# Does the existence of only right neutral mean that there can only exist right inverse?

Let $(G,\circ$) be groupoid. If there exists only right neutral, does that mean that there can exist only right inverse?
To put it in another way, does the existence of only right neutral mean that there can not exist left inverse?
I assumed this from an example where $\circ$ is operation of exponentiation and is defined by: $$(\forall x,y\in G) x\circ y=x^y$$ So I want to make sure my assumption is correct.

Edit: Definition of groupoid I'm using:

Ordered pair $(G,\circ)$ where $G$ is a set, and $\circ$ is an internal binary operation is called groupoid.

• @ProfessorVector operation doesn't have to be associative to make it a grupoid, it only has to be internal Sep 23, 2017 at 13:05
• I see, and yes, "magma" is an apt name for that. Have fun!
– user436658
Sep 23, 2017 at 21:20
• @Professor Vector, if you check Wikipedia article on magmas, you can find that it says: "In abstract algebra, a magma (or groupoid [...])". If you are unaware of different terminologies used (which is perfectly fine), you can't expect others to be aware. What I mean is that the line: "If you use another definition of "groupoid" than the one commonly used, you should say so in your question." should be replaced with a kinder variant: "What definition are you using?" because OP might not be aware of different naming conventions. Sep 23, 2017 at 21:54
• @Ennar In this case I think the note on Wikipedia is misleading. The term groupoid as used here has essentially been replaced by the term magma in all modern contexts, since groupoid has taken on a new meaning. This all leads to a bunch of confusion between people who have not seen the discrepancy before, and Wikipedia just stating them as synonyms with no additional note does not help. Sep 24, 2017 at 9:08
• @Tobias, the term grupoid is still actively used on my university in introductory course in algebra for magmas. I'm not saying that this follows modern trends, but here you have three different sources that uses "grupoid" instead of "magma". Personally, I do prefer to use "grupoid" for category theory, and I hardly use magma (or any other term) at all. Still, my point was that OP was probably unaware of conflicting terminologies in the first place. Sep 24, 2017 at 9:46

After further research I found out that existance of only right neutral doesn't mean there can't exist left inverse. I found this out by observing operation of division $/$ on a set of positive rational numbers $Q^+$.
The operation of division has only right neutral, but has both left and right inverse.

Here is my proof:

Let $(G,\circ)$ be groupoid (or magma as suggested) where $G=Q^+$ and $\circ =/$. $$(\exists e\in G)(\forall x\in G) e\circ x=x\circ e=e$$
For left neutral: $$e\circ x=x$$ $$e\circ x=\frac{e}{x}$$ from these two we get $x=\frac{e}{x}$ which means that there is no left neutral.

For right neutral:
$$x\circ e=x$$ $$x\circ e=\frac{x}{e}$$ from these two we get $x=\frac{x}{e}$ which means that there is only right neutral $e=1$.

For inverse:
$$(\forall x\in G)(\exists x^{-1}\in G)x\circ x^{-1}=x^{-1}\circ x=e$$
For left inverse:
$$x^{-1}\circ x=e$$ $$x^{-1}\circ x=\frac{x^{-1}}{x}$$ from these two we get $\frac{x^{-1}}{x}=1\Rightarrow x^{-1}=x$

For right inverse:
$$x\circ x^{-1}=e$$ $$x\circ x^{-1}=\frac{x}{x^{-1}}$$
from these two we get $\frac{x}{x^{-1}}=1\Rightarrow x^{-1}=x$.

This proves there exist both left and right inverse $x^{-1}=x$ even though there exists only right neutral.
(At least I think this proves it, correct me if I'm wrong :) )