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Math Help - Subgroup coset problem

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    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.
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  3. #3
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    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|.
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