Answer to Question #350789 in Abstract Algebra for Eliz

Question

2.3. Let G be a nonempty set closed under an associative product, which in addition satisfies:

(a) There exists an e ∈ G such that ae = a for all a ∈ G.

(b) Given a ∈ G, there exists an element y(a) ∈ G such that ay(a) = e.

Prove that G must be a group under this product.

Answer 1

Given $a\in G$. Since right inverse exists, there exists $y(a)\in G$ such that $ay(a)=e$. Then, $y(a)=y(a)e=y(a)(ay(a))=(y(a)a)y(a)$. Also, there exists $t\in G$ such that $y(a)t=e$. This implies that $(y(a)a)y(a)t=e$ then $(y(a)a)=e$. Hence $y(a)a=e$. So every right inverse is also a left inverse.

Now for any $a\in G$ we have $ea=(ay(a))a=a(y(a)a)=ae=a$ as $e$ is a right identity. Hence $e$ is left identity.