Can a set that is closed under a binary operation have inverses for all of its elements.. without the existence of an identity element? Can a set that is closed under a binary operation have inverses for all of its elements.. without the existence  of an identity element?
My mind is telling me that in order to even have inverses, there needs to exist an identity element.
I am attempting to prove that the group
Axioms
1)associativity
2)identity element
3) exisistance of inverse elements
are independent of each other, meaning you can't have 2 and imply the third one.
 A: The usual definition of "inverse" requires an identity element to even mean anything.
But if you wanted to look at the independence of the various group axioms from one another, you could replace the statement saying that inverses exist with the statement
$$\forall a \forall b \big(\exists x (ax = b) \,\land\, \exists y (ya = b)\big)$$
or alternatively with the cancellation laws
$$\forall a \forall x \forall y \big((ax=ay \,\lor\, xa=ya)\rightarrow x=y\big)$$
(since those statements do not depend on the existence of an identity element) and see if either of those is independent of the other axioms.
A: As already explained, the standard definition of an inverse depends on the existence of an identity element, so from that perspective the question makes no sense.  However: In Semigroup Theory one sometimes explores a concept called inversive semigroups, in which the definition of an inverse element is replaced with the following axiom:
$\forall a \exists b ((aba = a) \,\land\, (bab = b))$
This is probably what you are looking for, and yes, in these structures there need not be an identity element.
A: I will adapt here another answer of mine to a similar question.
In associative structures, the inverse element is defined after the identity element. In a generic (non associative) groupoid, an inverse property can be defined without identity element instead.
A generic groupoid $G$ has the left inverse property if
$$\forall x\in G, \exists\ z\in G : z(xy)=y\ \forall y\in G\tag{1}$$
and has the right inverse property if
$$\forall x\in G, \exists\ t\in G : (yx)t=y\ \forall y\in G.\tag{2}$$
$G$ is also said to have the inverse property if it satisfies both the above equations.
Be careful because $z$ as in $(1)$ and $t$ as in $(2)$ may exist but not be unique.
Let now $G$ be a loop with identity element $e$. Then:
$$\forall x\in G, \exists\,!\,x_l\in G : x_l\,x = e$$
and
$$\forall x\in G, \exists\,!\,x_r\in G : x\,x_r = e.$$
Here $x_l$ and $x_r$ are unique because the loop is a quasigroup. We call $x_l$ left inverse of $x$ and $x_r$ right inverse of $x$.
In the absence of associativity, the only existence of $x_l$ and $x_r$ does not imply $(1)$ and $(2)$ to be verified, indeed not all the loops have the inverse property.
