Splitting field for every subgroup implies sufficiently large

Statement
Suppose $$G$$ is a finite group and $$k$$ is a field of characteristic not dividing the order of $$G$$ such that $$k$$ is a fact about::splitting field for every subgroup of $$G$$. Then, $$k$$ is a fact about::sufficiently large field of $$G$$: it contains all the primitive $$m^{th}$$ roots of unity where $$m$$ is the exponent of $$G$$.

Related facts
This statement is one of the directions in the equivalence of definitions of sufficiently large field. A converse statement is:

Sufficiently large implies splitting for every subquotient.

Other related facts

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

Proof
The key idea behind the proof is to look at cyclic subgroups of $$G$$. Specifically, the splitting field for a cyclic subgroup of order $$d$$ is obtained by adjoining the primitive $$d^{th}$$ roots of unity. Thus, if $$k$$ is a splitting field for all subgroups of $$G$$, it contains the primitive $$d^{th}$$ roots of unity for all $$d$$ that occur as orders of elements. These generate the primitive $$m^{th}$$ roots of unity, because $$m$$ is the least common multiple of the $$d$$s, and hence $$k$$ contains primitive $$m^{th}$$ roots of unity.