# Given $H,K$ finite index subgroups of $G$, prove that the order of $G/(H\cap K)$ obeys these bounds

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.

• I expect you meant to assume that $H$ and $K$ are finite index subgroups of $G$, not finite subgroups. Sep 15, 2017 at 21:58
• @DerekHolt I did mean that, and it hadn't occurred to me that these were two distinct concepts. I see it now though.
– Ceph
Sep 17, 2017 at 14:01
• 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. Sep 17, 2017 at 14:23
• @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?
– Ceph
Sep 17, 2017 at 14:45
• I found a proof that $|G:H\cap K| = |G:K||K:H\cap K|$ here.
– Ceph
Sep 17, 2017 at 14:55

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:

$$[G:H∩K]≤[G:H][G:K]$$

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

$$lcm([G:H],[G:K])≤[G:H∩K]≤[G:H][G:K]$$

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

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

• 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 :(
– Ceph
Feb 7, 2021 at 16:01
• 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?
– Ceph
Feb 7, 2021 at 16:02
• I don't think there is any mistake in this answer.
– AGH
Feb 8, 2021 at 4:59
• how can i prove the equality part? 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 $[G :K]$.

• 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.
– Ceph
Sep 17, 2017 at 14:00
• @Ceph You can still prove the last equalities in that case, but the proof is a bit trickier/ Sep 17, 2017 at 14:47