# Generalization of $|HK|=\frac{|H||K|}{|H∩K| }$, where $H,K$ are finite subgroups

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?

Thank you for your attention to this matter.

• I'm not sure that you can do much better than $|HIK| = \frac{|H||I||K|}{|H \cap IK||I \cap K|}$. – Sammy Black Jun 8 '14 at 6:52
• OK, I'll check this. Thanks for your answer. – user143993 Jun 8 '14 at 7:14
• If one of your subgroups is normal the product will again be a subgroup. So, any generalization for 3 subgroups will have to reduced to the 2 subgroup case when you have a normal subgroup. – John Machacek Jun 9 '14 at 10:35
• @JMac31 , Right. If subgroups are normal, – user143993 Jun 9 '14 at 11:49

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: $$|HgK| = |H| | \{ Hgk : k \in K \}| = \frac{|H||K|}{|K \cap H^g|}. \quad \square$$
• By the way if you use the fact that it holds for $g=1$, there is a quick proof namely $|HgK|=|gg^{-1}HgK|=|H^gK|=\frac{|H^g|\,|K|}{|H^g\cap K|}=\frac{|H|\,|K|}{|H^g\cap K|}$. – Myself Jan 16 '15 at 0:20