# 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 $H \le K \le G$ 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

## 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 $K/H$.