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

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


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