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, 2017 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, 2017 at 14:01
  • 1
    $\begingroup$ 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, 2017 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, 2017 at 14:45
  • $\begingroup$ I found a proof that $|G:H\cap K| = |G:K||K:H\cap K|$ here. $\endgroup$
    – Ceph
    Sep 17, 2017 at 14:55

2 Answers 2


Consider any $g∈a(H∩K)$, then $g=ah_1=ak_1$, so $g∈aH∩aK$. Every left coset of $H∩K $ is an intersection of a left coset of $H$ and a left coset of $K$. As the number of left cosets of $H$ and $K$ are finite, number of distinct left cosets of $H∩K$ must be finite.

As there are $[G:H][G:K]$ possible intersection of left cosets of $H$ and $K$, we get:


Also $H∩K≤H,H∩K≤K$, so by tower law of subgroups: $$[G:H∩K]=[G:H][H:H∩K]$$ $$[G:H∩K]=[G:K][K:H∩K]$$ so $[G:H],[G:K]$ both divide $[G:H∩K]$, hence


Equality occurs if $[G:H]$ and $[G:K]$ are coprime.

Notation: $[G:H]$ is the index of subgroup $H$ in group $G$

  • $\begingroup$ I regret to report that it has been so long since I studied this material that I no longer am able to understand my own question, nor this answer, and so cannot judge whether this answer succeeds in answering my question :( $\endgroup$
    – Ceph
    Feb 7, 2021 at 16:01
  • $\begingroup$ I am uncertain of the correct StackExchange etiquette in this case. Should I accept the answer on the charitable assumption that it does answer my question? $\endgroup$
    – Ceph
    Feb 7, 2021 at 16:02
  • $\begingroup$ I don't think there is any mistake in this answer. $\endgroup$
    – AGH
    Feb 8, 2021 at 4:59
  • $\begingroup$ how can i prove the equality part? $\endgroup$
    – R_Squared
    Feb 10, 2023 at 10:32

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, 2017 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, 2017 at 14:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .