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 $[H':H] = p$?

  • $\begingroup$ Yes, let me correct that, thanks! $\endgroup$
    – Asvin
    Commented Dec 7, 2018 at 23:05
  • 1
    $\begingroup$ This is possible if $L/K$ is Galois. Note that your Galois theory translation only works if $L/K$ is separable. $\endgroup$ Commented Dec 7, 2018 at 23:24
  • 1
    $\begingroup$ If the extension is Galois, though, the answer is “yes”, since every group of order $p^n$ has normal subgroups of order $p$ $\endgroup$ Commented Dec 7, 2018 at 23:34
  • $\begingroup$ 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. $\endgroup$
    – C Monsour
    Commented Dec 8, 2018 at 3:20

1 Answer 1


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 $2$.

  • $\begingroup$ 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. $\endgroup$ Commented Dec 8, 2018 at 6:45

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .