Order of a product of subgroups. Prove that $o(HK) = \frac{o(H)o(K)}{o(H \cap K)}$.

The group $H \times K$ acts on the set $HK \subseteq G$ via $(h,k) x := hxk^{-1}$. Cleary the action is transitive. The stabilizer of $1 \in HK$ is easily seen to be isomorphic to $H \cap K$. The orbit-stabilizer "theorem" implies $|HK| \cdot |H \cap K| = |H \times K| = |H| \cdot |K|$.

By the way, this proof also works when $H,K$ are infinite.

Here is LaTex-ed version of the proof posted in BBred's comment. I've tried to add details of one place of the proof. If the OP explains which part of the proof is the problem, perhaps that part can be explained in more detail. I've made this answer a CW - anyone, feel free to contribute.

Certainly the set $HK$ has $|H||K|$ symbols. However,not all symbols need represent distinct group elements. That is, we may have $hk=h'k'$ although $h\ne h'$ and $k\ne k'$. We must determine the extent to which this happens.

For every $t\in H\cap K$, $hk =(ht)(t^{-1} k)$, so each group element in $HK$ is represented by at least $|H\cap K|$ products in $HK$.

But $hk = h'k'$ implies $t = h^{-1} h' = k(k')^{-1}\in H\cap K$ so that $h'=ht$ and $k' = t^{-1} k$. Thus each element in $HK$ is represented by exactly $|H\cap K|$ products. So, $$|HK|= \frac{|H||K|}{|H\cap K|}.$$

If we have $hk=h'k'$ and we multiply this by $h^{-1}$ from left and by ${k'}^{-1}$ from right, we get $$k{k'}^{-1}=h^{-1}h.$$ Maybe it should be stressed that $t\in H$, since $t=h^{-1}h'$; and $t\in K$ since $t=k{k'}^{-1}$. (Which means $t\in H\cap K$.)

We know that $$HK=\bigcup_{h\in H} hK$$ and each $hK$ has the same cardinality $|hK|=|K|$. (See ProofWiki.)

We also know that for any $h,h'\in G$ either $hK\cap h'K=\emptyset$ or $hK=h'K$.

So the only problem is to find out how many of the cosets $hK$, $h\in H$, are distinct.

Since $$hK=h'K \Leftrightarrow h^{-1}h'\in K$$ (see ProofWiki) we see that for each $k\in K$, the elements $h'=hk$ represent the same set. (We have $k=h^{-1}h'$.) We also see that if $k=h^{-1}h'$ then $k$ must belong to $H$.

Since the number of elements that represent the same coset is $|H\cap K|$, we have $|H|/|H\cap K|$ distinct cosets and $\frac{|H||K|}{|H\cap K|}$ elements in the union.