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

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*$ \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?

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:


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

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

*Sharma: Group Theory, p.14
