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Math Help - Generating a group

  1. #1
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    Generating a group

    Please oh please help me finish this

    1,000,000 < 2^20

    Know that until you have a generating set, you can show that you keep adding elements so the prodcuts of each subset are different.

    Prove by induction that you can generate any group of at most 2^k elements w/ k generators. Assume that you ahve a group G w/ a subgroup H, generated by k elements.

    If g is an element of G not in H, the subgroup generated by the generators of H and g contains all the products gh for all h in H.

    Try to prove all elements are distinct, and none exist in H. THis means that starting w/ a subgroup H generated by k elements you can add a generator and get a subgroup twice (at least) as large as H.


    Your help will be GREATLY appreciated, thank you.
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  2. #2
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    Quote Originally Posted by EyeHeartAdrianT View Post
    Please oh please help me finish this

    1,000,000 < 2^20

    Know that until you have a generating set, you can show that you keep adding elements so the prodcuts of each subset are different.

    Prove by induction that you can generate any group of at most 2^k elements w/ k generators. Assume that you ahve a group G w/ a subgroup H, generated by k elements.

    If g is an element of G not in H, the subgroup generated by the generators of H and g contains all the products gh for all h in H.

    Try to prove all elements are distinct, and none exist in H. THis means that starting w/ a subgroup H generated by k elements you can add a generator and get a subgroup twice (at least) as large as H.


    Your help will be GREATLY appreciated, thank you.
    let G be a group with |G| \leq 2^k, \ k \in \mathbb{N}. let 1 \neq g_1 \in G and put G_1=\langle g_1 \rangle. then |G_1| \geq 2. now, if possible, choose g_2 \in G \setminus G_1 and put G_2=\langle g_1,g_2 \rangle. note that G_1 \cap g_2G_1 = \emptyset, because

    g_2 \notin G_1, and G_2 \supseteq G_1 \cup g_2G_1. thus |G_2| \geq |G_1 \cup g_2G_1|=|G_1| + |g_2G_1|=2|G_1| \geq 2^2. again, if possible, choose g_3 \in G \setminus G_2 and put G_3=\langle g_1,g_2,g_3 \rangle. then again G_2 \cap g_3G_2 = \emptyset and

    |G_3| \supseteq |G_2 \cup g_3G_2|=2|G_2| \geq 2^3. continuing this way we will eventually find some r \in \mathbb{N} and G_r=\langle g_1, g_2, \cdots , g_r \rangle with |G_r| \geq 2^r and it is no longer possible to choose g \in G \setminus G_r. that

    means \langle g_1,g_2, \cdots , g_r \rangle = G_r=G. but then 2^k \geq |G|=|G_r| \geq 2^r and so r \leq k. thus G can be genrated by r \leq k elements.
    Last edited by NonCommAlg; May 11th 2010 at 11:35 AM.
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