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Thread: Subgroup proof

  1. #1
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    Subgroup proof

    Assume that H and K are subgroups of a group G and that K is normal in G. Then
    HK is a subgroup of G, where

    HK = {hk | h e H, k e K}


    Thanks for any help
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by jzellt View Post
    Assume that H and K are subgroups of a group G and that K is normal in G. Then
    HK is a subgroup of G, where

    HK = {hk | h e H, k e K}


    Thanks for any help
    First note that $\displaystyle e\in HK$ (easy to verify).

    Let $\displaystyle h_1k_1, h_2k_2\in HK$, we seek to show that $\displaystyle (h_1k_1)(h_2k_2)^{-1}\in HK$

    So $\displaystyle (h_1k_1)(h_2k_2)^{-1}=h_1k_1k_2^{-1}h_2^{-1}$.

    Now, we see that $\displaystyle h_1k_1k_2^{-1}h_2^{-1}=(h_1h_2^{-1})(h_2 k_1k_2^{-1}h_2^{-1})$. Observe that $\displaystyle h_1h_2^{-1}\in H$ and since $\displaystyle K\lhd G$, $\displaystyle h_2(k_1k_2^{-1})h_2^{-1}\in K$.

    Thus $\displaystyle (h_1k_1)(h_2k_2)^{-1}\in HK\implies HK\leq G$.

    Does this make sense?
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