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In the discussion of this question users WillG and Yakk ask whether the following conditions define a group:

Let $A$ be a set and $*$ be a binary operation on $A$ satisfying:

  1. $*$ is associative.
  2. There exists an $e\in A$ such that for all $a\in A$, either (1) $e*a=a$ and there exists an $a'\in A$ such that $a'*a=e$, or (2) $a*e=a$ and there exists an $a'\in A$ such that $a*a'=e$.

I created this question to track the answer separately.

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Yes, these axioms are enough to define a group

Let $a \in A$ be a "left" element, so that $ea = a$ and $a' a = e$ for some $a' \in A$. We need to show $a a' = e$. Once this is done, the proof applies to "right" elements $a$ as well by symmetry, and we have shown that every element has a two-sided inverse. Since there is a superweak identity $e$, the claim then follows from this answer.

We will prove this by casework.

Case 1: $a'$ is a left element

In this case we have $a'' \in A$ with $a'' a' = e$, and $e a' = a'$. So we have $a = e a = (a'' a') a = a'' (a' a) = a'' e$. This means that $a a' = (a'' e) a' = a'' (e a') = a'' a' = e$, which we wanted to show.

Case 2: $a'$ is a right element

In this case, we only learn that $a' e = a'$. Define $b := a a'$, we need to show $b = e$.

Case 2.1: $b$ is a left element

Note that $eb = b$ and take $c \in A$ with $cb = e$. Note that $a = e a = (cb) a = c (ba) = c (a a' a) = c a e$, which implies that $ae = (c a e) e = c a e = a$. Thus, we have $b^2 = (a a') (a a') = a (a' a) a' = a e a' = a a' = b$. Apply $c$ on the left to yield $e b = e$ and thus $b = e$.

Case 2.2: $b$ is a right element

This is the same proof as case 2.1 mirrored, but I will spell it out to convince you and myself.

Note that $be = b$ and take $c \in A$ with $bc = e$. Note that $a' = a' e = a' (bc) = (a' b) c = (a' a a') c = e a' c$, which implies that $e a' = e ( e a' c) = e a' c = a'$. Thus, we have again $b^2 = (a a') (a a') = a (a' a) a' = a e a' = a a' = b$. Apply $c$ on the right to yield $b e = e$ and thus $b = e$.

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  • $\begingroup$ The group $A$ must be closed under $\ast$, that is, $a \ast b \in A$ for all $a, b \in A$. Is that the case here? $\endgroup$
    – Ramanujan
    Sep 17, 2021 at 9:01
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    $\begingroup$ @Ramanujan Yes, in the question we already assume that $\ast$ is a binary operation on $A$. This means that $A$ is closed under $\ast$. $\endgroup$
    – Vincent
    Sep 17, 2021 at 9:03
  • $\begingroup$ This is great! And here I was starting to think these weaker conditions were hopeless... $\endgroup$
    – WillG
    Sep 17, 2021 at 15:02
  • $\begingroup$ I've asked a follow-up question here $\endgroup$
    – WillG
    Sep 17, 2021 at 15:57

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