Right identity and Right inverse in a semigroup imply it is a group Let $(G, *)$ be a semigroup. Suppose

*

*$ \exists e \in G$ such that $\forall a \in G,\  ae = a$;

*$\forall a \in G, \exists a^{-1} \in G$ such that $aa^{-1} = e$. 
How can we prove that $(G,*)$ is a group?
 A: It is conceptually very simple that a right inverse is also a left inverse (when there is also a right identity). It follows from the axioms above in two steps:
1) Any element $a$ with the property $aa = a$ [i.e. idempotent] must be equal to the identity $e$ in the axioms, since in that case:
$$a = ae = a(aa^{-1}) = (aa)a^{-1} = aa^{-1} = e$$
This already proves the uniqueness of the [right] identity, since any identity by definition has the property of being idempotent.
2) By the axioms, for every element $a$ there is at least one right inverse element $a^{-1}$ such that $aa^{-1}=e$. Now we form the product of the same two elements in reverse order, namely $a^{-1}a$, to see if that product also equals the identity. If so, this right inverse is also a left inverse.  We only need to show that $a^{-1}a$ is idempotent, and then its equality to $e$ follows from step 1:
$$[a^{-1}a][ a^{-1}a] = a^{-1}(a a^{-1})a = a^{-1}ea = a^{-1}a $$
3) It is now clear that the right identity is also a left identity. For any $a$:
$$ea = (aa^{-1})a = a(a^{-1}a) = ae = a$$
4) To show the uniqueness of the inverse:
Given any elements $a$ and $b$ such that $ab=e$, then
$$b = eb = a^{-1}ab = a^{-1}e = a^{-1}$$
Here, as above, the symbol $a^{-1}$ was first used to denote a representative right inverse of the element $a$. This inverse is now seen to be unique. Therefore, the symbol now signifies an operation of "inversion" which constitutes a single-valued function on the elements of the set.
See Richard A. Dean, “Elements of Abstract Algebra” (Wiley, 1967), pp 30-31.
A: I assume that (a) should read $\exists e\in G$ such that $ae=a$, $\forall a\in G$. For each $a \in G$ we have
$$\begin{align*}
(a^{-1})^{-1}a^{-1} &= e[(a^{-1})^{-1}a^{-1}]\\
&= (aa^{-1})[(a^{-1})^{-1}a^{-1}]\\
&= [(aa^{-1})(a^{-1})^{-1}]a^{-1}\\
&= (a[a^{-1}(a^{-1})^{-1}])a^{-1}\\
&= (ae)a^{-1}\\
&= aa^{-1}.
\end{align*}$$
Multiplying $(a^{-1})^{-1}a^{-1} = aa^{-1}$ on the right by $(a^{-1})^{-1}$ yields $$\begin{align*}
(a^{-1})^{-1} &= (a^{-1})^{-1}e\\
&= (a^{-1})^{-1}[a^{-1}(a^{-1})^{-1}]\\
&= [(a^{-1})^{-1}a^{-1}](a^{-1})^{-1}\\
&= (aa^{-1})(a^{-1})^{-1}\\
&= a[a^{-1}(a^{-1})^{-1}]\\
&= ae\\
&= a,
\end{align*}$$
so $a^{-1}a=e$ for all $a \in G$. 

Added: The foregoing obviously assumes that $e$ is a left identity, which was not given, and somehow none of us caught it at the time. Here is a corrected argument. For each $a\in G$ we have $$a^{-1}=a^{-1}e=a^{-1}(aa^{-1})=(a^{-1}a)a^{-1}\;,$$ so $$e=a^{-1}(a^{-1})^{-1}=\left((a^{-1}a)a^{-1}\right)(a^{-1})^{-1}=(a^{-1}a)\left(a^{-1}(a^{-1})^{-1}\right)=(a^{-1}a)e=a^{-1}a\;.$$

In other words, $a^{-1}$ is both a left as well as a right inverse for $a$. It follows that 
$$ea = (aa^{-1})a = a(a^{-1}a) = ae = a\;,$$ 
so $e$ is a left as well as a right identity for $G$. Now you can use the usual arguments to show that the identity and inverses are unique. (For example, if $e'$ were another identity, we’d have $e = ee' = e'$, because $e$ is a left identity and $e'$ is a right identity.)
A: This is stated with left identity and left inverse as Proposition 20.4 in the book Spindler: Abstract Algebra with Applications. Let me copy here the proof from this book (it should be easy for you to change it for the right instead of left):

Let $x\in G$  be arbitrary. We want to show that the left inverse $x^{-1}$ is in fact also a right inverse. Let $y:=xx^{-1}$. Then
  $$yy=(xx^{-1})(xx^{-1})=x(x^{-1}x)x^{-1}=x(ex^{-1})=xx^{-1}=y.$$
  Hence
  $$e=y^{-1}y=y^{-1}(yy)=(y^{-1}y)y=ey=y=xx^{-1},$$
  i.e. $xx^{-1}=e$ which was what we wanted to show.
Now we prove that the left-neutral element $e$ is also a right-neutral element. Let $x\in G$ be arbitrary; we want to establish that $xe=x$. Now
  $$xe=x(x^{-1}x)=(xx^{-1})x=ex=x. $$ 

I googled a little and found out that several authors take this in fact as a definition of group, here are some of the first hits from google books when searching for "left inverse" "left identity" group:


*

*Robinson: A course in the theory of groups, p.2

*Gelbaum, Olmsted: Theorems and counterexamples in mathematics, p.1

*Sharma: Group Theory, p.14
A: Most proofs I've seen contains many ingenious equations. The motivation behind these equations is not very clear. Here I present a more intuitive proof.
For any element $a$ in $G$, we can map it to a function $f_a:G \to G$, simply define $f_a(x) = xa$. We can see that every right identity element $e$ is mapped to the identity function $id$,
$$f_e=id,$$
and the operation on $G$ amounts to function composition:
$$ f_{ab} = f_b \circ f_a, $$
and the existence of right inverses means for any $a$ there is an $a^{-1}$ such that
$$f_{a^{-1}} \circ f_a = id.$$
That will imply that every function $f_a$ is injective, otherwise the composition of it, which is $id$, can't be injective either. Actually, it's bijective, but we don't have to use that strong conclusion here.
We want to show $a^{-1}a=e$. Because $f_{a^{-1}a} = f_a \circ f_{a^{-1}}$, we wonder whether $f_a \circ f_{a^{-1}} =id$. Intuitively that should be true since $f_{a^{-1}} \circ f_a = id$ and both $f_a$ and $f_{a^{-1}}$ are bijective. For a proof that only assumes that one of them is injective, see Lemma below.
Now that $f_{a^{-1}a} = f_e$, to prove $a^{-1}$ is a left inverse, we only need to show that for any $c \in G$ and $c' \in G$, $f_c = f_{c'}$ implies $c=c'$. We try to calculate $f_c(e)$:
$$f_c(e) = ec = (cc^{-1})c = c(c^{-1}c)=f_{c^{-1}c}(c)=id(c)=c.$$
For the same reason $f_{c'}(e) = c'$, so $c=c'$. Note here $ec=c$, so it has already proved every right identity is a left identity.
The uniqueness of the identity element is induced by the uniqueness of the identity function, for $f_e=f_{e'}$ implies $e=e'$, and also the uniqueness of inverse elements is induced by the uniqueness of inverse functions.
So we've proved $G$ is a group.

Lemma.  Let $X$ and $Y$ be two arbitrary sets. Let $f$ be a function $f: X \to Y$ and $g$ be another function $g: Y \to X$.
  Suppose $g$ is injective, and $g \circ f = id$, then $f \circ g = id$.
Proof. Let $y \in Y$, we want to show that $(f \circ g)(y) = y$. Let $y' = (f \circ g)(y)$. Apply $g$ to both side, $$g(y')=(g \circ f \circ g)(y) = (id \circ g)(y) = g(y).$$ So $y'=y$.

A: Martin's answer, but using a notation that may be easier to follow.
Let $G = (S,\cdot)$ be a semi-group such that it has a right identity $1_{r}$, and that each of its elements has a right inverse; i.e.,


*

*$\forall x \in G: x1_{r} = x$

*$\forall x \in G: \exists x^{-1}_{r}: xx^{-1}_{r} = 1_{r}$
We show that $G$ is in fact a group:
Let $y = x^{-1}_{r}x$.
$$yy = (x^{-1}_{r}x)(x_{r}^{-1}x) = x^{-1}_{r}(xx^{-1}_{r})x = x^{-1}_{r}(1_{r})x = (x^{-1}_{r}1_{r})x = x^{-1}_{r}x$$
Therefore, $y$ is idempotent, and we can use that to show that $x^{-1}$ is actually a two-sided inverse:
$$1_{r} = yy^{-1}_{r} = y(yy^{-1}_{r}) = y1_{r} = y = x^{-1}_{r}x$$
Then, that $1_{r}$ is also two-sided:
$$1_{r}x = (xx^{-1}_{r})x = x(x^{-1}_{r}x) = x1_{r} = x$$
This suffices to show that $G$ is a group.
Bonus
This allows us to quickly demonstrate that any semi-group in which $ax = b$ and $ya = b$ are uniquely determined is also a group:
Let $G = (S, \cdot)$ be a semi-group with such proprieties, and $a$ be an element of that group. We know $ax = a$ must have a unique solution, lets call it $e$. Then for any $b \in G$:
$$be = yae = ya = b$$
Where the first equality is also due to $ya = b$ being uniquely determined. We now have everything we need: $e$ is a right identity, and each element of $G$ has an inverse.
A: Thinking of the elements of a semigroup as functions makes the proof of this fact much easier to understand.
For $a \in G$, we can define a function $\sigma_a : x \mapsto x * a$. By the existence of right inverses, we have:
\begin{align*}
    \sigma_{a^{-1}} &\circ \sigma_a = \operatorname{id}_G \\
    \sigma_{(a^{-1})^{-1}} &\circ \sigma_{a^{-1}} = \operatorname{id}_G
\end{align*}
This shows that $\sigma_{a^{-1}}$ has both left and right inverses, so both $\sigma_{a^{-1}}$ and its inverse $\sigma_{a}$ are bijective.
Having showed that $\sigma_a$ is bijective, it is very easy to deduce that $e$ is also a left identity. There exists an $x \in G$ such that $y * a = a$. Canceling the $a$'s by multiplying with $a^{-1}$ from right, we conclude that $x = e$, so $e * a = a$ for all $a \in G$.
Moreover, showing that right inverses are also left inverse is easy as well. There exists a $y \in G$ such that $x * a = e$. Again by multiplying with $a^{-1}$ from right, we get $x = a^{-1}$, so $a^{-1} a = e$ for all $a \in G$.
In summary, $e$ is actually two sided identity, and $a^{-1}$ is actually two sided inverse of $a$.
