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Math Help - Order of product of elements of two subgroups

  1. #1
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    Order of product of elements of two subgroups

    Let G be a finite group. Let H \leq G, K \leq G, and HK \leq G where \leq denotes a subgroup and HK := \left\{hk | h \in H, k \in K\right\}. Let o(x) denote the (group or element) order of x.
    (a) Show that for every h \in H, k \in K, o(hk) | o(H)o(K), where o(x) is the order of x.
    (b) Let N \leq G such that o(N) is coprime to o(H)o(K). Prove that if HN = KN, then H = K.

    I'm not quite sure how to start on this problem. I don't want a full solution, just some help in figuring it out.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: Order of product of elements of two subgroups

    Quote Originally Posted by thehobbit View Post
    Let G be a finite group. Let H \leq G, K \leq G, and HK \leq G where \leq denotes a subgroup and HK := \left\{hk | h \in H, k \in K\right\}. Let o(x) denote the (group or element) order of x.
    (a) Show that for every h \in H, k \in K, o(hk) | o(H)o(K), where o(x) is the order of x.
    Here's my hint, all you have to do is show that (hk)^{|H||K|}=e, can you do this?

    (b) Let N \leq G such that o(N) is coprime to o(H)o(K). Prove that if HN = KN, then H = K.

    I'm not quite sure how to start on this problem. I don't want a full solution, just some help in figuring it out.
    Let h\in H then h\in KN and so h=kn for [tex]k\in K[/tx] and n\in N and so hk^{-1}=n and so you know that o(n)=o(hk^{-1})\mid\cdot...
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  3. #3
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    Re: Order of product of elements of two subgroups

    I thought about this one while shopping and now the answer seems quite simple.

    For (a), since HK \leq G and hk \in HK, we have |HK| = \frac{|H| \cdot |K|}{|H \cap K|}, thus (hk)^{|H| \cdot |K|} = (hk)^{|HK| \cdot |H \cap K|} = e^{|H \cap K|} = e

    For (b), picking up where you left off, o(n) = o(k^{-1}h)\:\big|\:|H| \cdot |K|. Since o(n) is coprime to |H| \cdot |K|, we must have n=1, hence N is the trivial subgroup, yielding the desired result.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Re: Order of product of elements of two subgroups

    Quote Originally Posted by thehobbit View Post
    I thought about this one while shopping and now the answer seems quite simple.

    For (a), since HK \leq G and hk \in HK, we have |HK| = \frac{|H| \cdot |K|}{|H \cap K|}, thus (hk)^{|H| \cdot |K|} = (hk)^{|HK| \cdot |H \cap K|} = e^{|H \cap K|} = e

    For (b), picking up where you left off, o(n) = o(k^{-1}h)\:\big|\:|H| \cdot |K|. Since o(n) is coprime to |H| \cdot |K|, we must have n=1, hence N is the trivial subgroup, yielding the desired result.
    Right!
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    Re: Order of product of elements of two subgroups

    Thank you for your help!
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