# Given a finite Group G, with A, B subgroups prove the order of AB [closed]

How do you prove:

Given a finite group $G$, with $A,B$ subgroups then $$|AB|=\frac{|A||B|}{|A \cap B|}.$$

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## closed as off-topic by user26857, R_D, RecklessReckoner, Martin Sleziak, 91500Jul 19 at 8:50

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See this post and other questions linked there – Martin Sleziak Jul 19 at 8:48

You can prove that $|AB||A\cap B|=|A||B|$ directly.

There is a natural map $p$ from $A\times B$ to $AB$ by $(a,b)\mapsto ab$, which is onto. The cardinality of $A\times B$ is therefore equal to $$\sum_{x\in AB}|p^{-1}(g)|.$$

Given an element $g\in AB$, let $(a,b)\in p^{-1}(g)$. For each $x\in A\cap B$, we obtain a second pair $(ax,x^{-1}b)$ that also maps to $ab$; thus, each element of $AB$ has at least $|A\cap B|$ preimages.

If $(a,b)$ and $(a',b')$ have the same image, then $ab=a'b'$, hence $bb'^{-1}= a^{-1}a'\in A\cap B$. Letting $x=a^{-1}a'$ we have that $(a',b') = (ax,x^{-1}b)$. That is, for each element $g$ of $AB$, there is a bijection between the preimages of $g$ in $A\times B$ and the set $A\cap B$. Therefore, $$|A\times B| = |A||B| = \sum_{x\in AB}|p^{-1}(g)| = \sum_{x\in AB}|A\cap B| = |AB||A\cap B|,$$ and this holds in the sense of cardinality, even if the sets are infinite.

In the case where $A\cap B$ is finite, we get the desired equality.

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This is the orbit-stabilizer theorem.

Let $X=\lbrace aB\ |\ a\in A\rbrace$ be a subset of the cosets of $B$ in $G$. Then $A$ acts transitively on $X$, so $|X|$ is equal to the index of a stabilizer in $A$. So let $C\leq A$ be the subgroup stabilizing the coset $B\in X$; this is simply the elements $z\in A$ such that $zB=B$. This just means $z\in B$, so the stabilizer $C=A\cap B$. So we have $|X|=[A:C]=\dfrac{|A|}{|A\cap B|}$. Since $|AB|=|B|\cdot |X|$ (each coset has $|B|$ elements), we get the formula you wrote.

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This can also prove the count for the double cosets of an element, right?Thanks for the post. – awllower May 30 '12 at 7:51
@awllower: Yes it can. – user641 May 30 '12 at 21:12