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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.

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    $\begingroup$ If both conditions hold, then $e_R = e_L * e_R = e_L$ $\endgroup$
    – AlvinL
    Commented Sep 17, 2021 at 16:11
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    $\begingroup$ @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. $\endgroup$
    – WillG
    Commented Sep 17, 2021 at 16:14
  • $\begingroup$ In general we might only have $e_L*e_L=e_L$ and $e_R*e_R=e_R$. $\endgroup$
    – WillG
    Commented Sep 17, 2021 at 16:15

2 Answers 2

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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 $x' = 0$.

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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 $e_L=e_G$, $e_R=e_H$.

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    $\begingroup$ 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. $\endgroup$
    – amrsa
    Commented Sep 22, 2021 at 16:31
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    $\begingroup$ Thank you @amrsa. Maybe I should have commented on this point. But I wanted to make the example as concise as possible. $\endgroup$
    – kabenyuk
    Commented Sep 22, 2021 at 16:43
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    $\begingroup$ 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. $\endgroup$
    – amrsa
    Commented Sep 22, 2021 at 18:16

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