Galois group of an irreducible , separable polynomial be abelian , then each of the roots of the polynomial generates the splitting field?

The proof goes as follows:

Since $\operatorname{Gal}(E/k)$ is abelian, any subgroup is normal, hence any intermediate field of $E/k$ is normal. In particular $k(a)/k$ is normal and by the definition of a normal extension, it follows that all roots of $f$ are contained in $k(a)$. Hence $E=k(a)$.

Since $f$ is irreducible, the Galois group is transitive on the roots.
In a transitive commutative permutation group, for any $x$, $Stab_G(x) = \{id\}$ :

Let $x$ be a root. if some element $g$ fixes $x$ and $y$ is another root, then there is an $h$ such that $h(x)=y$ and so $g(y) = gh(x) = hg(x) = h(x) = y$. Therefore if $g$ fixes $x$, it fixes every root, and so $g$ must be the identity.

This shows that $k(x) = E^{Stab_G(x)} = E^{\{id\}} = E$