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Math Help - Find a non subgroup out of subgroups cosets

  1. #1
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    Find a non subgroup out of subgroups cosets

    Find a group G with subgroups A and B such that AB is not a subgroup.

    I think I may be overthinking this, let G be the set of integers, then let A be {1} and B = {0}, then both A and B are subgroup, but there would be no inverse for 0 in the set AB.

    Is this right?

    Thank you.
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    Find a group G with subgroups A and B such that AB is not a subgroup.

    I think I may be overthinking this, let G be the set of integers, then let A be {1} and B = {0}, then both A and B are subgroup, but there would be no inverse for 0 in the set AB.

    Is this right? no! the set of integers is an additive group and so {1} is not a subgroup.
    you know that if one of A and B is normal, then AB will be a subgroup. so the group G that you're looking for has to be non-abelian. there are many examples. here are two of them:

    Example 1 G finite: let G=S_3, \ A=\{(1),(1,2)\}, \ B=\{(1),(1,3)\}. then |AB|=4, which doesn't divide |S_3|=6. so AB is not a subgroup.

    Example 2 G infinite: let G=GL(2,\mathbb{Q}). let A=\{\begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}: \ n \in \mathbb{Z} \}, \ \ B=\{\begin{pmatrix} 1 & 0 \\ n & 1 \end{pmatrix}: \ n \in \mathbb{Z} \}. then AB=\{\begin{pmatrix} nm+1 & n \\ m & 1 \end{pmatrix}: \ n,m \in \mathbb{Z} \}. now x=\begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \in AB, \ \ y=\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \in AB,

    but xy=\begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix} \notin AB. so AB is not a subgroup.
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  3. #3
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    If you are looking for an infinite case counterexample it is easier just to consider \mathbb{Z}\times S_3.
    Now choose A = \mathbb{Z}\times \left< (12) \right> and B = \mathbb{Z}\times \left< (13) \right> .
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