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Suppose $K/F$ and $L/F$ are extensions of $F$ (contained in some common field) of degree $p$, where $p$ is prime. Standard arguments show that $[KL:F]$ must be in $\{p,2p,\ldots,p^2\}$. But are all these multiples attained for every $p$? I.e. for all primes $p$ and for all integers $1\leq r\leq p$, do there exist $K$, $L$, and $F$ as above with $[KL:F]=rp$?

Stating this in terms of Galois theory (using the fact that every prime-degree extension is separable), does there exist a group $G$ with index-$p$ subgroups $H_1,H_2$ of $G$ such that $|G:H_1\cap H_2|=rp$?

Edit (I'm not sure whether this should be in an answer or not): It's straightforward to show that if $KL/F$ is Galois, so that $G$ is its Galois group, then either $r=p$ or $(r,p-1)\neq 1$:

Assume $r\neq p$. By Sylow's theorems, $n_p:=\text{number of Sylow }p\text{-groups}$ is congruent to $1$ mod $p$ and divides $r$. But $r<p$, so $n_p=1$. Thus, there's a unique (and therefore normal) Sylow $p$-subgroup $P\cong Z_p$. So, $G= P\rtimes H_1=P\rtimes H_2$. If $(r,p-1)=1$, then there's no nontrivial homomorphism from either $H_1$ or $H_2$ to $\operatorname{Aut}(Z_p)\cong Z_{p-1}$, and so the semidirect products are in fact direct. But then (since $r\neq p$) $H_1$ and $H_2$ are normal Hall subgroups of the same order and therefore equal, which can only happen if $r=1$. Thus, $r=p$ or $(r,p-1)\neq 1$.

So, to solve the general problem when $(r,p-1)=1$, we can't make the simplifying assumption that $H_1$ and $H_2$ have trivial intersection, i.e. that $KL/F$ is Galois.

Avi Steiner
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  • Have you done a case-by-case check for small values of $p$? – YCor Jul 03 '13 at 09:09
  • When p=2 and 3, they all work. I haven't checked higher yet. – Avi Steiner Jul 03 '13 at 14:03
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    yes, more generally the cases $r=1,2,p-1,p$ always work. (for $r=1,p$, take $G$ to be cyclic of order $p$ or the direct product of two such groups; for $r=p-1$ take $G$ be be the symmetric group $S_p$; for $r=2$ take $G$ to be dihedral of order $2p$). More generally if $r$ divides $p-1$ then $r$ is OK using a similar semidirect product. The first case to look at is thus $(p,r)=(5,3)$. – YCor Jul 03 '13 at 20:51
  • @YvesCornulier See my edit for some additional progress. – Avi Steiner Jul 04 '13 at 02:37
  • I was going to write up an answer for the divisors of $p-1$, using cyclotomic fields, but I believe that that’s what @Yves Cornulier had in mind. So I agree that $p=5$, $r=3$ is the first crucial case. – Lubin Jul 04 '13 at 02:49

1 Answers1

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I did a computer calculation which I believe shows that there is no such example with $p=5$, $r=3$. If we had such an example $G$ then consider the transitive permutation representation of $G$ on the 15 cosets of $H_1 \cap H_2$. This would be a transitive subgroup of $S_{15}$, and the subgroups $H_1$ and $H_2$ would be the stabilizers of distinct systems of imprimitivity of 5 blocks of size 3.

I let $K = S_3 \wr S_5$ be the full stabilizer of such a system in $S_{15}$. There are 1401400 such systems, all conjugate, and I computed the intersection of $K$ with each of its 1401400 conjugates in $S_{15}$, and found that none of these interseections (except for $K$ itself) was transitive. So I believe that rules out such an example.

Derek Holt
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    $G$ acts on the disjoint union of $G/H_1$ and $G/H_2$ and thus (after modding out the irrelevant kernel) is a subgroup of $S_5\times S_5$ (or even $S_5$ if $H_1$ and $H_2$ are conjugate), could this give rise to an easier verification? – YCor Jul 04 '13 at 07:40
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    Yes, good idea! I checked all 181 (conjugacy classes of) subgroups of $S_5 \times S_5$ of order divisible by 15, and quickly found that there are no examples. – Derek Holt Jul 04 '13 at 08:53
  • @DerekHolt This definitely answers my question, but I'm going to wait a bit before I accept it, on the off chance that someone comes along with a more general solution. – Avi Steiner Jul 04 '13 at 16:22
  • A very pleasing, interesting, and surprising result. – Lubin Jul 06 '13 at 03:09