Let $G$ be a group and let $H,K \le G$ be subgroups of finite index, say $|G:H| = m$ and $|G:K|=n$. Prove that $\mathrm{lcm}(m,n) \le |G:H \cap K| \le mn$.
I was able to establish the upper bound, but I am having difficulty establishing the lower bound, so I consulted this. However, I am having trouble following the author's reasoning. Here is the relevant part I am referring to:
Now…we have $H \cap K \le H \le G$. Thus, $m$ divides $|G:H \cap K|$ and $n$ divides $|G:H \cap K|$, so that $\mathrm{lcm}(m,n)$ divides $|G:H \cap K|$.
Exactly what theorem is being used to make this conclusion?
Best Answer
HINT: $$[G:H \cap K] = [G:H][H:H \cap K]$$