# Subgroup coset problem

• Sep 12th 2008, 10:07 AM
Subgroup coset problem
Let G be a finite group, and suppose that A and B are subgroups of G, prove that $|AB| = \frac { |A||B|} { |A \bigcap B |}$

Proof so far:

Now, by the second iso theorem, $\frac {|AN|}{|N|} = \frac {|A|}{|A \bigcap N|}$ 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, 11:30 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 $G=S_3$ and $A = \left< (123) \right>$ and $B = \left<(12)\right>$. Then $AB = S_3$ while $B$ is not a normal subgroup.
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Let $S$ to be the left cosets of $B$. Let $A$ act on $S$ by left translation i.e. $a(xB) = (ax)B$. Now consider the orbit of $B$ 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, 10:38 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 $|AB| = \frac { |A||B|} { |A \bigcap B |}$

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

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

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

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

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