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Full disclosure: this is a homework problem, but it is not assigned to turn in for credit. The problem is from Dummit and Foote, Chapter 3.2:

Suppose $H, K$ are subgroups of finite index a group (not necessarily finite) group $G$ with $[G:H] = m$, $[G:K] = n$. Prove that LCM($m,n$) $\le [G:H\cap K] \le mn $.

I looked at this post, but I'm still somewhat confused. I don't understand how to apply the Orbit-Stabilizer Theorem.

Letting $G$ act on $G/H \times G/K$ by left multiplication, the stabilizer of $(H,K)$ is clearly $(H\cap K)$, I see that much. Thus, $[G:H \cap K] = |Orb(H,K)|$, but how do I determine the length of the orbit?

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William Henry Langhoff
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    Why do you need to know the size of the orbit? It is contained in $G/H\times G/K$ which has size $mn$. So, the orbit has size at most $mn$. So, one of the required inequalities follows. Can you show the other inequality by noting that the orbit maps onto both the factors? – Mohan Oct 16 '15 at 03:05
  • Wow I can't believe I didn't see that. Thank you – William Henry Langhoff Oct 16 '15 at 03:08

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After a very helpful hint from Mohan I saw what I was missing and the answer is rather simple.

For the second inequality, $Orb(H,K) \subset G/H \times G/K$, so $|Orb(H,K)| \le mn$.

For the first inequality, since:

$$ [G:H \cap K] = [G:H][H:H \cap K] = [G:K][K:K \cap H] $$

So $m,n$ both divide $[G:H \cap K]$.

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William Henry Langhoff
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