# Thread: List elements of order 10

1. ## List elements of order 10

List all elements of $Z_{40}$ with order 10.

My solution: By a theorem, I know that the unique subgroup of order 10 is <40/10> = <4> = {0,4,8,12,16,20,24,28,32,36>

So are those the answers?

2. Originally Posted by tttcomrader
List all elements of $Z_{40}$ with order 10.

My solution: By a theorem, I know that the unique subgroup of order 10 is <40/10> = <4> = {0,4,8,12,16,20,24,28,32,36>

So are those the answers?
Think about it. It has to have ten members and it has to be closed and it has to include 0. That's going to be awfully hard to do with a different subset than the one you listed.

That's the only one.

-Dan

3. Originally Posted by tttcomrader
List all elements of $Z_{40}$ with order 10.

My solution: By a theorem, I know that the unique subgroup of order 10 is <40/10> = <4> = {0,4,8,12,16,20,24,28,32,36>

So are those the answers?
Suppose $k$ is an element of $Z_{40}$ of order $10$, then:

$
10k \equiv 0 \mod 40
$

or there exists a $\lambda \in \bold{N}$ such that:

$
10k=\lambda 40
$

which imples that $k$ is a multiple of $4$.

But not all multiples of $4$ are of order $10$, for example $8$ is of order $5$.

RonL

4. Given $G=\mathbb{Z}_{40}$ if $a\in G$ then $\mbox{ord}(a) = \frac{40}{\gcd(a,40)}$. You can take it from here.

5. So does that means the elements are {4,12,28,36}?

But I use a theorem in the book, in which says ord(a^k) = n/gcd(k,n).

Then I get k = 1, 3, 7, 9.

So shouldn't the answers be {4, 4^3, 4^7, 4^9}?

6. Originally Posted by tttcomrader
So does that means the elements are {4,12,28,36}?

But I use a theorem in the book, in which says ord(a^k) = n/gcd(k,n).

Then I get k = 1, 3, 7, 9.

So shouldn't the answers be {4, 4^3, 4^7, 4^9}?
So what I posted (theorem) is basically the same.

So we have $\mbox{ord}(a) = \frac{40}{\gcd(a,40)}$.

Since we have the order 10. Thus, $\gcd(a,40)=10$. Now find all such so that this is 10.