G, H, e, \cdot \rangle$, where $G$ is the set of forward elements and $H$ is the set of reverse elements, and $e$ is the identity element. The binary operation $\cdot$ is:


*

*closed independently for $G$ and $H$. That is, if $f\in G$ and $g \in G$ then $f\cdot g \in G$, and if $f\in H$ and $g \in H$ then $f\cdot g \in H$.

*is associative (again independently for $G$ and $H$)

*has the invertibility property that for every $f\in G$ there exists $f^{-1}\in H$ such that $ff^{-1}=f^{-1}f=e$, and similarly for every  $h\in H$ there exists $h^{-1}\in G$ such that $hh^{-1}=h^{-1}h=e$.

*$(fg)^{-1} = g^{-1}f^{-1}$ for all $f, g \in G$, and similarly $(fg)^{-1} = g^{-1}f^{-1}$ for all $f, g \in H$.


It's very similar to a group, but where the elements are partitioned into (possibly overlapping) "forward" and "reverse" sets that might not contain the identity element. Note that if $g\in G$ and $h\in H$ then $fh$ might not be in either set.
In the example above, $G$ is the set of translations in the positive direction and $H$ is the set of translations in the negative direction.
I'm not sure what consequences this definition would have, but it seems like it might be a useful thing to define in the context of reversible dynamical systems. Does this concept already have a name, and if so where can I read about it?
 A: Let $G$ be any semigroup. Define $H=G^{op}$ to be the semigroup with underlying set $G \times \{2\}$ but with the opposite operation, $(g \cdot_G h, 2) = (h,2) \cdot_H (g,2)$.
Define $(g,2)^{-1} = g$ and $g^{-1} = (g,2)$. Let $e$ be an element not in $G \cup H$, and define $(h,2) \cdot g = g \cdot (h,2) = e$. Then $\langle G,H,e,\cdot\rangle$ satisfies the axioms.
In particular, there is no restriction on the possibilities for $G$: it can be any semigroup, even those semigroups that cannot be embedded into any group.

I suspect what you want is to partially order your group, and define $G$ to be those elements greater than or equal to the identity, and $H$ to be those less than or equal to the identity. There will be some elements that are not comparable to the identity (all non-identity elements of finite order). For example, let $E = \{ x \mapsto Ax + b : AA^T = I, b \in \mathbb{R}^n \}$ be the group of linear isometries of $\mathbb{R}^n$, and define $Ax + b \geq 0$ iff $A=I$ and $b \geq 0$ (meaning each entry of $b$ is greater than or equal to $0$). Then $G$ is the group of first orthant translations, $H$ are their opposites, and $G \cup H$ is only a small fraction of the total group (all rotations missing, all "mixed" translations missing). 
A: Let $G$ be a set with binary operation satisfying associativity. 
Then $G$ is called as semigroup. If $G$ contains idendity then it is called as monoid.
And semigroups need not to contain inverse of an element. But you want something more; $g^{-1}$ exist but may not be contained in $G$.
There is an example of such monoid. 
Let $G=Q[x]-\{0\}$ be all nonzer0-polynomial over  $\text{ }  \mathbb{ Q}$. Then $G$ is monoid with respect to multiplication since multiplication of polynomials are associative and it contains $1$. But it does not contain inverse of $x^2+1$ as an example. But we know that inverse of it exist 
$$1\over x^2+1$$ which is not element of $Q[x]$. And if we add all inverse of elements to $Q[x]$ we will get a group.
