[SOLVED] order of elements in multiplicative group

**yvonnehr** · Apr 27th 2009, 08:29 PM

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!

**Jhevon** · Apr 27th 2009, 08:37 PM

Originally Posted by yvonnehr

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?

**yvonnehr** · Apr 28th 2009, 08:42 AM

Originally Posted by Jhevon

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

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.

**Jhevon** · Apr 28th 2009, 09:34 AM

Originally Posted by yvonnehr

$\mathbb{U}_p$ 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

**yvonnehr** · Apr 28th 2009, 10:04 AM

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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Thanks!

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