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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

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Hints:

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

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$ 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 $.

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