Suppose $G$ is a semigroup and it holds both left and right cancellation.

Question

Suppose $G$ is a semigroup and it holds both left and right cancellation. Also for each $a,b\in G$, $xa=b$ has solution in $G$. Prove G is a group.

I know this question looks very "old" style. First I think it's easy for me too since I already practice probelems like "semigroup, $xa=b$ and $ax=b$ have solution , then group" Or "finite semigroup, both left and right cancellation hold, then group ". But I still get stuck...
My try: First of course, for $xa=a$ , there exists $e_a$ such that $e_aa=a$. Now we need to proof $e_a$ is left identity for all elements. Then left inverse exists hence $G$ is a group. But the key is to proof $e_ab=b$ for all $b\in G$.

Answer 1

First note that once you found that $e_aa=a$ you can multiply both sides by $a$ from the right side and get $(ae_a)a=a^2$. By the cancellation rule you get $ae_a=a$. Now take any $b\in G$. You know there is an element $x\in G$ such that $xa=b$. Multiply both sides by $e_a$ on the left side and you will get:

$be_a=xae_a=x(ae_a)=xa=b$

Now you know that $be_a=b$. Multiply both sides by $b$ from the left side to get $b(e_ab)=b^2$ and by the cancellation rule you get that also $e_ab=b$.

Answer 2

Since $e_aa=a$, then for any $b$, $be_aa=ba$. Now by right cancellation $be_a=b$. Next, $be_ab=b^2$, and by left cancellation we get $e_ab=b$.