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Math Help - Group, Cosets, Normal Group help

  1. #1
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    Group, Cosets, Normal Group help

    Hello,

    Could you help me in solving this problem?

    ---- Let G group, and H, K subgroups of G. Prove that if [G:H]=h and
    [G:K]=k then lcm(h,k)|[G:H \capK] <= h.k and if H or K are
    normal subgroups of G then
    [G:H intersection K] | h.k ----



    I could prove that [G:H \capK] <= h.k but i cannot prove
    lcm(h,k)|[G:H intersection K] and if H or K are normal subgroups of G
    then
    [G:H \capK] | h.k

    Pleas i really appreciate very much your help and i'm sorry for my bad english

    Cheers,
    RP
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  2. #2
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    Quote Originally Posted by roporte View Post
    Hello,

    Could you help me in solving this problem?

    ---- Let G group, and H, K subgroups of G. Prove that if [G:H]=h and
    [G:K]=k then lcm(h,k)|[G:H \capK] <= h.k and if H or K are
    normal subgroups of G then
    [G:H intersection K] | h.k ----
    You prove that if H,K have finite index then H\cap K have finite index.
    It follows that H\cap K is finite index in H.
    Now note H\cap K \subseteq H \subseteq G.
    This means (G:H\cap K) = (G:H)(H:H\cap K).
    Therefore, (G:H)|(G: H \cap K).
    Now note H\cap K \subseteq K\subseteq G.
    This means (G:H\cap K) = (G:K)(K:H\cap K).
    Therefore, (G:K) | (G:H\cap K).
    Thus, \text{lcm}(h,k) must divide (G: H\cap K).
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  3. #3
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    Thanks a lot!
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  4. #4
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    If H,K are normal subgroups then H\cap K is a normal subgroup. This means we can form the factor group G/(H\cap K).

    For the second part we wil use Lagrange's theorem. We will embed* G/(H\cap K) in G/H\times G/K. Then it will follow that the index of H\cap K in G must divide |G/H\times G/K | = hk and that would complete the proof. Note that a(H\cap K) \mapsto (aH,aK) is an embedding (you need to show it is well-defined an a homomorphism).


    *)Given a group G_1 and G_2 to embed G_1 in G_2 means to identity G_1 as an isomorphic subgroup of G_2 i.e. find a one-to-one homorphism \theta: G_1\to G_2. Then G_1\simeq \theta (G_1) and \theta(G_1) is a subgroup of G_2 because a homomorphic image of a subgroup remains a subgroup.
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