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walters

  • 3 years ago

subgroup question let G be an abelian group and H be a non-empty closed subset of G prove that

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  1. walters
    • 3 years ago
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    \[H ^{*}=\left\{ xy ^{-1}:x,y \in H \right\}\le G\]

  2. walters
    • 3 years ago
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    @terenzreignz

  3. walters
    • 3 years ago
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    @myko

  4. myko
    • 3 years ago
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    what the asterics H* means?

  5. walters
    • 3 years ago
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    it sort of another H (it can be any later ie Q)

  6. myko
    • 3 years ago
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    and by ≤ you mean that H* is a subset of G?

  7. walters
    • 3 years ago
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    i mean it is a subgroup

  8. myko
    • 3 years ago
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    it's vean long time since I did this, but I would try to prove the axioms for the elements of H* to be a abelian group: Closure: For all a, b in H*, the result of the operation a • b is also in H*. Associativity For all a, b and c in H*, the equation (a • b) • c = a • (b • c) holds. Identity element There exists an element e in H*, such that for all elements a in H*, the equation e • a = a • e = a holds. Inverse element For each a in H*, there exists an element b in H* such that a • b = b • a = e, where e is the identity element. CommutativityFor all a, b in H*, a • b = b • a.

  9. myko
    • 3 years ago
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    by the way, which operation respect to is this an abelian group?

  10. myko
    • 3 years ago
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    i gues multiplication?

  11. walters
    • 3 years ago
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    yes that one i get i want to show that 1 H* is not empty ie e element of H* 2 pq^-1 element H* for every p,q element H*

  12. walters
    • 3 years ago
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    1 H*\[\neq \] 2 let P,Q \[\in H ^{*}\]we want to show that PQ^-1 element H*

  13. walters
    • 3 years ago
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    then PQ^-1 =\[(xy ^{-1})((xy ^{-1}))^{-1}\]

  14. walters
    • 3 years ago
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    =\[(xy ^{-1})(x ^{-1}y)=(xx ^{-1})(yy ^{-1})\]

  15. walters
    • 3 years ago
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    @terenzreignz pls help

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