Problem about Galois Extension.

Question

$Gal(\Bbb K/\Bbb F)\cong S$. Where $\Bbb K$ is extension of $\Bbb F$. And $S= G_1\times G_2\times\cdots\times G_K$ is a solvable group. Where each $G_i$ are groups of prime power order and $o(S)= n.$

How will we show that for any $d$ such that $d|n$, $\exists $ field $\Bbb L$ so that $\Bbb F\subset \Bbb L\subset\Bbb K$ and $(\Bbb L/\Bbb F)=d$

And What will be the counterexample to above for S solvable?

Answer 1

fact: Any group with order $p^n$ has normal subgroup of order $p^i$ for all $i\in\{0,1,...,n\}$

You can prove above fact by using induction and by using above fact you can conclude that your group $S$ has normal subgroup for each $d|n$. (since direct product of normal subgroups of $G_i$ is normal in $S$)

Now, Let $N$ be normal subgroup of S with index $d$. Then there is an corresponding field $L$ to $N$ and $Gal(L/F)\cong G/N \implies (L/F)=|G/N|=d$.

notes: I hope you are familiar with the fact that every subgroup of $Gal(K/F)$ corresponds to an intermediate field.