I am given that $H,K$ are finite index subgroups of a group $G$ (which may not be finite), and that $|G:H|=m,\ |G:K|=n$. I am asked to show that $$\text{lcm}(m,n)\leq |G:H\cap K|\leq mn$$

I'm a bit at a loss as to how to proceed, because all the relevant tools I'm aware of (Like LaGrange's theorem) require $G$ to be finite, which we don't have here. So I've tried a reductio in which I assume that $G$ can be decomposed into $p>mn$ disjoint sets $g_i(H\cap K)$, but this doesn't seem to go anywhere; I can't find any contradiction.

  • $\begingroup$ I expect you meant to assume that $H$ and $K$ are finite index subgroups of $G$, not finite subgroups. $\endgroup$ – Derek Holt Sep 15 '17 at 21:58
  • $\begingroup$ @DerekHolt I did mean that, and it hadn't occurred to me that these were two distinct concepts. I see it now though. $\endgroup$ – Ceph Sep 17 '17 at 14:01
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    $\begingroup$ Elements in distinct cosets of $H \cap K$ in $K$ lie in distinct cosets of $H$ in $G$, so $|K:H \cap K| \le m$. Also $|G:H \cap K| = |G:K||K:H \cap K|$, so $|G:H \cap K| \le mn$ and $n$ divides $|G:H \cap K|$. Similarly $m$ divides $|G:H \cap K|$, giving the first inequality. $\endgroup$ – Derek Holt Sep 17 '17 at 14:23
  • $\begingroup$ @DerekHolt this is very helpful; but how do you have that $|G:H\cap K|=|G:K||K:H\cap K|$? Doesn't this require that $|G:H\cap K|$ be known to be finite? $\endgroup$ – Ceph Sep 17 '17 at 14:45
  • $\begingroup$ I found a proof that $|G:H\cap K| = |G:K||K:H\cap K|$ here. $\endgroup$ – Ceph Sep 17 '17 at 14:55

First, note that since $H$ is finite and $G:H$ is finite, then $|G|=|H| \cdot [G:H]< \infty$ so $G$ is finite.

Now, you have $$|G|=|H| \cdot [G:H]=[G:H] \cdot [H : H \cap K] \cdot |H \cap K| \\ |G|=[G : H\cap K] \cdot |H \cap K|$$ from where you get $$[G : H\cap K]=[G:H] \cdot [H : H \cap K]=m \cdot [H : H \cap K]$$

Same way you get $$[G : H\cap K]=[G:K] \cdot [K : H \cap K]=n \cdot [K : H \cap K]$$

To complete the proof, compare the equivalence classes in $[H : H \cap K]$ to the equivalence classes in $[G :K]$.

  • $\begingroup$ Thank you, but as DerekHolt suggested in a comment, my description was erroneous. $H$ and $K$ are finite index subgroups of $G$, not finite subgroups -- so the first line of this answer does not apply (right?). My apologies. $\endgroup$ – Ceph Sep 17 '17 at 14:00
  • $\begingroup$ @Ceph You can still prove the last equalities in that case, but the proof is a bit trickier/ $\endgroup$ – N. S. Sep 17 '17 at 14:47

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