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Math Help - cyclic groups

  1. #1
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    cyclic groups

    Why is the set of integers under addition cyclic with generator 1? How can negative number be reached by continuosly adding 1?
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  2. #2
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    You can check your definition of "cyclic group", but I'm pretty sure that when you claim a group is cyclic, you mean that every element of the group is (in the additive case) a multiple of one element or its inverse. So that's how you get your negative integers.
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  3. #3
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    Quote Originally Posted by Ackbeet View Post
    You can check your definition of "cyclic group", but I'm pretty sure that when you claim a group is cyclic, you mean that every element of the group is (in the additive case) a multiple of one element or its inverse. So that's how you get your negative integers.
    Thanks for that. I wonder if you could help me out with the proof that subgroups of cyclic groups are themselves cyclic.

    working; g=<a> , H is a subgroup of G
    if H={e}, it is cyclic
    if not, then a^n E H for some n

    i am thinking i have to show a^n can generate every element of H .
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  4. #4
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    It's been too long, my friend, since I took Abstract Algebra. Let me get another user to help you out here, one that knows algebra better than me!

    Cheers.
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  5. #5
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    Let G be a cyclic group, say G = <g>, and H\leq G. Since H\leq G, all elements of H are of the form g^k for some integer k. If H = \{ e\}, then H is cyclic, so assume \{ e\} < H and let k be the least positive integer such that g^k is in H. Let g^s be any other element of H. By the division algorithm we can write s = qk + r with 0\leq  r < k. g^r = g^{s-qk} = g^s(g^{-k})^q is in H since g^k and g^s are. By the definition of k, r must be 0. So g^s = (g^k)^q. and H = <g^k>.
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  6. #6
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    Quote Originally Posted by DrSteve View Post
    Let G be a cyclic group, say G = <g>, and H\leq G. Since H\leq G, all elements of H are of the form g^k for some integer k. If H = \{ e\}, then H is cyclic, so assume \{ e\} < H and let k be the least positive integer such that g^k is in H. Let g^s be any other element of H. By the division algorithm we can write s = qk + r with 0\leq  r < k. g^r = g^{s-qk} = g^s(g^{-k})^q is in H since g^k and g^s are. By the definition of k, r must be 0. So g^s = (g^k)^q. and H = <g^k>.
    (g^-k)^q isn't this a negative power?
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  7. #7
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    I'm not sure what you mean by that. Yes, the exponent is negative. The important thing is that it's a memeber of H.
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