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Math Help - Abstract Algebra (commutative property)

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    Abstract Algebra (commutative property)

    Suppose H and K are Normal Subgroups of G and ( H intersection K is the identity). Prove that hk=kh for all h in H and k in K
    Thanks from jbloomfeld
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    Re: Abstract Algebra (commutative property)

    Clearly, hk = kh iff hkh-1k-1 = e. Prove this by showing that hkh-1k-1 ∈ H ∩ K.
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    Re: Abstract Algebra (commutative property)

    Please I want would like to see more steps/details
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    Re: Abstract Algebra (commutative property)

    Quote Originally Posted by emakarov View Post
    Clearly, hk = kh iff hkh-1k-1 = e. Prove this by showing that hkh-1k-1 ∈ H ∩ K.
    Do you agree that showing hkh-1k-1 ∈ H ∩ K is sufficient? To prove that hkh-1k-1 ∈ H, break hkh-1k-1 into two factors both of which are in H, and do similarly for K.
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    Re: Abstract Algebra (commutative property)

    How do you know that hkh^{-1}k^{-1} \in\ H intersection K ?
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    Re: Abstract Algebra (commutative property)

    Quote Originally Posted by mohammedlabeeb View Post
    How do you know that hkh^{-1}k^{-1} \in\ H intersection K ?
    Do you know that hkh^{-1}\in K~? HOW?

    Then, do you know that \left[hkh^{-1}\right]k^{-1}\in K~? HOW?.

    Now do as you have been told: do the same wrt H.
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    Re: Abstract Algebra (commutative property)

    I don't know How?!. Can you tell me How?
    I attached my solution I think it is not correct but you might be able to see what I did and tell me how to correct it
    CCI04202013_00003.bmp
    Quote Originally Posted by Plato View Post
    Do you know that hkh^{-1}\in K~? HOW?

    Then, do you know that \left[hkh^{-1}\right]k^{-1}\in K~? HOW?.

    Now do as you have been told: do the same wrt H.
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    Re: Abstract Algebra (commutative property)

    Quote Originally Posted by Plato View Post
    Do you know that hkh^{-1}\in K~? HOW?
    Quote Originally Posted by mohammedlabeeb View Post
    I don't know How?!. Can you tell me How?
    In another thread we have already discussed the definition of a normal subgroup. Don't you recognize it in Plato's quote?
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    Re: Abstract Algebra (commutative property)

    Quote Originally Posted by mohammedlabeeb View Post
    I attached my solution I think it is not correct but you might be able to see what I did and tell me how to correct it
    I don't think it works. First, you did not properly introduce a and b: what exactly are they? More seriously, you started with hk = kh and arrived at h2k2 = k2h2. You assumed what you are supposed to prove.

    The idea is to note that hk = kh ⇔ hkh-1k-1 = e ⇔ (hkh-1k-1 ∈ H and hkh-1k-1 ∈ K). This has been described several times in this thread.
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