Index of an Intersection of two Subgroups

Question

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?

Answer 1

HINT: $$[G:H \cap K] = [G:H][H:H \cap K]$$

Answer 2

The theorem you are looking for is the following: If $G_1\geq G_2\geq G_3$, then $[G_1:G_3]=[G_1:G_2][G_2:G_3]$. The proof of this theorem can be found in most introductory texts on abstract algebra.

Recall that the least common multiple of two integers $a$ and $b$ is the smallest positive integer that is divisible by both $a$ and $b$. Using the above theorem, one can see that both $m$ and $n$ divides $[G:H\cap K]$. Thus $lcm(m,n)\leq [G:H\cap K]$.

Answer 3

If $ \prod p^{m_p} $ and $ \prod p^{n_p} $ are the prime factorizations of $ m $ and $ n $, then their LCM is $ w = \prod p^{max({m_p}, {n_p})} $. (https://en.wikipedia.org/wiki/Least_common_multiple#Fundamental_theorem_of_arithmetic).

As $ p^{max({m_p}, {n_p})} | [G: H \cap K]$ for each $p$ and as such factors of $w$ are relatively prime, $w | [G: H \cap K]$.