Subgroup coset problem

Printable View

September 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 $|AB| = \frac { |A||B|} { |A \bigcap B |}$

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.
September 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 $G=S_3$ and $A = \left< (123) \right>$ and $B = \left<(12)\right>$ . Then $AB = S_3$ while $B$ is not a normal subgroup.
----
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.
September 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 $|AB| = \frac { |A||B|} { |A \bigcap B |}$

$\boxed{1}$ 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