Exercise 26c.

Let $S$ be a semigroup with an element $e$ such that $ea = a$ for all $a \in S$ and such that for every $a \in S$ there exists $a^{-1} \in S$ for which $a^{-1}a = e$. Prove that $S$ is a group.

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Proof

We need to show that the usual group axioms are true for $S$, that is:

1. Closure
2. Associativity
3. Existence of the identity element
4. Existence of an inverse for all elements of $S$

Closure and associativity we get for free, since $S$ is a semigroup, and semigroups by definition have these properties.

We are also given the existence of the left identity and the left inverse, so what we really need to show is that there exists the right identity element for all elements in $S$ and the right inverse of all elements of $S$.

The updated goals are:
(i) Existence of the right identity element
(ii) Existence of the right inverse element

Let's prove (ii) first. Also, we will be using the associative property freely, without explicitly mentioning that we used it later on.

Proof of (ii):

We need to show that for all $a \in S$ there exists $a^{-1}$ such that $aa^{-1}=e$. Let $a \in S$. Then, by assumption we know that there exists $a^{-1} \in S$ that is the left inverse of $a$. Now, since $a^{-1} \in S$, using the same assumption, we get that there exists the inverse of this element in $S$, namely, there exists $(a^{-1})^{-1} \in S \colon (a^{-1})^{-1}a^{-1}=e$, and let's define $b \colon = (a^{-1})^{-1}$. Now we can write the last equality compactly as $ba^{-1}=e$. Consider now the equation $a^{-1}a = e$ that we are given, and let's multiply it with $b$ from the left, we get the following: $$\begin{align} a^{-1}a &= e &\iff\\ b(a^{-1}a) &= be &\iff \\ (ba^{-1})a &= be &\iff \\ ea &= be &\iff \\ a &= be \end{align}$$ Now we use the same reasoning as before and deduce that, since $b \in S$, that means that there is $b^{-1} \in S \colon b^{-1}b = e$. Combining this with what we have derived so far we get $$ba^{-1} = e = b^{-1}b \implies ba^{-1}=b^{-1}b$$ Let's manipulate this last equality using everything we know so far: $$\begin{align} ba^{-1} &= b^{-1}b &\iff \\ b(ea^{-1}) &= e &\iff \\ (be)a^{-1} &= e &\iff \\ aa^{-1} &= e \end{align}$$

Thus we obtain the desired existence of the right inverse, (ii) is proven.

Proof of (i):

For this part, we need to show that there exists $e \in S: ae=a$ for all $a \in S$. Consider the product $ae$, since $S$ is a semigroup, we have that $ae \in S$. Let's rewrite $ae$ using the given assumption about the existence of the left identity element, and let's proceed from there using the results we obtained in (ii): $$\begin{align} ae &= a(a^{-1}a) &\iff \\ ae &= (aa^{-1})a &\iff \\ ae &= ea \end{align}$$

The last equality being what we wanted to show, thus (i) is proven.

We proved both (i) and (ii), $S$ is indeed a group.

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Additional thoughts

As you surely noted, I abused and overloaded $e$ and $a^{-1}$, the proper way to do it was to show that there exist $e'$ and $a'$ with the desired properties, and then show that they are equal to $e$ and $a^{-1}$, respectively. If I find time I'll fix this.

It feels like there should be an easier proof of this, but after struggling for hours, I was happy to settle for this proof. Initially, I failed to prove this and skipped it. When I returned to it, it took me probably 3 hours to solve it, split across multiple sessions. Not sure why I struggled so much. In hindsight, the idea is straight forward (as always): just work with what you are given: you have the left inverse, what you need to realize is that you also have the left inverse of the left inverse, and in turn, that that too, has a left inverse. The rest is the usual symbol manipulation.