# Let $A$ and $B$ be two finite subgroups of a group $G$. Show that $|AB|=\frac{|A| \times |B|}{|A \cap B|}$ [duplicate]

Let $A$ and $B$ be two finite subgroups of a group $G$. Show that $$|AB|=\frac{|A| \times |B|}{|A \cap B|}$$ I have no idea how to start. Anyone can help ? I think of divisibility to prove this, but I got nowhere.

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## marked as duplicate by Martin Brandenburg, Hagen von Eitzen, Henry T. Horton, 5PM, Brian M. ScottFeb 9 '13 at 19:43

One way to prove it is by showing that $$\lvert AB : B \rvert = \lvert A : A \cap B \rvert.$$ So you should set up a (well defined,) 1-1 correspondence between the cosets of $B$ in (the subset) $AB$ and those of $A\cap B$ in $A$.

Spoiler

Try $aB \mapsto a(A \cap B)$. Remember to show it's well defined first of all.

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so I need to prove that $B$ is a normal subgroup of $AB$ and $A \cap B$ is a normal sbugroup of $A$ ? – Idonknow Feb 9 '13 at 6:38
This is valid without any normality assumption, and from your statement you are required to prove it without those assumptions. $AB$ is not a subgroup in general, but still it's the union of the cosets $aB$, for $a \in A$. Just give a map between the set of the cosets $aB$ and the set of cosets $a(A\cap B)$, for $a \in A$, and prove it is well-defined and bijective. – Andreas Caranti Feb 9 '13 at 6:41
are we going to use Lagrange theorem after we prove the map is well-defined and bijective? – Idonknow Feb 9 '13 at 6:46
Precisely, so you count cosets on both sides. In the case of cosets of $B$ inside $AB$, it's not quite Lagrange, as $AB$ is not a subgroup in general, but still $AB$ is a union of cosets of $B$, and distinct cosets are disjoint, so it is still true that the number $\lvert AB : B \rvert$ of the cosets of $B$ in $AB$ is $\lvert AB \rvert / \lvert B \rvert$. – Andreas Caranti Feb 9 '13 at 6:51

Let $a_1,a_2\in A$ and $b_1,b_2\in B$ and $$a_1.b_1=a_2.b_2...........(1)$$ $$\Rightarrow a_2^{-1}a_1=b_2.b_1^{-1}$$ $$\Rightarrow a_2^{-1}a_1\in (A\cap B)\text{ and }b_2.b_1^{-1}\in (A\cap B)$$ $$\Rightarrow a_1(A\cap B)=a_2(A\cap B) \text{ and } b_1(A\cap B)=b_2(A\cap B)$$ And similarly the other way round (i.e. if they belong to the same coset then (1) follows).

From this reasoning the formula follows.This is because if for some $a_1,a_2\in A$ and $b_1,b_2\in B$, $a_1.b_1=a_2.b_2$ then $a_1,a_2$ must be in the same coset of $(A\cap B)$ so here for finding the total no. of elements in $AB$ we have to count the total no. of elements of the form $ab, a\in A,b\in B$ we have to devide it by the total no. of its repeted representations. $$|AB|=(|A| \times |B|)\times \frac{1}{\text{no of repeated reresentation of each}}={(|A| \times |B|)}\times \frac{1}{|A\cap B|}$$

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what about the $|A| \times |B|$? – Idonknow Feb 9 '13 at 6:51
Now do you get it???? @Idonknow – Abhra Abir Kundu Feb 9 '13 at 6:57
okay, i get it now – Idonknow Feb 9 '13 at 7:08