Proving that $pq$ divides $[G\!:\!H \cap K]$. Problem: Let $H$ and $K$ be subgroups of a finite group $G$ such that $[G\!:\!H] = p$ and $[G\!:\!K] = q$, with $p$ and $q$ distinct primes. Prove that $pq$ divides $[G\!:\!H \cap K]$.
Here's what I've come up with informally. So given the Lagrange Theorem, $|G| = |H| \cdot p$ and $|G| = |K|\cdot q$. The only way this works is if $|H| = q$ and $|K| = p$. Also because of Lagrange, $H \cap K = \{e\}$ since every element of $H$ and $K$ is either 1 or the prime of the group order, and since only $e$ has order 1, thats the only element in the intersection.
The index of $H \cap K$ is all the cosets which in this case is going to be $ea, \forall a \in G$, so its going to be the size of $G$ which we know is $pq$ and obviously $pq \mid pq$
Is this correct? I feel uneasy about it, so wanted some help. Thank you
 A: Hint: use two facts: $|G:H|$ divides $|G: H \cap K|$ (and symmetrically, also $|G:K|$ divides $|G: H \cap K|$). Secondly, if two positive integers $a$ and $b$ divide the integer $n$, then lcm$(a,b)$ divides $n$.
A: I'll make some specific comments about your lines, then some general comments about how to get some traction on this problem.

So given Legrange theorem, $|G| = |H| \cdot p$ and $|G| = |K|\cdot q$. 

This is correct.  This tells us $p$ divides $|G|$ and $q$ divides $|G|$.  So $\mathrm{lcm}(p,q)$ divides $|G|$.  Since $p$ and $q$ are distinct primes, $\mathrm{lcm}(p,q) = pq$.  Therefore, $pq$ divides $|G|$.

The only way this works is if $|H| = q$ and $|K| = p$. 

Not quite.  We know $p$ and $q$ divide $|G|$ at least once each, but we don't know if higher powers of these divide $|G|$ and we don't know anything about other primes dividing $|G|$.  In a comment, I pointed out that $|G| = p^2 q^2$ is a bit of a problem.  Also $|G| = p q r$, where $r>1$ is not necessarily distinct from $p$ and $q$ causes trouble for this line.

Also because of Legrange, $H \cap K = \{e\}$ since every element of $H$ and $K$ is either 1 or the prime of the group order, and since only $e$ has order 1, thats the only element in the intersection.

Though this would follow from your previous line, your previous line is a little stronger than we can actually show.  We only know that the indices of $H$ and $K$ in $G$ are prime, which is not enough to know the orders of $H$ and $K$ are prime, so we do not know that $H \cap K = \{e\}$.  Your next line uses this, so I'll stop here.
You seem to have the right picture:  split $G$ up into a sort of grid of intersections of cosets of $H$ and cosets of $K$.  In slow motion, you're thinking about two sorts of partitions into cosets; first, $G$ into $p$ cosets of $H$ and then, second, $eH$ into $q$ cosets of $eH \cap K$ (and then, duplicating that partitioning into the other cosets of $H$).  But $eH \cap K = H \cap K$ is the intersection of two subgroups, so is a subgroup.  This means $|H \cap K|$ divides all of $|H|$, $|K|$, and $|G|$.
This problem talks about indices, so we need to translate these facts from relations among orders to equivalent relations among indices.  Above, we showed $[G:H]$ divides $[G:H \cap K]$.  If we had split into $K$ cosets first, we would have seen $[G:K]$ divides $[G:H \cap K]$.  Then the $\mathrm{lcm}$ result from way back between your first and second lines finishes this off.
