# Do these very weak axioms define a group? (Every element has left identity + left inverse or right identity + right inverse) [duplicate]

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.

## 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 $$b = e$$.

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