Sufficiently large implies splitting for every subquotient

Verbal statement
A fact about::sufficiently large field for a finite group is a fact about::splitting field for every fact about::subquotient of the group. In particular, it is a splitting field for every subgroup as well as for every quotient group.

Statement with symbols
Suppose $$G$$ is a finite group and $$k$$ is a sufficiently large field for $$G$$. Then, if $$H \le K \le G$$ and $$H$$ is a normal subgroup of $$K$$, then $$k$$ is a splitting field for the quotient group $$K/H$$.

Converse
The converse of the statement is true. In fact, if a field is a splitting field for every subgroup of a finite group, then it is sufficiently large.

Other related facts

 * Splitting not implies sufficiently large
 * Splitting field for a group not implies splitting field for every subgroup
 * Splitting field for a group implies splitting field for every quotient

Facts used

 * 1) uses::Sufficiently large implies splitting
 * 2) uses::Exponent of subgroup divides exponent of group
 * 3) uses::Exponent of quotient group divides exponent of group

Proof
Given: A group $$G$$, a sufficiently large field $$k$$ for $$G$$. Subgroups $$H \le K \le G$$ with $$H$$ normal in $$K$$.

To prove: $$k$$ is a splitting field for $$K/H$$.

Proof: By facts (2) and (3), the exponent of $$K/H$$ divides the exponent of $$G$$. By the definition of sufficiently large, the polynomial $$x^d - 1$$ splits completely into distinct linear factors over $$k$$, where $$d$$ is the exponent of $$G$$. Let $$d'$$ be the exponent of $$K/H$$. Then the polynomial $$x^{d'} - 1$$, being a factor of $$x^d - 1$$, also splits completely into linear factors over $$k$$. (A simpler way of saying this is that a field containing primitive $$d^{th}$$ roots of unity also contains primitive $$d'^{th}$$ roots of unity for $$d' | d$$).

Thus, $$k$$ is sufficiently large for $$K/H$$. Hence, by fact (1), $$k$$ is a splitting field for $$K/H$$.