For sets $C, D \subset Q$ why is $|CD|=\dfrac{|C||D|}{|C\cap D|}$ Could someone please clearly explain why the following set cardinality is true? For subgroups $C, D \subset Q$ why is $|CD|=\dfrac{|C||D|}{|C\cap D|}$ Where $CD= \{cd|c\in C, d\in D\}$ and $C,D$ are subgroups of $Q$.


*

*I think the trouble for me is that I can't really see what $CD$ is. I know that every element in $CD \in Q$, but that is about it.


Thanks for the help.
 A: Inspect the set $ C \times D := \{(c,d) | c \in C, d \in D\}$. You can already see that $|C\times D|=|C|\,|D|$.
You can now prove that the map $\phi: C \times D \rightarrow CD: (c,d) \rightarrow cd$ is surjective. Each element will be obtained $|C \cap D|$ times by $\phi$. From this reasoning the cardinal equality should follow.
Detailed:
$\phi$ is surjective, because each element of $CD$ can be obtained by this map. Just take an arbitrary element from $CD$, which we can write as $cd$ (with $c \in C$ and $d \in D$).
Now we prove that each element will be obtained $|C \cap D|$ times by $\phi$.

*

*Take an element from $CD$, let's call it $a=cd$. For each element $b \in C \cap D$ it now holds that $(c b^{-1})(bd)=a$ (because we can just remove $b^{-1}b$). This reasoning gives us now $|C \cap D|$ different couples of elements from $C \times D$ which are mapped to $a$ by $\phi$.


*Now we prove the other direction. If $\phi (x,y)=a$, then $cd=xy$ and thus $x^{-1}c=yd^{-1}$. Let's call this last element $h$. Now, $h \in |C \cap D|$. But then $x=c h^{-1}$ and $y=hd$. So the element $(x,y)$ exactly looks like the way we wanted in our proof.

With the help of this subproof we can get to the final result:
$|C \times D| = |C||D|= \frac{|CD|}{|C \cap D|}$

