# Do these very VERY weak axioms guarantee a group? (Every element has left identity $e_L$ or right identity $e_R$, with same-sided inverse)

I can't help but ask, after we've come so far weakening the group axioms in these two posts, whether we can get even weaker?

Let $$A$$ be a set with an associative binary operation $$*$$, and suppose there exist two elements $$e_L,e_R\in A$$ such that, for all $$x\in A$$, at least one of the following two conditions holds:

1. $$e_L*x=x$$ and there exists an $$x'\in A$$ such that $$x'*x=e_L$$;
2. $$x*e_R=x$$ and there exists an $$x'\in A$$ such that $$x*x'=e_R$$.

Must $$(A,*)$$ be a group?

The second linked post proves that the answer is “yes” in the case where $$e_L=e_R$$. This generalized version was posed by @Yakk in the comments to @Vincent's answer in the first linked post.

• If both conditions hold, then $e_R = e_L * e_R = e_L$ Commented Sep 17, 2021 at 16:11
• @AlvinL Yes, but only if both conditions hold. Otherwise we do not know that $e_L$ is a left inverse for $e_R$ and vice versa. Commented Sep 17, 2021 at 16:14
• In general we might only have $e_L*e_L=e_L$ and $e_R*e_R=e_R$. Commented Sep 17, 2021 at 16:15

Take $$A = \{1, 0\}$$ with the usual multiplication. Let $$e_L = 1$$ and $$e_R = 0$$. If $$x = 1$$, then $$e_Lx = x$$ and $$x'x = e_L$$ for $$x' = 1$$. If $$x = 0$$, then $$xe_R = x$$ and $$xx' = e_R$$ for $$x' = 0$$.
If I'm not mistaken. Here's a counterexample. Let $$G$$ and $$H$$ be arbitrary groups, $$A=G\cup H$$. Define multiplication on the set $$A$$ by the rule
$$xy= \left\{% \begin{array}{ll} xy, & \hbox{if x,y\in G or x,y\in H;} \\ y, & \hbox{if x\in G and y\in H;} \\ x, & \hbox{if x\in H and y\in G.} \\ \end{array}% \right.$$ Here $$e_L=e_G$$, $$e_R=e_H$$.
• Nice non-commutative example (+1). Perhaps you have noted that for $x\in G$ and $y\in H$, we have $xy=yx=y$, that is, the product is always the element from $H$; this could save one branch in the definition, but perhaps it would be less clear. Commented Sep 22, 2021 at 16:31
• Yes, it's more concise than saying if one belongs to $G$ and another to $H$ then the product is the one from $H$. Although it may be a relevant observation, the definition as you stated is indeed more concise; it would actually be somewhat cumbersome to define it as I did above. I agree that it is better as it is. Commented Sep 22, 2021 at 18:16