# Proving that a groupoid is a group, knowing the following properties

(G, ·) is a groupoid. Prove that if it has the following properties it is also a group: $$1) (a · b) · c = a · (b · c), (\forall)a, b, c \in G;$$ $$2) (∃)u ∈ G : u · a = a · u = a, (∀)a ∈ G;$$ $$3) (∀)a ∈ G(∃)a ∈ G : a · a' = u \lor a' · a = u$$

I'm confused. Aren't the above mentioned properties all we need to prove that it's a group? What else am I supposed to do?

• What definition of "group" are you using? Dec 30, 2020 at 10:42
• @user3482749 The one from here: en.wikipedia.org/wiki/Group_(mathematics) Dec 30, 2020 at 10:47

Take a look to the property 3: it states take every element has a left or a right inverse, while in your definition every element must have a two-sided inverse.

In general, in a (non-commutative, of course) monoid, if an element has a right inverse, this is not also its left inverse (see In a non-commutative monoid, is the left inverse of an element also the right inverse?), but in you case it is different, as every element has a left or a right inverse (some a right inverse, some a left inverse).

Just for fun: the following listing can be used in SPASS (a theorem prover) to prove that actually properties $$1-3$$ imply that every element has a two-sided inverse.

begin_problem(monoid).

list_of_descriptions.
name({**}).
author({**}).
status(unsatisfiable).
description({**}).
end_of_list.

list_of_symbols.
functions[(b,0), (P,2)].
end_of_list.

list_of_formulae(axioms).

formula(forall([x],equal(P(x,b),x))).
formula(forall([x],equal(P(b,x),x))).
formula(forall([x],forall ([y],forall([z],equal(P(P(x,y),z),P(x,P(y,z))))))).
formula(forall([x],exists([y],or(equal(P(x,y),b),equal(P(y,x),b))))).

end_of_list.

list_of_formulae(conjectures).

formula(forall([x],exists([y],and(equal(P(x,y),b),equal(P(y,x),b))))).
end_of_list.

end_problem.


Now, let's try to solve your problem. Let $$a\in G$$ and suppose $$a$$ is right-invertible, i.e., exists $$a'\in G$$ such that $$aa'=u$$. Clearly, $$aa'$$ is left- and right-invertible, now, let us consider $$a'a$$. From $$3$$, $$a'a$$ is either left- or right-invertible. So suppose $$a'a$$ right-invertible, and let $$b_l$$ and $$b_r$$ a left and a right inverse of $$aa'$$ and $$a'a$$, respectively. Then

$$u=b_l(aa')=(b_l a)a'$$ and $$u=(a'a)b_r=a'(ab_r)$$

so $$a'$$ has both left (i.e., $$b_l a$$) and right (i.e., $$ab_r$$) inverse, and then it is invertible (this is a standard fact in a monoid) and $$b_l a=ab_r$$. Clearly, if $$a'a$$ is actually left-invertible, the argument is similar. Now, back to $$a$$.

From above, we have

$$u=(b_l a)a'=(ab_r)a'=a(b_r a')$$

and

$$u=a'(ab_r)=a'(b_l a)=(a' b_l) a$$

and then also $$a$$ has both left (i.e., $$a'b_l$$) and right (i.e., $$b_r a'$$) inverse, and then it is (two-sided) invertible. As $$a$$ is generic, we have proved that $$G$$ is a group (of course, you have to work by yourself the case in which $$a'a$$ is left-invertible).

• I donn't think so, consider that "grupoid" also means "magma". Dec 30, 2020 at 10:53