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Math Help - Normal subgroups

  1. #1
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    Normal subgroups

    Let H and K be distinct subgroups of G. Then the set HK = \{hk : h \in H, k \in K\} is a subset of G by the closure of a group. Prove that HK is a subgroup of G if and only if at least one of H or K is a normal subgroup of G. I'm just kind of stuck with this - I know that a subgroup H is a normal subgroup of G iff ghg^{-1} \in H for all g in G and h in H. I also know that a group H is a subgroup iff xy^{-1} \in H for all x,y in H. But I'm having trouble using the definition of HK to show these properties, both in the forward and backward direction of the "if and only if". Can someone just point me in the right direction?
    Last edited by dancavallaro; February 19th 2009 at 05:17 AM. Reason: H and K need to be distinct subgroups
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  2. #2
    Newbie Halmos Rules's Avatar
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    "Then the set HK = \{hk : h \in H, k \in K\} is a subgroup of G by the closure of a group."

    What do you mean by this? Isn't this the very thing you are trying to prove?
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  3. #3
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    Quote Originally Posted by Halmos Rules View Post
    "Then the set HK = \{hk : h \in H, k \in K\} is a subgroup of G by the closure of a group."

    What do you mean by this? Isn't this the very thing you are trying to prove?
    Oops, I meant HK is a subset of G. I fixed it in the original post.
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  4. #4
    Newbie Halmos Rules's Avatar
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    Lets see if this helps..... Suppose that H is normal.
    If x,yHK then x=hk and y=h'k', where h,h'H and k,k'K.

    xy= (hk)(h'k')= h(kh')(k.k')

    What can you say about ?
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  5. #5
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    Ok, here's my proof for the right-to-left direction of the statement ("if at least one of H or K is a normal subgroup of G, then HK is a subgroup of G"):

    Assume H is a normal subgroup of G, and let h_1,h_2 \in H and k_1,k_2 \in K, so that h_1k_1,h_2k_2 \in HK. Then (h_1k_1)(h_2k_2) = h_1k_1h_2k_1^{-1}k_1k_2 = h_1(k_1h_2k_1^{-1})k_1k_2. But since H is a normal subgroup, then k_1h_2k_1^{-1} \in H, so (h_1)(k_1h_2k_1^{-1}) \in H by closure of a group, and k_1k_2 \in K by closure of a group, so (h_1k_1h_2k_1^{-1})(k_1k_2) \in HK, and HK is closed under the group operation. An analogous argument can be made if K is assumed to be a normal subgroup of G. Now let h \in H and k \in K so that hk \in HK. Then (hk)^{-1} = k^{-1}h^{-1}, and since K is a subgroup of G, then k^{-1} \in K, and since H is a subgroup, h^{-1} \in H, so k^{-1}h^{-1} \in HK and HK is closed under inverses. Since HK is closed under the group operation and inverses, HK is a subgroup of G.


    I think that's pretty sound. I'm not sure how to go in the other direction though: If HK is a subgroup of G, then at least one of H or K is a normal subgroup of G.
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  6. #6
    Newbie Halmos Rules's Avatar
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    Quote Originally Posted by dancavallaro View Post
    Now let h \in H and k \in K so that hk \in HK. Then (hk)^{-1} = k^{-1}h^{-1}, and since K is a subgroup of G, then k^{-1} \in K, and since H is a subgroup, h^{-1} \in H, so k^{-1}h^{-1} \in HK and HK is closed under inverses.

    That isn't correct..... , not HK!

    Try using the same trick you used for closure under the group operation....
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  7. #7
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    Quote Originally Posted by Halmos Rules View Post
    That isn't correct..... , not HK!

    Try using the same trick you used for closure under the group operation....
    Oops, good catch. How about,

    Assume H is normal: (hk)^{-1} = k^{-1}h^{-1} = k^{-1}h^{-1}kk^{-1} = (k^{-1}h^{-1}k)k^{-1}, and since H is normal, and h^{-1} \in H, then k^{-1}h^{-1}k \in H, and k^{-1} \in K, so (k^{-1}h^{-1}k)k^{-1} \in HK. Likewise if K is assumed normal. So HK is closed under inverses.
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  8. #8
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    Ah, it turns out the other direction of the problem statement isn't true in general. So I'm done with the problem. Thanks for the help!
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