# 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.

• What would "inverse" even mean without an identity element? – Henning Makholm Nov 11 '16 at 21:35
• How do you define an inverse? Does the definition not include the identity element? – user384138 Nov 11 '16 at 21:35
• @HenningMakholm that's exactly what I am thinking. I think inverses elements imply that an identity element must exist! – mjo Nov 11 '16 at 21:36
• @OpenBall it does in fact include the identity in the definition of inverses. – mjo Nov 11 '16 at 21:37

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.

• One of the standard textbook definitions of a group is a set with an associative operation satisfying your displayed statement. Of course that statement, together with associativity, easily implies the existence of an identity element and inverses. – bof Nov 11 '16 at 21:59
• @bof I think OP's question was really how to even talk about the existence of inverses in the absence of an identity element, which is why my answer is, in some sense, just a revision of the problem -- I think that's what's called for here. – Mitchell Spector Nov 11 '16 at 22:02
• The first of these proposals will guarantee inverses but also implies that there is an identity. The second does not necessarily give you a group; it is satisfied by $(\mathbb N_0,+)$, for example. – Henning Makholm Nov 12 '16 at 0:15
• @HenningMakholm Yes, that's exactly the kind of analysis that I was suggesting OP could do -- but the key point was how one might try to formulate something similar to the existence of inverses without using the existence of an identity element. – Mitchell Spector Nov 12 '16 at 2:02

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.