Group theory subgroups problem Hello I am trying to prove the following problem. Let G be a group $$A\le G , B\le G, A\cdot B = \{\;ab\;|\;a\in A, b\in B\;\}$$ I have to prove that $$\vert {A}\cdot{B}|=\frac{\vert{A}\vert\cdot\vert{B}\vert}{A\cap B}$$ So I am trying this approach $$A\times B \frac{\varphi}{\mapsto}A.B$$ this is a Descartes product sorry for the bad format I didn't find the right codes for mathjax. $$|A\times B|=|A|\cdot |B|$$ So I take an element that belongs to the intersection of the groups A and B $$p\in A \cap B<G$$ so I am taking elements from groups $A$ and $B$ and the $p $ element from $A\cap B$ and map them the following way $$\varphi(ap,p^{-1}b)=a(pp^{-1})b =aeb=ab $$ by this I am trying to show that for every element $$p\in A\cap B < G$$ there is an element such that$c=ab$ from here I want to show that there is an  element $p$ such that $$p\in B\cup A$$ so $A\cup B < B$ and $\varphi(\dot a,\dot b)=\varphi(a,b) <=>ab=\dot a\dot b <=>\dot a=ab(\dot b)^{-1}=a(bb^{-1})=ap$ and from here we can see the element $bb^{-1}\in B\cap A$ and $bb^{-1}$ is also $p \in B\cap A$ so from here I cannot proceed with the solution if someone can help me I'll be thankful.
 A: The map $\phi\colon A\times B\to A\cdot B$, $(a,b)\mapsto ab$ is clearly onto.
For $y\in A\cdot B$, we want to find a bijection $\psi_y$ between $\phi^{-1}(y)$ and $A\cap B$. If we manage to do so, we end up with a bijection
$$\begin{align}A\times B&\to (A\cdot B)\times (A\cap B)\\
(a,b)&\mapsto(ab,\psi_y(\phi^{-1}(ab))) \end{align}$$ 
and this then shows the desired result $|A\times B|=|A\cdot B|\cdot |A\cap B|$.
So let's fix $y$ and find a bijection $\psi_y\colon\phi^{-1}(y)\to A\cap B$. 
By surjectivity of $\phi$, we can find $(a_y,b_y)$ with $\phi(a_y,b_y)=y$. 
Now if $(a,b)$ is any pair with $\phi(a,b)=y$, then $ab=a_yb_y$, so $a_y^{-1}a=bb_y^{-1}$. As the left side is in $A$ and the right side is in $B$, we have $a_y^{-1}a\in A\cap B$. So we can define $$\begin{align}\psi_y\colon \phi^{-1}(y)&\to A\cap B\\ (a,b)&\mapsto a_y^{-1}a\end{align}$$
Then $\psi_y$ is injective: Suppose that $(a,b),(a',b')\in\phi^{-1}(y)$ and $\psi_y(a,b)=\psi_y(a',b')$, then $a_y^{-1}a=a_y^{-1}a'$ implies $a=a'$ and after that $ab=a'b'=ab'$ implies $b=b'$.
And $\psi_y$ is surjective: If $c\in A\cap B$, then $a_yc\in A$ and $c^{-1}b_y\in b$ so that $(a_yc,c^{-1}b_y)$ is an element of $\phi^{-1}(y)$ and maps to $c$ under $\psi_y$. We conclude that $\psi_y$ is a bijection, as desired.
A: I'll try the following approach using your notation. For $\;a,a'\in A\,,\,\,b,\,b'\in B\;$ :
$$ab=a'b'\iff x:=a'^{-1}a=b'b^{-1}\in A\cap B$$
and on the other hand:
$$x\in A\cap B\implies \forall\,a\in A,\,b\in B\;,\;\;ab=:\phi(a,b)=\phi\left(ax,\,x^{-1}b\right)$$
Thus,  each $\;ab\in AB\;$ appears as image of $\;|A\cap B|\;$ elements in $\;A\times B\;$
