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Let $G$ be a group and $n\geq 2$ an integer. Let $H_1$ and $H_2$ be two distinct subgroups of $G$ that satisfy

$$[G:H_1]=[G:H_2]=n\text{ and } [G:H_1\cap H_2]=n(n-1).$$

I've been asked to prove that $H_1$ and $H_2$ are conjugate.

My thoughts

This is the same as saying that some coset of $H_1$ coincides with some coset of $H_2$, so I think the argument must be in the line of concluding that there cannot be so many different cosets. From the index theorem I've concluded that $[H_1: H_1\cap H_2]=[H_2: H_1\cap H_2]=n-1$ but I don't know if this is useful.

I've thought of considering all possible intersections between a coset of $H_1$ and a coset of $H_2$. There are $n^2$ such intersections and only $n(n-1)$ cosets of $H_1\cap H_2$. By this question I know that these intersections are either empty or a coset of the intersection, so there are exactly $n$ intersections that are either empty or repeated. I have no idea how to continue from this, any help would be appreciated.

EDIT: I've also tried to treat the particular case $H_1\cap H_2=1$, in which $[G:H_1\cap H_2]=|G|=n(n-1)$, but I don't really see any shortcut in this case.

Javi
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  • I've never seen this result before, at least as I can remember. My first guess would be to look at the permutation action of $G$ on the cosets of $H_1$. What are the conjugates of $H_1$ in such an action? – David A. Craven Aug 01 '20 at 20:24
  • @DavidA.Craven I have seen it on a Discord group, so I cannot be 100% sure that the result is true, but thanks for the advice. – Javi Aug 01 '20 at 20:25
  • Seems to work, using my hint. – David A. Craven Aug 01 '20 at 20:35
  • I've written a complete answer, and deleted it again. I can undelete it when you think you have a proof. – David A. Craven Aug 01 '20 at 20:39
  • @DavidA.Craven let's see. There are the actions $\phi:G\to S_n$ induced by $H_1$, $\psi:G\to S_n$ induced by $H_2$ and $\theta:G\to S_{n(n-1)}$ induced by $H_1\cap H_2$. The last action should restrict to $\theta_1:H_1\to S_{n-1}$ and $\theta_2:H_2\to S_{n-1}$. I don't know... I know the kernel of any of those actions is the intersection of all conjugacy classes of the domain, but that wouldn't tell me if two of them are equal because they are intersecting other things. – Javi Aug 01 '20 at 21:22
  • I will give you the first few sentences of my answer, which should help. It's easier to phrase this in terms of permutation groups. Let $G$ act on the cosets of $H_1$. Then an alternative formulation is:

    Let $G$ be a transitive permutation group on $X$, and let $H$ be a point stabilizer of $x\in X$. Let $K$ be another subgroup of the same order as $H$, and such that $|H:H\cap K|=n-1$. Prove that $K$ is a point stabilizer.

     – David A. Craven Aug 01 '20 at 21:23
  • @DavidA.Craven thank you, I'll be thinking about that, but in the meantime I going to ask you something that might bee too elementary but I'm a bit rusty in group theory. I think you're implying that $H_1$ is a stabilizer of any of its cosets. But I don't quite see it. If $xH_1$ is a coset, then an element $h\in H_1$ sends it to $hxH_1$ For this to be equal to $xH_1$ there should be elements $k,l\in H_1$ such that $hxk=xl$, but this implies $hxkl^{-1}=x$. And I don't know how this can be true. – Javi Aug 01 '20 at 21:38
  • Ah, you will have trouble if you don't know this stuff. $H$ is the stabilizer of one coset, namely itself, under the action. The conjugates of $H$ stabilize other points, namely $H^g$ stabilizes $xg$, if $x$ is the point stabilized by $H$. – David A. Craven Aug 01 '20 at 21:41
  • @DavidA.Craven ah ok thanks, I think I misunderstood the path that the answer was going to take. I'll tell you my thoughts tomorrow because it is a bit late here. – Javi Aug 01 '20 at 21:47
  • @DavidA.Craven Hi, I was trying what you said yesterday. I had to show that $K$ is a point stabilizer of some $y\in X$. I guess that either $y$ is somehow determined by $x$ or I need to make an argument by contradiction. Since the action is transitive we have $Gx=X$, so I don't think I can follow the first path since every element is somehow determined by $x$. For the second path I have to assume that for all $y\in X$ there is some $k\in K$ s.t. $ky\neq y$. – Javi Aug 02 '20 at 20:20
  • Let us continue this discussion in chat. – David A. Craven Aug 02 '20 at 20:23
  • This is Problem 7 from the IMC 2020 https://www.imc-math.org.uk/?year=2020§ion=problems&item=prob7q – Alexdanut Aug 19 '20 at 22:57

1 Answers1

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It is easier to phrase this in terms of permutation groups. Let $G$ act on the cosets of $H_1$. Then an alternative formulation is:

Let $G$ be a transitive permutation group on $X$, and let $H$ be a point stabilizer of $x\in X$. Let $K$ be another subgroup of the same order as $H$, and such that $|H:H\cap K|=n-1$. Prove that $K$ is a point stabilizer.

The proof is now clear. If $H=G_x$ then $K_x=H\cap K$ has index $n-1$. Thus by orbit-stabilizer, the orbit of $K$ containing $x$ has length $n-1$. This leaves a single orbit of length $1$, thus $K$ is a point stabilizer (as it has the correct order).

David A. Craven
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