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.


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.

  • 1
    $\begingroup$ 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

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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