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Math Help - Prove normal subgroup

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    Prove normal subgroup

    Let H be a subgroup of G with index 2.

    a. Prove that H is a normal subgroup of G.

    b. Prove that g^(2) E H for all g E G.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by rainyice View Post
    Let H be a subgroup of G with index 2.

    a. Prove that H is a normal subgroup of G.
    There is a fairly well-known theorem which says that "if all the right cosets of a subgroup are left cosets of the subgroup then the subgroup is normal". So, notice that if L_1,L_2 are the left cosets and R_1,R_2 the right then L_1\amalg L_2=R_1\amalg R_2 (where that funny symbol just means union but the sets are pairwise disjoint). What now?


    b. Prove that g^(2) E H for all g E G.
    A more general theorem says that if N\unlhd G then g^{\left[G:N\right]}\in N for all g\in G. To see this think about G/N. What's it's order? what is \left(gN\right)^{[G:N]}=?
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    Quote Originally Posted by rainyice View Post
    Let H be a subgroup of G with index 2.

    a. Prove that H is a normal subgroup of G.

    b. Prove that g^(2) E H for all g E G.
    For the first part:

    for g E G

    if g E H , then gH = H = Hg. Therefore H is a normal subgroup.

    if g is not an element of H, G = eH U gH because there are only 2 cosets.
    also G = He U Hg.

    So G = H U Hg and G = H U gH (H = eH = He)

    Therefore, gH = Hg. H is a normal subgroup.

    I don't know how to prove part b. Sorry
    Last edited by glasssocks; April 27th 2010 at 07:55 PM.
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    Quote Originally Posted by Drexel28 View Post
    There is a fairly well-known theorem which says that "if all the right cosets of a subgroup are left cosets of the subgroup then the subgroup is normal". So, notice that if L_1,L_2 are the left cosets and R_1,R_2 the right then L_1\amalg L_2=R_1\amalg R_2 (where that funny symbol just means union but the sets are pairwise disjoint). What now?




    A more general theorem says that if N\unlhd G then g^{\left[G:N\right]}\in N for all g\in G. To see this think about G/N. What's it's order? what is \left(gN\right)^{[G:N]}=?



    about the Quotient subgroup, I haven't learn that yet
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by rainyice View Post
    about the Quotient subgroup, I haven't learn that yet
    That's lame. I wrote up a proof. Here's the basic idea. Break it into two cases

    1. If g\in N it's obvious what to do.

    2. If g\notin N then it has to be in the other coset, call it hN. And, so if g^2\notin N\implies\cdots g\notin g^{-1}N\implies\cdots
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    Quote Originally Posted by Drexel28 View Post
    That's lame. I wrote up a proof. Here's the basic idea. Break it into two cases

    1. If g\in N it's obvious what to do.

    2. If g\notin N then it has to be in the other coset, call it hN. And, so if g^2\notin N\implies\cdots g\notin g^{-1}N\implies\cdots

    i don't get it ....
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  7. #7
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    Quote Originally Posted by glasssocks View Post

    if g is not an element of H, G = eH U gH because there are only 2 cosets.
    also G = He U Hg.
    Doesn't this only work if G is a normal group?
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