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

abstract-algebragroup-theory

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.

Best Answer

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 $G$