# $L/k$ finite Galois extension with solvable Galois group divisible by a prime $p$ . Does there exist field $k\subseteq F \subseteq L$ s.t. $p=[F:k]$?

Let $L/k$ be a finite Galois extension with solvable Galois group and let $p$ be a prime such that $p|Gal(L/k)$ , then does there exist an intermediate field $F$ ($k\subseteq F \subseteq L$ ) such that $p=[F:k]$ ? So I can see that it is equivalent to ask whether $G:=Gal(L/k)$ has a subgroup $H$ of index $p$ . But I can't proceed from here . Please help . Thanks in advance

Hints:

Let $\;L/\Bbb Q\;$ be such that Gal$(L/\Bbb Q)\cong A_4\;$ . And now take $\;p=2\;$ ...

$G = A_4$ is a solvable group of order $12$, but choosing $p = 2$, it has no subgroup of index $p$. Indeed, a subgroup of index $2$ would be normal in $A_4$, and it would have to contain all $3$-cycles (why?), of which there are $8$. This is a contradiction, therefore, $A_4$ has no subgroup of index $2$.

Now, all we need is to show that $\mathbf Q$ admits an extension with Galois group $A_4$. It is left as an exercise to the reader to verify that $X^4 + 4X^3 + 28$ is a polynomial with Galois group $A_4$ over $\mathbf Q$.