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Thread: normal

  1. #1
    Junior Member mathemanyak's Avatar
    Jul 2008


    Let G be a group and H,K is a normal subgroup of G
    Prove that <HUK>=HK
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  2. #2
    Senior Member JaneBennet's Avatar
    Dec 2007
    $\displaystyle x\in\left<H\cup K\right>\ \Rightarrow\ x=y^n$ for some $\displaystyle y\in H\cup K,\ n\in\mathbb{Z}$

    $\displaystyle y\in H\ \Rightarrow\ x=y^n\cdot e\in HK$
    $\displaystyle y\in K\ \Rightarrow\ x=e\cdot y^n\in HK$

    $\displaystyle \therefore\ \left<H\cup K\right>\subseteq HK$

    For the reverse inclusion, use the fact that $\displaystyle H$ and $\displaystyle K$ are normal in $\displaystyle G$.

    $\displaystyle hk\in HK\ \Rightarrow\ hk=(hkh^{-1})h\in H$ and $\displaystyle hk=k(k^{-1}hk)\in K$

    $\displaystyle \therefore\ HK\subseteq \left<H\cup K\right>$
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