Are two subgroups of prime index with a large intersection necessarily conjugate?

Question

Assume that $G$ is a finite group, $p$ a prime number, and $H_1,H_2\le G$ such that $[G:H_1]=p=[G:H_2]$. Assume further that $[G:H_1\cap H_2]<p^2$. Does it follow that $H_2=gH_1g^{-1}$ for some $g\in G$?

As an alternative goal: if the main "guess" fails, will it still follow that $H_1$ and $H_2$ share the same core in $G$ (in other words: $\bigcap_{g\in G}gH_1g^{-1}=\bigcap_{g\in G}gH_2g^{-1}$).

Either a proof or a counterexample is welcome!

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My preliminary thoughts:

• It is impossible to have $H_1\unlhd G$, as then we would have $G=H_1H_2$, and the parallelogram law would imply $[G:H_1]=[H_2:H_1\cap H_2]$. Obviously the same holds for $H_2$ as well, so the two subgroups are self-normalizing in $G$.
• The examples I can think of support this. For example, $H_1$ and $H_2$ can be point stabilizers of (the natural action of) $S_p$, and as the action of $G$ is transitive, they are always conjugate. The intersection $H_1\cap H_2$ has index $p(p-1)<p^2$.
• The exact same thing happens, when $G$ is the holomorph $C_p\rtimes C_{p-1}$ ($C_{p-1}$ identified with $Aut(C_p)$).
• Another example is the five Sylow $2$-subgroups $H_1,H_2,\ldots, H_5$ of the dihedral group $D_{20}$ of symmetries of a regular $20$-gon. The Sylow $2$s are the groups of symmetries of the five embedded squares (see here for a picture). The four rotations by multiples of 90 degrees are in all the $H_i,i=1,2,3,4,5$, so the intersections are "large". Of course, in that case conjugacy also follows from the fact that $H_1$ and $H_2$ are Sylow $2$-subgroups.
• If $X$ stands for the set $G/H_1 \times G/H_2$ with $G$ acting by left multiplication, then the assumption on the size of $H_1\cap H_2$ implies that the action is not transitive. Clearly all the orbits of that action have sizes that are multiples of $p$ (the point stabilizers are of the form $xH_1x^{-1}\cap yH_2y^{-1}$). If we could show that the smallest orbit must have size $p$ exactly, we would be done. For the purposes of the alternative goal we would need to conclude that the kernel of this "product action" is equal to kernel of the action on either component, but I don't see a way forward.

Background:

I started thinking about this question when trying to settle this claim. You immediately see that my alternative goal is equivalent to that claim, translated to the language of groups by Galois correspondence. I think I'm missing something simple :-(

Searching the site:

• While I was typing this question, the site engine found this nice thread, where the main question is answered in the affirmative, if we know that $[G:H_1\cap H_2]=p(p-1)$. The elegant solution is related to my thoughts in the last bullet, but I cannot extend the method to cover this case as well.
• On the field theory side we have this intriguing old query.

Answer 1

${\rm PSL} (2,7)\ (\cong {\rm PSL}(3,2))$ is a counterexample with two non-conjugate subgroups of index $7$, and intersection of index $28$.

Other examples are ${\rm PSL}(2,11)$ with $p=11$ and ${\rm PSL}(3,3)$ with $p=13$.

More generally, for a prime power $q$, if $(q^n-1)/(q-1)$ is prime with $n \ge 3$, then ${\rm PSL}(n,q)$ is a counterexample.

These examples are all simple, so the two subgroups both have trivial core.

In fact under your hypotheses $H_1$ and $H_2$ always have the same core. To see this, let $C_2$ be the core of $H_2$ in $G$. If $C_2$ is not contained in $H_1$ then, since $H_1$ is maximal in $G$, we must have $G=C_2H_1$, so $|C_2H_1:H_1| = |C_2:C_2 \cap H_1| = p$.

But $C_2 \le H_2$, so $|C_2:C_2 \cap H_1| \le |H_2:H_2 \cap H_1|$, which contradicts $|G:H_1 \cap H_2| < p^2$.