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For $x \in X$, let $[x] = \{y \in X | f(y) = f(x)\}$ and let $U_n = \{[a] \in \mathbb{Z}_n | \gcd(a, n) = 1\}$ (the group of units of $\mathbb{Z}_n$). Determine the order of $[4] \in U_{23}$.

What exactly is this question asking? (What does it mean to find the order of $[4]$?) Am I to find $|U_n|$? Am I to find $\phi(n)$ (Euler phi function)? I am trying to go over my notes and find some relevant example, but no luck. A clarification of this problem would be very useful.

Edit: additional information originally left out

For any sets $X,Y$ and any function $f:X→Y$, there exists an equivalence relation defined on $X$ called $Ker(f)=\{(x1,x2)|f(x1)=f(x2)\}$. Also, $X/Ker(f)=\{[x]|x∈X\}$ where for $x∈X, [x]=\{y∈X|f(y)=f(x)\}$.

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What in the world of ill-posed questions is $\,f\,$?! – DonAntonio Feb 6 '13 at 4:57
The order of an element $a$ is $$ord(a) =\min \{ n \in \mathbb{N} \ : \ a^n=1\}$$ – andybenji Feb 6 '13 at 5:00
Sorry, I was trying to define as much as I can while still keeping things terse. "For any sets $X, Y$ and any function $f: X \to Y$, there exists an equivalence relation defined on $X$, $Ker(f)$. Also, $X / Ker(f) = \{[x] | x \in X\}$ where for $x \in X$, $[x] = \{y \in X | f(y) = f(x)\}$". – user41419 Feb 6 '13 at 5:01
And what exactly is $\,\ker f\,$ when we talk of a function between sets?... – DonAntonio Feb 6 '13 at 5:04
$Ker(f) = \{(x_1, x_2) | f(x_1) = f(x_2)\}$. – user41419 Feb 6 '13 at 5:05
up vote 3 down vote accepted

The problem asks you to find the first positive integer $n$ such that


If you learned about $\phi(23)$ this makes the problem simpler: this $n$ must divide $\phi(23)=22$. Thus there are only four choices for $n$: 1, 2, 11, 22.

$n=1$ clearly doesn't work, you now have to decide for $n=2$ and $n=11$. You already know that 22 works, but have to figure which is the first one to work...

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Thank you so much for the clarification! I think I understand this problem now. – user41419 Feb 6 '13 at 5:45
Alright, so I picked up where you left off, and I found the following. $[4]^2 = [16]$, so $n = 2$ does not work. However, $[4]^{11} = [1]$, so $n = 11$ and since $11$ is least, then the order of $[4]$ is 11. Am I correct? – user41419 Feb 6 '13 at 6:03
Yes. Note that the reason why you know that one of the 4 works is because by Euler Theorem (or Fermat Little Theorem) $[4]^{22}=[1]$, and one of the properties of the order sais that whenever $[a]^n=[1]$ the order must divide $n$.... – N. S. Feb 6 '13 at 7:10

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