Concerning

When not, are the centers of the division rings Galois over $k$

A finite dim'l $k$-algebra $A$ is split provided $\operatorname{End}_A(S) = k$ for every simple $A$-module $S$.

[This terminology is consistent with that used in other contexts -- $A$ is split just in case the reductive quotient of the "unit group" $A^\times$ is a split reductive algebraic group over $k$.]

Suppose that $k$ is perfect and that $A \otimes_k \ell$ is a split $\ell$-algebra for a finite, separable extension $\ell \supset k$. If $S$ is a simple $A$-module, the center $Z$ of the division $k$-algebra $\operatorname{End}_A(S)$ is a subfield of $\ell$ (indeed, since $k \subset \ell$ is separable and since $A \otimes_k \ell$ is split, $\operatorname{End}_A(S) \otimes_k \ell$ is a split semisimple $\ell$-algebra whose center is thus a split commutative etale $\ell$-algbra $\ell \times \cdots \times \ell$).

Apply this to $A = kG$ for a finite group $G$. By Torsten's comment (following Emerton's answer) we may suppose $k$ to be perfect. Then $A \otimes_k \ell$ is split for an Abelian extension $\ell$ of $k$ -- a suitable $\ell$ can be obtained by adjoining to $k$ enough roots of unity. It follows that $Z$ is always Galois over $k$ (for $A = kG$).

Concerning

are the centers of the division rings Galois over $k$

A finite dim'l $k$-algebra $A$ is split provided $\operatorname{End}_A(S) = k$ for every simple $A$-module $S$.

[This terminology is consistent with that used in other contexts -- $A$ is split just in case the reductive quotient of the "unit group" $A^\times$ is a split reductive algebraic group over $k$.]

Suppose that $k$ is perfect and that $A_\ell = A \otimes_k \ell$ is a split $\ell$-algebra for a finite, separable extension $\ell \supset k$. Then we have the following:

$(*)$ If $S$ is a simple $A$-module, the center $Z$ of the division $k$-algebra $\operatorname{End}_A(S)$ is a subfield of $\ell$.

To see this, I claim first that $\operatorname{End}_A(S) \otimes_k \ell$ is a split semisimple $\ell$-algebra. Indeed, since $k \subset \ell$ is separable, the $A_\ell$-module $S \otimes_k \ell = S_\ell$ is semisimple, say $S_\ell = \bigoplus_i S_i$ as $A_\ell$-module, where $S_i$ is the $T_i$-isotypic component of $S_\ell$ and where $T_i$ are distinct simple $A_\ell$-modules. Since by assumption $\operatorname{End}_{A_\ell}(T_i) = \ell$, we see that $$\operatorname{End}_A(S) \otimes_k \ell \simeq \operatorname{End}_{A_\ell}(S_\ell) = \prod_i \operatorname{End}_{A_\ell}(S_i)$$ is a product of full matrix algebras over $\ell$. Now observe that the center of a split semisimple $\ell$-algebra is a a split commutative etale $\ell$-algbra $\ell \times \cdots \times \ell$. Assertion $(*)$ now follows.

Apply this to $A = kG$ for a finite group $G$. By Torsten's comment (following Emerton's answer) we may suppose $k$ to be perfect. Then $A \otimes_k \ell$ is split for an Abelian extension $\ell$ of $k$ -- a suitable $\ell$ can be obtained by adjoining to $k$ enough roots of unity. It follows that $Z$ is always Galois over $k$ (for $A = kG$).

Concerning

When not, are the centers of the division rings Galois over $k$

A finite dim'l $k$-algebra $A$ is split provided $\operatorname{End}_k(S) = k$ for every simple $A$-module $S$.

[This terminology is consistent with that used in other contexts -- $A$ is split just in case the reductive quotient of the "unit group" $A^\times$ is a split reductive algebraic group over $k$.]

Suppose that $k$ is perfect and that $A \otimes_k \ell$ is split for a finite, separable extension $\ell \supset k$. If $S$ is a simple $A$-module, the center $Z$ of the division $k$-algebra $\operatorname{End}_A(S)$ is a subfield of $\ell$ (indeed, since $k \subset \ell$ is separable and since $A \otimes_k \ell$ is split, $\operatorname{End}_A(S) \otimes_k \ell$ is a split semisimple $\ell$-algebra whose center is thus a split commutative etale $\ell$-algbra $\ell \times \cdots \times \ell$).

Apply this to $A = kG$ for a finite group $G$. By Torsten's comment (following Emerton's answer) we may suppose $k$ to be perfect. Then $A \otimes_k \ell$ is split for an Abelian extension $\ell$ of $k$ -- a suitable $\ell$ can be obtained by adjoining to $k$ enough roots of unity. It follows that $Z$ is always Galois over $k$ (for $A = kG$).

Concerning

When not, are the centers of the division rings Galois over $k$

A finite dim'l $k$-algebra $A$ is split provided $\operatorname{End}_A(S) = k$ for every simple $A$-module $S$.

[This terminology is consistent with that used in other contexts -- $A$ is split just in case the reductive quotient of the "unit group" $A^\times$ is a split reductive algebraic group over $k$.]

Suppose that $k$ is perfect and that $A \otimes_k \ell$ is a split $\ell$-algebra for a finite, separable extension $\ell \supset k$. If $S$ is a simple $A$-module, the center $Z$ of the division $k$-algebra $\operatorname{End}_A(S)$ is a subfield of $\ell$ (indeed, since $k \subset \ell$ is separable and since $A \otimes_k \ell$ is split, $\operatorname{End}_A(S) \otimes_k \ell$ is a split semisimple $\ell$-algebra whose center is thus a split commutative etale $\ell$-algbra $\ell \times \cdots \times \ell$).

Apply this to $A = kG$ for a finite group $G$. By Torsten's comment (following Emerton's answer) we may suppose $k$ to be perfect. Then $A \otimes_k \ell$ is split for an Abelian extension $\ell$ of $k$ -- a suitable $\ell$ can be obtained by adjoining to $k$ enough roots of unity. It follows that $Z$ is always Galois over $k$ (for $A = kG$).