Do these “ultraweak” one-sided group axioms guarantee a group?

Solution 1:

Consider any set $X$ equipped with the left projection operation $$*:(a,b)\mapsto a.$$ Associativity is trivial. Meanwhile, every element is a (right) identity: fixing $e\in X$ we have $x*e=x$ for all $x\in X$. Similarly, fixing $e\in X$ we can think of $e$ itself as the "ultraweak (left) inverse of $x$ with respect to $e$" - since it satisfies $e*x=e$.

So $Proj(X):=(X,*)$ satisfies a very strong version of the axioms you list. But as soon as $X$ has more than one element, $Proj(X)$ is very far from a group.

Re: the gap between this notion and full grouphood, note that crucially in the above we have a contrast between the ways the second and third axioms are satisfied - every element is a right identity, but we only have left inverses. The post linked in the OP meanwhile shows that if we demand right identity + right inverses, or left identity + left inverses, we do get full grouphood. So it's not so much that any single type of "sidedness" can vary from element to element, but rather that the relevant "sidednesses" are not required to be the same, which gives a weak notion.

Solution 2:

What axioms are enough to guarantee a group?

Assuming associativity. "Two-sided inverses" only requires that there is a left inverse and a right inverse for every element, they don't have to be the same.

• Left identity and left inverses: Enough $\color{green}{\checkmark}$
• Left identity and right inverses: Not enough (Noah's answer) $\color{red}{\mathcal{X}}$
• Ultraweak identity and two-sided inverses: Enough (See below) $\color{green}{\checkmark}$
• Two-sided identity and ultraweak inverses: Enough (See below) $\color{green}{\checkmark}$

In summary, once either the identities or the inverses are two-sided, we have a group. But if that is not the case, the only way to still guarantee a group is if the identity and inverses are both always on the same side.

Ultraweak identity and two-sided inverses are enough

We only require that there is a left inverse and a right inverse for every element, they don't have to be the same.

We show that $e$ is a left identity for every element. Since we have left inverses, the claim then follows from this answer. For an element $a$, the ultraweak identity yields $ea=a$ or $ae = a$. We only need to focus on the second case. $a$ has a right inverse $a'$, and $a'$ has a right inverse $a''$. Thus, $a = a e = a (a' a'') = (a a') a'' = e a''$. This shows that $ea = e(e a'') = e a'' = a$ as required.

Two-sided identity and ultraweak inverses are enough

Indeed, in this case $x^2=x$ implies $x=e$ for all $x$. Thus, if a has right-inverse $b$, we have $ab=e$ and thus $ba=b(ab)a=(ba)^2$, so that $ba=e$, therefore $b$ is also the left inverse.

Answer to REVISED UPDATE

I decided to track this in a seperate question. See this question and answer.

Solution 3:

No, these are not sufficient. Consider the set of the two matrices $$\begin{pmatrix}1&0\\0&0\end{pmatrix},\begin{pmatrix}1&1\\0&0\end{pmatrix}$$ with respect to matrix multiplication.