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Math Help - H and K normal subgroups, intersection of H and K=<e>

  1. #1
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    H and K normal subgroups, intersection of H and K=<e>

    Let H and K be normal subgroups of G such that H intersect K=<e>. Show that hk=kh for all h in H and k in K.
    H and K are normal so ghg^-1 is in H and gkg^-1 is in K.
    want to show hk=kh. So basically I'm showing this is abelian.
    Can I do ghg^-1=gkg^-1?
    ghg^-1g=gkg^-1g
    gh=gk
    so that works if g=h
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  2. #2
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    Consider the order of the Union of hK where h is an element of H (All the left cosets) This will be equal to the |H|*|K|/|H intersecct K|

    I think this will help especially since you know the intersection is of size one.
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  3. #3
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    Since intersection is size one, we are left with |H|*|K|
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by kathrynmath View Post
    Let H and K be normal subgroups of G such that H intersect K=<e>. Show that hk=kh for all h in H and k in K.
    H and K are normal so ghg^-1 is in H and gkg^-1 is in K.
    want to show hk=kh. So basically I'm showing this is abelian.
    Can I do ghg^-1=gkg^-1?
    ghg^-1g=gkg^-1g
    gh=gk
    so that works if g=h
    Note that,

    khk^{-1} = h_0 \in H and hk^{-1}h^{-1} = k_0 \in K as both subgroups are normal.

    Then,
    khk^{-1} = h_0h^{-1}h \Rightarrow khk^{-1}h^{-1} = h_0h^{-1} \in H
    and
    hk^{-1}h^{-1} = k^{-1}kk_0 \Rightarrow khk^{-1}h^{-1} = kk_0 \in K.

    Can you finish it from there? You need to apply the fact that the intersection is trivial.
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