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Thread: finding generator of a cyclic group

  1. #1
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    finding generator of a cyclic group

    finding generator of a cyclic group-screenshot-77-_li.jpg How do i find the generator for this cyclic group? Im thinking about <5m+7n> but im not sure if its correct. Would gladly like to know if there are any methods to find the generator of such cyclic group. On a side note i guess it should be m,n in Z instead of m.n
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  2. #2
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    Re: finding generator of a cyclic group

    Quote Originally Posted by noobpronoobpro View Post
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    How do i find the generator for this cyclic group? Im thinking about <5m+7n> but im not sure if its correct. Would gladly like to know if there are any methods to find the generator of such cyclic group. On a side note i guess it should be m,n in Z instead of m.n
    Has it occurred to you that the question gives the set underling the group, but nothing about the group operation? Now a reasonable guess is that the group operation is simple addition. Is that correct?

    In any case, that group contains $\{0,\pm 5, \pm7,\pm 35,\pm 36,\pm 36,\pm 37,\cdots\}$ can you show that?

    Now assume that addition, $\oplus $ is the operation.
    1) What is the identity of the operation ?
    2) If $t$ is an element is that set, what is the operational inverse of $t~?$.
    3) If $t~\&~s$ are elements is that set, can you show that $(-t)\oplus s$ is in the set?

    Having done 1) & 2), if you can do 3) then you have proved that is a group(subgroup).

    Can you find a generator?
    Thanks from topsquark
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    Re: finding generator of a cyclic group

    im confused with how you arrived at the various elements, can you give me some help?
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    Re: finding generator of a cyclic group

    If you are referring to "{0,5,7,35,36,36,37,⋯}", those are just 5m+ 7n with m= n= 0, then m= 1, n= 0, m= 0, n= 1, etc. (Having "36" twice is a typo of course).
    Is it true that the group operation is addition? What is the definition of "generator" of a group?
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    Re: finding generator of a cyclic group

    A generator is able to produce the whole set of elements in a group upon repeated application on itself. So, it doesnt matter if the generator produces elements that are not included in the group? The generator i'm referring to are 1 and -1.
    Last edited by noobpronoobpro; Oct 18th 2017 at 06:59 AM.
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    Re: finding generator of a cyclic group

    H=Z

    since

    x=5(3x)+7(-2x)
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    Re: finding generator of a cyclic group

    i see... thats really helpful in solving my queries :>
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