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Math Help - Is the subgroup of an abelian group always abelian?

  1. #1
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    Is the subgroup of an abelian group always abelian?

    That's my question. I think it is, but can you proof this?

    And is it both ways: and if a group has a abelian subgroup, is the group itself always abelian?
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    Yes, no.

    Think about it! If a group is abelian, then xy=yx \forall x,y \in G, and so this definitely holds for any subset S \subset G.

    On the other hand, if you take any group G, then the trivial subgroup \{1\} is abelian, but G isn't necessarily.
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  3. #3
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    For the second one it gets better: A group can have all its subgroups abelian without being abelian itself (Quaternion group)
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    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by Jose27 View Post
    For the second one it gets better: A group can have all its (proper) subgroups abelian without being abelian itself (Quaternion group)
    Yes, much better example!
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    Quote Originally Posted by Bruno J. View Post
    Yes, no.

    Think about it! If a group is abelian, then xy=yx \forall x,y \in G, and so this definitely holds for any subset S \subset G.

    On the other hand, if you take any group G, then the trivial subgroup \{1\} is abelian, but G isn't necessarily.
    So, If a group is abelian, then xy=yx \forall x,y \in G, and so this definitely holds for any subset S \subset G is a proof?
    Thanks both!
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  6. #6
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by MaryB View Post
    So, If a group is abelian, then xy=yx \forall x,y \in G, and so this definitely holds for any subset S \subset G is a proof?
    Thanks both!
    What do you think? Does it convince you?
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  7. #7
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by MaryB View Post
    That's my question. I think it is, but can you proof this?
    Also, if a group is abelian then so are all its homomorphic images (equivalently, quotients):

    (a\phi) (b \phi) = (ab)\phi = (ba)\phi = (b\phi) (a \phi)
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