Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How do you prove: Given a finite Group G, with A,B subgroups, that: $|AB|=\frac{|A||B|}{|A \cap B|}$

share|cite|improve this question

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.

share|cite|improve this answer

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.