Can a degree $p^n$ field extension always be factored in a sequence of prime extensions?

Suppose $$L/K$$ is a field extension of degree $$p^n$$ for some prime $$p$$ (if necessary, assume the characteristic of $$K$$ is not $$p$$).

Then, is it always possible to find a sequence of extensions $$K = K_0 \subset K_1 \subset K_2 \dots \subset K_n = L$$ such that $$[K_r:K_{r-1}] = p$$?

Using Galois theory, this problem translates into the following:

Suppose $$G$$ is a finite group with a subgroup $$H$$ such that $$[G:H] = p^n$$. Is it always possible to find a subgroup $$G \supset H' \supset H$$ so that $$[H':H] = p$$?

• Yes, let me correct that, thanks! Commented Dec 7, 2018 at 23:05
• This is possible if $L/K$ is Galois. Note that your Galois theory translation only works if $L/K$ is separable. Commented Dec 7, 2018 at 23:24
• If the extension is Galois, though, the answer is “yes”, since every group of order $p^n$ has normal subgroups of order $p$ Commented Dec 7, 2018 at 23:34
• If the extension is purely inseparable the answer is "Yes" since we have for every $x\in L$ that $x^{p^j}\in K$ for some $j$. Take the $x\in L\setminus K$ and take $k$ such that $x^{p^k}\notin K$ but $x^{p^{k+1}}\in K$. Then $K(x^{p^k})$ has dimension $p$ over $K$ and now induct. Commented Dec 8, 2018 at 3:20

No, this is not always possible. For instance, consider $$G=A_4$$. Then $$G$$ has a subgroup of index $$2^2$$ (any subgroup generated by a $$3$$-cycle) but has no subgroup of index $$2$$.
• Trying to thwart an eventual follow-up question with the edit :-) The key really is the lack of intermediate groups of prescribed order. For example $G=S_4$ has a subgroup of order $12$, but no intermediate subgroups of order $12$ between $G$ and a point stabilizer $H=S_3$ of index four. Commented Dec 8, 2018 at 6:45