# Subgroup coset problem

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• Sep 12th 2008, 09:07 AM
tttcomrader
Subgroup coset problem
Let G be a finite group, and suppose that A and B are subgroups of G, prove that $\displaystyle |AB| = \frac { |A||B|} { |A \bigcap B |}$

Proof so far:

Now, by the second iso theorem, $\displaystyle \frac {|AN|}{|N|} = \frac {|A|}{|A \bigcap N|}$ if N is a normal subgroup of AN.

So since B is a normal subgroup of AB, this would be true, right?

This thing look too simple, so I probably did something wrong, please check..
Thank you.
• Sep 12th 2008, 10:30 AM
ThePerfectHacker
Quote:

Originally Posted by tttcomrader
So since B is a normal subgroup of AB, this would be true, right?

This is false. Here is an counterexample.
Let $\displaystyle G=S_3$ and $\displaystyle A = \left< (123) \right>$ and $\displaystyle B = \left<(12)\right>$. Then $\displaystyle AB = S_3$ while $\displaystyle B$ is not a normal subgroup.
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Let $\displaystyle S$ to be the left cosets of $\displaystyle B$. Let $\displaystyle A$ act on $\displaystyle S$ by left translation i.e. $\displaystyle a(xB) = (ax)B$. Now consider the orbit of $\displaystyle B$ under this action. Use the theorem which says the number of elements in an orbit is equal to the index of the stabilzer in the group to get this number.
• Sep 12th 2008, 09:38 PM
NonCommAlg
Quote:

Originally Posted by tttcomrader
Let G be a finite group, and suppose that A and B are subgroups of G, prove that $\displaystyle |AB| = \frac { |A||B|} { |A \bigcap B |}$

$\displaystyle \boxed{1}$ let $\displaystyle a \in A, \ b \in B,$ and $\displaystyle I(a,b)=\{(x,y) \in A \times B: \ ab=xy\}.$ then: $\displaystyle |I(a,b)|=|A \cap B|.$

Proof: define $\displaystyle f: I(a,b) \longrightarrow A \cap B$ by: $\displaystyle f(x,y)=a^{-1}x.$ since $\displaystyle xy=ab,$ we have $\displaystyle a^{-1}x=by^{-1} \in B.$ thus: $\displaystyle a^{-1}x \in A \cap B.$ so $\displaystyle f$ is well-defined. now if $\displaystyle f(x_1,x_2)=f(x_2,y_2),$ then $\displaystyle a^{-1}x_1=a^{-1}x_2.$

thus $\displaystyle x_1=x_2$ and because $\displaystyle x_1y_1=x_2y_2=ab,$ we'll get $\displaystyle y_1=y_2.$ so $\displaystyle f$ is one-to-one. to prove that $\displaystyle f$ is onto, choose $\displaystyle c \in A \cap B$ and put $\displaystyle x=ac, \ y=c^{-1}b.$ obviously $\displaystyle x \in A, \ y \in B$ and $\displaystyle xy=ab.$

thus $\displaystyle (x,y) \in I(a,b)$ and $\displaystyle f(x,y)=a^{-1}x=c.$ so $\displaystyle f$ is a bijection and hence $\displaystyle |I(a,b)|=|A \cap B|. \ \ \ \square$

now suppose $\displaystyle |AB|=n$ and $\displaystyle AB=\{a_jb_j : 1 \leq j \leq n \}.$ clearly $\displaystyle A \times B = \bigcup_{j=1}^n I(a_j,b_j),$ and so using the result in $\displaystyle \boxed{1}$ we have: $\displaystyle |A||B|=|A \times B|=\sum_{j=1}^n|I(a_j,b_j)|=n|A \cap B|=|AB||A \cap B|.$ http://www.mathhelpforum.com/math-he...ags/Canada.gif