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Related Link: Right identity and Right inverse implies a group

Reference: Fraleigh p. 49 Question 4.38 in A First Course in Abstract Algebra

I will present my proof (distinct from those in the link) for critique and then ask my question. $G$ is a set and $ \times $ is an associative binary operation. Suppose that there exists a $e \in G$ such that, for all $a \in G$, $ea = a$ and $a^{-1}a = e$ for some $a^{-1} \in G$. Show that for the same $e$, $ae = a$ and $aa^{-1} = e$.

$a^{-1}(aa^{-1})a = (a^{-1}a)(a^{-1}a) = ee = e = a^{-1}a = a^{-1}(a^{-1}a)a$
Since $a^{-1} \in G$, it has a left inverse; apply it to both ends, and we have $(aa^{-1})a = (a^{-1}a)a$.
As a result, $ae$ = $a(a^{-1}a) = (aa^{-1})a = (a^{-1}a)a = ea$.

For the right inverse, start with $aa^{-1} = a(a^{-1}a)a^{-1} = (aa^{-1})(aa^{-1})$.
Since $\times$ is a binary operation, $aa^{-1} \in G$ and has a left inverse; apply it to both ends, and we have $e = aa^{-1}$.

In second comment following the question in the link, Mr. Derek Holt pointed out that the requester did not word his/her question correctly. Specifically, the identity in the second axiom is not well-defined.

Let $(G, *)$ be a semi-group. Suppose
1. $ \exists e \in G$ such that $\forall a \in G,\ ae = a$;
2. $\forall a \in G, \exists a^{-1} \in G$ such that $aa^{-1} = e$.
How can we prove that $(G,*)$ is a group?

This formulation makes the same technical error as many textbooks. The $e$ in your second axiom is not well-defined. "But obviously it's intended to be the same $e$ as in the first axiom" you reply. But the first axiom does not necessarily specify a unique element $e$. So should we interpret the second axiom as meaning "for some $e$ as in 1" or "for all $e$ as in 1"? – Derek Holt Sep 17 '11 at 15:31

Was he saying that if, in axiom 1, we have $ae_1 = a, ae_2 = a$, but $e_1 \neq e_2$,
when we get to axiom 2, do we have $aa^{-1} = e_1, aa^{-1} = e_2$, or two different inverses so that $aa_1^{-1} = e_1, aa^{-1}_2 = e_2$? I think my wording eliminated the ambiguity. It does not imply that $e$ is unique, but if $e$ is a left identity and produces left inverses, then it is also a right identity and produces right inverses. I tried really hard on this one; please kindly point out my mistakes.

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Your proof seems correct to me, and it also seems that you understood what is the problem with the axioms. The usual proof works just fine in the following case:

Let $(G, *)$ be a semi-group. Suppose
(1) $\exists e \in G$ such that $\forall a \in G,\ ae = a$;
(2) $\forall a \in G, \exists a^{-1} \in G$ such that for all $e\in G$ satisfying 1, $aa^{-1} = e$.

It is then obvious, in this case, that the element $e$ in (1) is unique: Indeed, since $G$ is nonempty (by (1)), let $g\in G$ be arbitrary. Then, if $e_1$ and $e_2$ satisfy (1), we have $e_1=gg^{-1}=e_2$.

The next case is more interesting:

Let $(G, *)$ be a semi-group. Suppose
(1) $\exists e \in G$ such that $\forall a \in G,\ ae = a$;
(2) $\forall e\in G$ satisfying (1) and $\forall a \in G, \exists a_e^{-1} \in G$ such that $aa_e^{-1} = e$.

The problem is to actually prove the uniqueness of the unit. Let's prove it:

Let $e$ and $f$ satisfy (1). Then $$f=ee_f^{-1}=(ee)e_f^{-1}=e(ee_f^{-1})=ef=e$$ Therefore, $e=f$, and we're actually in the first (and simpler) case.

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    $\begingroup$ Thanks for commenting on Mr. Holt's response. I should have kept going and verify that $e$ is unique. $\endgroup$ – Andy Tam Oct 26 '13 at 23:41
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Here is a short yet unintuitive proof (sorry):

Notice that $$((b^{-1})^{-1} \ b^{-1}) \ (b \ b^{-1}) = e \ (b \ b^{-1}) = b \ b^{-1}$$ and also because of associativity $$((b^{-1})^{-1} \ b^{-1}) \ (b \ b^{-1})$$$$=(b^{-1})^{-1} \ ((b^{-1}) \ b) \ b^{-1}$$$$ = (b^{-1})^{-1} \ e \ b^{-1} = (b^{-1})^{-1} \ (e \ b^{-1}) = (b^{-1})^{-1} \ b^{-1} = e$$Therefore $b \ b^{-1} = e$, in addition to $b^{-1}b=e$. (In other words, left inverse is also a right inverse).

Also notice that $b \ e = b \ (b^{-1} \ b) = (b \ b^{-1}) \ b = e \ b = b$. Therefore left identity is also the right identity.

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