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.
 A: 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 $G$
A: 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 $[G :K]$. 
