Are there groups except differing right and left inverse.

Question

A group is a monoid for which also:

For each element $a\in G$ there exists an element $a^{-1}$ such that $a^{-1}a = aa^{-1} = e$.

Now instead consider a slightly looser requirement:

For each element $a\in G$ there exists an element $/a$ such that $a/a=e$ and an element $a^{-1}$ such that $a^{-1}a=e$

Let's call such a structure a paragroup.

The question is if there's paragroups that are not groups? I'd guess so, because it would mean that one should have chosen the looser formulation of the invertibility axiom. I guess there's an example of it.

If we loosen the requirement to only reqire one sided inverses, would that really loosen the requirement for paragroups? Or can one prove the existence of RH-inverse from the existence of LH-inverse or vice versa?

Answer 1

Consider the element $a^{-1}a/a$. If we still have associativity, then: $$a^{-1}a/a=(a^{-1}a)/a=/a$$ and $$a^{-1}a/a=a^{-1}(a/a)=a^{-1}$$ Hence $a^{-1}=/a$.
The relation between left and right inverses depends on how you define the identity. Suppose we have a set $G$ with an associative operation on it. If $G$ contains an element $e$ with the following property:

For all $g \in G$, $eg=ge=g$

Then the statement

For all $x \in G$ there exists an element $x^{-1}$ such that $x^{-1}x=e$

implies the statement

For all $x \in G$ there exists an element $/x$ such that $x/x=e$

Indeed, they are the same element: let $x \in G$; then $$xx^{-1}xx^{-1}=x(x^{-1}x)x^{-1}=xx^{-1}$$ Now apply the (left!) inverse of $xx^{-1}$ to see that $xx^{-1}=e$.

A natural question to ask, then, is:

If $G$ is a set with an associative binary operation, and a left/right identity; what is the relationship between left/right inverses of $G$?

Let's decompose this a little further. Suppose $G$ has two elements, $e$ and $f$, with the following properties:

For all $g \in G$, $eg=g$ and $gf=g$

Then $ef=f$ (left identity) and $ef=e$ (right identity), so $e=f$. So if $G$ has a left identity and a right identity, they are the same (this does not even require associativity!).

Suppose $G$ only has a one-sided identity, i.e. it contains an $e$ such that

For all $g \in G$, $eg=g$.

Must it be the case that $G$ has a right identity? Not in general; consider any set $X$, with the multiplication $xy=y$. This is associative (check!) and every $x$ is a left identity. But there certainly isn't a right identity! On the other hand, suppose that $G$ has this property:

For every $g \in G$, there exists $g^{-1}$ such that $g^{-1}g=gg^{-1}=e$

i.e. we have a left identity and normal inverses (inverse in the sense of the left identity). Then

$$geg^{-1}=g(eg^{-1})=gg^{-1}=e$$ The applying $g$ on the right shows $ge=g$. So $e$ is a right identity.

What about a left identity with only left inverses? Well, writing $g^{-1}$ for the left identity of $g$,

$$gg^{-1}gg^{-1}=g(g^{-1}g)g^{-1}=gg^{-1}$$ Applying the left inverse of $gg^{-1}$ shows $gg^{-1}=e$.
Taking $g=e$ shows that $e^{-1}=e$. Moreover, $$ge=g(g^{-1}g)=(gg^{-1})g=eg=g$$ So $e$ is a right identity, and $G$ is a group.

Finally, for the case of left identity and right inverses, consider our counterexample from before. Every $x$ is a left identity and a right inverse for every other element. But it certainly isn't a group.

Answer 2

I am not sure whether this answer your question, but it can be shown that if

1. $G$ is a non-empty set,
2. $\cdot$ is an associative operation on $G$,

and the following conditions hold

3. there is $e \in G$ such that $a \cdot e = a$ for all $a \in G$, and
4. for each $a \in G$ there is $b \in G$ such that $a \cdot b = e$,

then $G$ is a group.

See this MSE post, where left identity and left inverses are used.