Sufficiently large implies splitting for every subquotient

=Verbal statement

A Sufficiently large field (?) for a finite group is a Splitting field (?) for every 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 $H\leq K\leq G$ and $H$ is a normal subgroup of $K$, then $k$ is a splitting field for the quotient group $K/H$.

Related facts

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. For full proof, refer: Splitting field for every subgroup implies 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. Sufficiently large implies splitting
2. Exponent of subgroup divides exponent of group
3. Exponent of quotient group divides exponent of group

Proof

Given: A group $G$, a sufficiently large field $k$ for $G$. Subgroups $H\leq K\leq G$ 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 $K/H$.