# [SOLVED] order of elements in multiplicative group

• April 27th 2009, 08:29 PM
yvonnehr
[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!
• April 27th 2009, 08:37 PM
Jhevon
Quote:

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?
• April 28th 2009, 08:42 AM
yvonnehr
Quote:

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.
• April 28th 2009, 09:34 AM
Jhevon
Quote:

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 $\mathbb{Z}_p^*$

Quote:

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.

Quote:

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 ;)

Quote:

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

Quote:

Thanks. Can someone please clarify? I really want to solve this homework problem.
you should be able to get somewhere now :)
• April 28th 2009, 10:04 AM
yvonnehr
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!!!