Assume that $L/K$ is an extension of fields and $[L:K]=n$, with $n$ composite. Assume that $p\mid n$, can we always produce a subextension of degree $p$ and if not under what conditions can it be done? I would guess this is very false, but I couldn't come up with any trivial counterexamples.
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We know that equations of degree 2 and 3 have solutions in radicals, but most equations of degree 6 (or any degree 5 or greater) don't. So let $f$ be a polynomial over $K$ of degree 6 not solvable in radicals, let $\alpha$ be a root of $f$ in some extension, let $L=K(\alpha)$. If there were an intermediate field $E$ of degree 2 or 3, then you could express $\alpha$ in radicals over $E$, and those radicals in radicals over $K$, contradicting the assumption that you couldn't solve $f$ in radicals over $K$.
You are very right when you write "I would guess this is very false": here is a precise statement.
An analogous problem
Suppose $L/K$ is Galois with group $G$. Then subextensions of degree $m$ over $K$ correspond to normal subgroups of $G$ of index $m$. Given an abelian group you can always find subgroups of order $p$, where $p$ is a prime divisor of $\vert G \vert =n$. So to find a counterexample, you should (with $L/K$ Galois) start with non-abelian Galois groups. I'm pretty sure there exists a group of order $n$ which doesn't have a normal subgroup of index $m$ for all $m\vert n$. (Note that if $m$ is prime, such a (not necesarily normal) subgroup exists).