# I Proving an inverse of a groupoid is unique

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1. Jan 1, 2017

### Matejxx1

Hello
I have a question about the uniqueness of the inverse element in a groupoid. When we were in class our profesor wrote $\text{Let} (M,*) \,\text{be a monoid then the inverse (if it exists) is unique}$. He then went off to prove that and I understood it, however I got curious and started thinking if it is possible to prove that there is only one unique inverse without taking into account the associative property of the semigroup. So then I started trying to prove it but I didn't really get too far and I tried looking online and also didn't find much about it. Could anybody tell me how this would be done ?
thanks

Last edited: Jan 1, 2017
2. Jan 1, 2017

### Staff: Mentor

If you have only a binary operation and a unit, you can define whatever you want. E.g.
$$\begin{bmatrix}*&e&a&b&c\\e&e&a&b&c\\a&a&b&e&e\\b&b&e&c&b\\c&c&e&b&a\\\end{bmatrix}$$

3. Jan 1, 2017

### Stephen Tashi

If it is a false statement then it can't be done.

The general form of your question is "How do I prove a theorem about an algebraic structure that has certain properties without using some of those properties? ". The usual interpretation of that type of question is that we consider a different algebraic structure that is formed by removing some properties of the original algebraic structure. Then we try to prove the theorem for this new structure. (Of course "algebraic structure" refers to the collection of possible examples that satisfy the definition of that structure. So if we remove properties from the definition of a structure we enlarge the number of examples that we must consider. If we enlarge the number of examples then we incur the risk of allowing an example where the statement of our theorem is false.)

If we take the definition of monoid and remove the requirement that it be associative then we create a definition of a new algebraic structure. Even if we keep the requirement that an identity element exists in this new structure it is not clear that the identity element is unique. If we don't have a unique identity element, then how do we define an inverse of an element x in this new algebraic structure? We would have to look at the definition for "the inverse of x" in a monoid and see if that definition relies on the uniqueness of the identity in a monoid.

4. Jan 18, 2017

### Martin Rattigan

If you mean a two sided identity how could it fail to be unique? 1a1b=???.

5. Jan 18, 2017

### Martin Rattigan

Stephen Tashi wrote:

"If we take the definition of monoid and remove the requirement that it be associative then we create a definition of a new algebraic structure."

I propose the the name oneoid after J. Milton Hayes.