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Math Help - [SOLVED] order of elements in multiplicative group

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    Talking [SOLVED] order of elements in multiplicative group

    13) Find the order of each of the following elements in the multiplicative group of units U(sub)p.

    a. [2] for p=13
    b. [5] for p=13
    c. [3] for p=17
    d. [8] for p=17

    If you can tell me what units mean, and how to solve at least one of the problems, I think I should be able to do the rest. At this point, I don't know where to begin.

    Thanks!
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by yvonnehr View Post
    13) Find the order of each of the following elements in the multiplicative group of units U(sub)p.

    a. [2] for p=13
    b. [5] for p=13
    c. [3] for p=17
    d. [8] for p=17

    If you can tell me what units mean, and how to solve at least one of the problems, I think I should be able to do the rest. At this point, I don't know where to begin.

    Thanks!
    \mathbb{U}_n = \{ [k] : 1 \le k < n \text{ and } (k,n) = 1 \}

    that is, the set of all equivalence classes in \mathbb{Z}_n, where the representatives are less than n and relatively prime to n (including 1)

    example, \mathbb{U}_{12} = \{ [1], [5], [7], [11] \}

    since 5,7, and 11 are the integers between 1 and 12 that are relatively prime to 12.

    \mathbb{U}_n is an Abelian group with respect to \odot (you know what \odot means, right?).

    so what would \mathbb{U}_p look like for some prime p?

    can you find the order of the elements? do you remember how order is defined?
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    Quote Originally Posted by Jhevon View Post
    \mathbb{U}_n = \{ [k] : 1 \le k < n \text{ and } (k,n) = 1 \}

    that is, the set of all equivalence classes in \mathbb{Z}_n, where the representatives are less than n and relatively prime to n (including 1)

    example, \mathbb{U}_{12} = \{ [1], [5], [7], [11] \}

    since 5,7, and 11 are the integers between 1 and 12 that are relatively prime to 12.

    \mathbb{U}_n is an Abelian group with respect to \odot (you know what \odot means, right?).

    so what would \mathbb{U}_p look like for some prime p?

    can you find the order of the elements? do you remember how order is defined?
    \mathbb{U}_p would equal \mathbb{Z}_p since each element would be relatively prime to p. Right?

    If I am not mistaken, the order of a group is the number of elements in the group. I don't know how to find the order of an element. And I don't know how being an Abelian group relates to finding the solution.

    Thanks. Can someone please clarify? I really want to solve this homework problem.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by yvonnehr View Post
    \mathbb{U}_p would equal \mathbb{Z}_p since each element would be relatively prime to p. Right?
    yes. well, not including [0]. you want \mathbb{Z}_p^*

    If I am not mistaken, the order of a group is the number of elements in the group.
    yes, for finite groups. but you want the order of the given element.


    I don't know how to find the order of an element.
    Let (G,*) be a group and a \in G. The order of a, denoted \circ (a), is defined as the smallest positive integer n such that a^n = e ( e is the identity element. a^n = \underbrace {a*a* \cdots *a}_{\text{product of }n~a\text{'s}})

    so going back to \mathbb{U}_{12}. Lets find the order of [7]. Since, as i said, the operation is \odot, lets do this operation with [7] and itself until we get the identity (which is [1]).

    [7] \odot [7] = [7 \cdot 7] = [49] = [1]_{12}.

    thus, [7]^2 = [1], and so \circ ([7]) = 2


    what about [5]?

    [5] \odot [5] = [25] = [1], so again, \circ ([5]) = 2

    now do your problems. the fact that you are working with \mathbb{Z}_p^* helps

    And I don't know how being an Abelian group relates to finding the solution.
    you asked what "units" mean, so i gave you the definition. whether all parts of the definition come in handy in solving the problem is another story

    Thanks. Can someone please clarify? I really want to solve this homework problem.
    you should be able to get somewhere now
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    Thanks!

    Thank you so much. After seeing your example, I realized that I was generating for an additive and not a multiplicative group. <:-)

    Thank you so much!!!
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