If $\forall a,b \in G, \exists x \in G: a * x = b$ and $\forall a,b \in G, \exists x \in G: x * a = b$ then $G$ is a group

Question

I'm having some trouble with the following exercise:

Let $G$ be a non-empty set and let $*:G\times G \to G$ such that:

1. $*$ is associative
2. $\forall a,b \in G, \exists x \in G: a * x = b$
3. $\forall a,b \in G, \exists x \in G: x * a = b$

Prove that $(G,*)$ is a group

We already know that $*$ is associative but I'm having some trouble proving that there exists an identity element. My plan is the following:

1. For any $a \in G$, show that if $a * x = a$ and $x'*a = a$, then $x = x'$
2. for any $a,b \in G$, if $a * x_a = a$ and $b * x_b = b$, then $x_a = x_b$

But I'm not being able to prove 1. How can this be done?

Answer 1

Let $g$ be any element of $G$ and take $a = b = g$ in assumption 2.
It follows that there exists $e \in G$ such that $g * e = g$.

It follows that $h*g*e = h*g$ for any $h\in G$.
By assumption 3, any element of $G$ can be written as $h*g$ for some $h \in G$.
Thus we have $x*e = x$ for any $x \in G$.

This shows that there exists a right identity $e$ of $G$.
The same proof, with left and right switched, shows that there exists a left identity $e'$ of $G$.

We then have $e' = e'e = e$, which means that $e$ is an identity of $G$.

It remains to show that every element has an inverse.
For any $g \in G$, assumptions 2 and 3 say that there exist $h$ and $h'$ such that $g*h = e = h'*g$.
As $h' = h'*e = h'*g*h = e*h = h$, we see that $h$ is an inverse of $g$.