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

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

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

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

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

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

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

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

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

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

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

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

    can you find the order of the elements? do you remember how order is defined?
    $\displaystyle \mathbb{U}_p$ would equal $\displaystyle \mathbb{Z}_p$ 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
    $\displaystyle \mathbb{U}_p$ would equal $\displaystyle \mathbb{Z}_p$ since each element would be relatively prime to p. Right?
    yes. well, not including [0]. you want $\displaystyle \mathbb{Z}_p^*$

    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 $\displaystyle (G,*)$ be a group and $\displaystyle a \in G$. The order of $\displaystyle a$, denoted $\displaystyle \circ (a)$, is defined as the smallest positive integer $\displaystyle n$ such that $\displaystyle a^n = e$ ($\displaystyle e$ is the identity element. $\displaystyle a^n = \underbrace {a*a* \cdots *a}_{\text{product of }n~a\text{'s}}$)

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

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

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


    what about $\displaystyle [5]$?

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

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