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In preparation of a next, I was reviewing topics in class that I still didn't hold a good grasp on, one of them was the $Z_n$ operation:

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So with this, we did some examples:

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From the above, it seemed clear, so doing an example to test my understanding, for $Z_{60}$, I tried listing all elements of order 2 and 15 and got the following:

For the elements of order 15, I look at how many multiples of 15 there are in 60, there are 4: 0, 15, 30, 45

For the elements of order 2, I look at how many multiples of 2 there are in 60, there are 31: 0, 2, 4, 6, 8, ...

Is there something I'm not getting and or is there a formula of some sort to save time for possibly long element listings?

Thanks for reading, greatly appreciate it!

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    @amWhy thanks, edited post with fix – 2017-01-30

2 Answers 2

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It turns out there is a formula, but I don't think it works like you think it does.

In $\Bbb Z_n$, the order of $[k]$ (the equivalence class modulo $n$ of the integer $k$) is:

$\dfrac{n}{\gcd(k,n)}$.

In your example, we have $n = 10$ and $k = 6$, so the order of $[6]$ is:

$\dfrac{10}{\gcd(6,10)} = \dfrac{10}{2} = 5$.

So, to continue, if $n = 60$, to find elements of order $15$, we need to find $k$ such that:

$\dfrac{60}{\gcd(k,60)} = 15$.

Thus $\gcd(k,60) = 4$. Obviously $k = 4$ works, and so we want multiples $4t$ such that $\gcd(t,15) = 1$ (for if $\gcd(t,15) = d > 1$, then $4d$ divides $60$ and $4t$, and thus $\gcd(4t,60) > 15$).

For example, with $t = 3$, we have $k = 4t = 12$, and $[12]$ has order:

$\dfrac{60}{\gcd(12,60)} = \dfrac{60}{12} = 5 < 15$.

It's easy to find numbers $t$ such that $\gcd(t,15) = 1$, there are:

$\phi(15) = \phi(3)\phi(5) = 2\cdot 4 = 8$ of them (here, $\phi$ is the Euler totient function, which is handy for problems like this), and they are:

$1,2,4,7,8,11,13,14$, so the elements of order $15$ in $\Bbb Z_{60}$ are:

$[1],[8],[16],[28],[32],[44],[52],[56]$

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    Very well explained! Really appreciate the explanation. For the check you did for [12], is that testing in the case where one might be asked what order a particular number might have? Also, how was the 1 from gcd(t, 15) = 1 derived? – 2017-01-31
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    One way of saying $\gcd(a,b) = d$ is to say that $d$ is the smallest positive integer so that there exists $r,s \in \Bbb Z$ with: $ra + sb = d$. It follows that $\gcd(a/d,b/d) = 1$ (after we divide out $d$, we have no (positive) divisors in common). Note that $4t$ and $60$ have at least $4$ as a common divisor, so if $u$ (for example) divides $t$, it must also divide $15$. It's not hard to show then, that $4u$ divides $60$. If $u$ is bigger than one, than $4u > 4$, and when we divide $60$ by $4u$, we get an upper bound on our element's order less than $15$. – 2017-01-31
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You're reasoning is not correct. Let's take the example of order 2. You are asking "What are all the elements of $\mathbb{Z}_{60}$ of order 2". These are all the elements $x\in\mathbb{Z}_{60}\setminus\{0\}$ such that $x+x\simeq 0$ (mod 60). This means that $x+x\in\{0,60,120,...\}$. Since $x+x\leq 118$ we have must have that $x+x=60$ and thus $x=30$. So there is only 1 such element.

This is very different from the number of multiples of 2 in $\mathbb{Z}_{60}$.