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 Splitting field (?) for every subgroup of $G$. Then, $k$ is a 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.