TITLE: Generalization of $|HK|=\frac{|H||K|}{|H∩K| }$, where $H,K$ are finite subgroups
QUESTION [8 upvotes]: I know (and proved) the theorem 
$$
|HK| = \frac{|H| |K|}{|H \cap K|}, \text{where $H,K$ are finite subgroups of $G$}.
$$
NOW I'm wondering about a generalization of this statement.
In my 1st attempt:  if $H, I, K$ are finite subgroups of $G$, then
$$
|HIK| = \frac{|H||I||K|}{|H \cap I \cap K|}
$$
BUT this is false for $G=$Klein 4 group, $H = \{e, a\}$, $I = \{e,b\}$, $K = \{e,c\}$
and 2nd attempt, 3rd, ... are not true...

Is this generalization NOT worth wondering about?

Please give me some advice.
Thank you for your attention to this matter.

REPLY [5 votes]: One generalisation I know (which appears as an exersice in I.M. Isaacs book Finite Group Theory):
Let $G$ be a group with $H,K \le G$ and $g \in G$. Define the set $HgK = \{ hgk : h \in H, k \in K \}$. If $H, K$ are finite, then
$$
 |HgK| = \frac{|H||K|}{|K\cap H^g|}.
$$
If you take $g = 1$ you get your formula.
Proof: First observe that
$$
 HgK = \bigcup_{k \in K} Hgk
$$
and because two right cosets are either equal or disjoint, and all have the same size $|H|$, we have to determine the number of distinct elements in the set $\{ Hgk : k \in K \}$ and multiple this number by $|H|$ to get $|HgK|$.
Consider the action  of $K$ on the right cosets of $H$ in $G$, i.e. on $\Omega = G / H = \{ Hg : g \in G \}$. Then the set $\{ Hgk : k \in K \}$ equals the orbit of $Hg$ under $K$. By an often used result of finite group theory, the size of an orbit equals the index of the stabilizer of some element from the orbit. So what is the stabilizer of $Hg$, it is
\begin{align*}
 \{ k \in K : (Hg)k = Hg \} & = \{ k \in K : gk \in Hg \} \\
                            & = \{ k \in K : k \in g^{-1}Hg \} \\
                            & = \{ k \in K : k \in H^g \} \\ 
                            & = K \cap H^g.
\end{align*}
So we have $|\{ Hgk : k \in K \}| = |K : K\cap H^G| = |K| / |K \cap H^G|$. And by the above remarks:
$$
 |HgK| = |H| | \{ Hgk : k \in K \}| = \frac{|H||K|}{|K \cap H^g|}.  \quad \square
$$