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Math Help - A subgroup problem.

  1. #1
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    A subgroup problem.

    Show that if H & K are subgroups of abelian group G, then {hk| h is an element of H and k is an element of K} is a subgroup of G.

    Please and thank you!
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  2. #2
    Senior Member roninpro's Avatar
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    Have you tried checking the group axioms?
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  3. #3
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by cookiejar View Post
    Show that if H & K are subgroups of abelian group G, then {hk| h is an element of H and k is an element of K} is a subgroup of G.

    Please and thank you!
    As Roninpro said, just check the group axioms. In fact, you just need to check the two conditions for something to be a subgroup, as HK will be a subgroup of G.

    This is actually a weaker form a quite a nice theorem. If one of H or K is normal then HK is a group. Clearly this holds here are you are in an abelian group so all subgroups are normal.
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  4. #4
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    I got what you two were trying to tell me by using the axioms to prove this. This is the proof that I came up.

    Since H and K are subgroups of G. There exist say h an element of H, such that (h)(h') = e
    => e is an element of H.
    => (e)(h') = h' is an element of H.

    Similarly, there exist say k an element of K, such that k(k') = e
    => e is an element of K
    => e(k') = k' is an element of K.

    This implies that (hk)(h'k') = e.
    => e is an element of of H and K => e is also an element of G.
    => e(kh)' = (kh)' = (h'k') is an element of G.

    Therefore hk is a group.

    Thoughts? I think it's not yet the best proof.
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  5. #5
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by cookiejar View Post
    ...This implies that (hk)(h'k') = e.
    Yes, but you need to use the fact that G is abelian!

    Right, you have two things to prove. You need to prove that if hk \in HK then there exists an element h_0k_0 such that hkh_0k_0=e. You are right that this element is h^{-1}k^{-1} but you need to mention WHY this is its inverse.

    You also need to prove that if h_1k_1 \in HK and h_2k_2 \in HK then h_1k_1h_2k_2 \in HK.
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