Regarding Enderton’s group axioms

Solution 1:

It's not entirely trivial. The two latter axioms include the existence of a right and left identity for each element $x$, if we take $y=x$. Then they read as

$$\forall x\exists e_l:e_lx=x,\\ \forall x\exists e_r:xe_r=x.$$

We can now prove that both are independent of $x$ in the sense that if $e_r$ is a right identity for $x$, it is a right identity for all group elements. Because for all $x$ and $y$ there is a $z$ such that $y=zx$. But then $ye_r=(zx)e_r=z(xe_r)=zx=y$. A similar argument works for the left identity.

Now we can show that every right identity is a left identity and vice versa, and that they are unique. The first statement is true because $e_l=e_le_r=e_r$ for any right identity $e_r$ and left identity $e_l$. For uniqueness consider two identities $e$ and $\tilde e$. Then $\tilde e=e\tilde e=e$, so the identity is unique.