Finding the order of an element in a group $\mathbb{Z}_{12}$ I'm trying to calculate the order of an element in a group. I saw a simple example: find the order of an element $3$ in $\mathbb{Z}_{12}$. First they wrote that $\langle 3 \rangle = \{0,3,6,9\}$ but I could not understand why. As I understand, we are looking for all the number $a\in \mathbb{Z}_{12}$ so there exists an $n\in  \mathbb{N}$ so $3^n=a$. So it should be $\langle 3 \rangle = \{3,9\}$ because there is no $n\in \mathbb{N}$ so $3^n=0$ or $3^n=6$. Am I missing something?
Also, is there a faster way to determine the order? What if I would like to calculate the order of $7$ when $\mathbb{Z}_{66}$?
 A: I think your mistake with the theory is notational. In some group theory settings, "$3^n$" would represent the group operation combining the element with itself n times. This is just for convenience, as it looks cleaner to talk about elements in an abstract group like "$a^2b^3$", sort of how you would imagine integers being factored like primes.
Therefore, when the operation over the set $\mathbb{Z}_k$ is + , "$3^n$" really means "$3n$," where you use the standard quotient to take remainders (over k).
You might try and use the greatest common divisor to figure out how large the set is. You want to see how large the set is in question, but the set should be linear. It only "wraps around" after a certain point, which I believe occurs at $\frac{k}{\text{gcd}(a, k)}$ for an element "a" and a modulus "k."
A: In a group we have an operation.  We don't know what the operation actually IS but it is an operation.  
Now maybe we shouldn't but we tend to use the notation of multiplication. So we'll often write $a\in G; b \in G$ and $a*b = c$ as $ab = c$.  That doesn't actually mean the operation is arithmetic multiplication.  It could be any group operation.
We use the notation $a^k; a \in G; k \in \mathbb N$ to mean $\underbrace{a*a*a*a....*a}_{k\text{ times}}$.  This is similar to the concept of multiplication but the operation isn't necessarily arthemetical multiplication.  It is the group operation.
Perhaps we shouldn't.  If $*$ is not multiplication the $a^k$ is not multiplicative exponents.  But we are doing ABSTRACT algebra.  $*$ is not multiplication; it is any operation we want.  Aid $a^k$ is not mmultiplicative exponents.  It is exponents on our operation whatever it is.
And if our operation is ADDITION, nothing changes.  $3^k = \underbrace{3 + 3+ .... + 3}_{k \text{ times}}$.
So
$3^2 = 3*3 = 3+3\pmod {12} = 6$.
And $3^3 = 3*3*3  = 3+3+3\pmod {12} = 9$.  And $3^4 = 3*3*3*3 = 3+3+3+3\pmod {12} = 0$.
In fact.  If $a*b = a + b\pmod {12}$ and $a^k = a+a+a..... + a \pmod {12} = k\times a \pmod {12}$.
The thing is... As for as the additive group $\mathbb Z_{12}$ goes the only operation we have is addition and multiplication does not exist.  So thinking of "powers" as exponents over multiplication just... doesn't make sense.  If anything they are "exponents of addition".
......

Also, is there a faster way to determine the order? What if I would like to calculate the order of 7 when Z66?

Okay $7^2 = 7+7 = 2\times 7 = 14$ and ....
we want to find the smallest $k$ where $7^k = k\times 7 \equiv 0 \pmod {66}$.
Well, trial and error shows us $7^9 = 9\times 7 =63$ and $7^{10} = 10\times 7 = 70 \equiv 4 \pmod {66}$ and.... and $7^{19} =19\times 7 19 = 133\equiv 1 \pmod{66}$ and $7^{20} = 20\times 7 = 140 \equiv 8\pmod{66}$ and... sheesh, how long do we have to go?!
Oh, heck... here's a secret. 
We want $7^k = k\times 7 \equiv 0 \pmod {66}$.
That means there is a $N$ so that $7k = 66N$.  So $k = \frac {66N}{7}$. Now $7$ is relatively prime to $66$ so $\frac m7$ must be an integer. so let $\frac N7 = v$.
Then $k = 66*v$.  So $\frac {k}{66} = v$.  The smallest possible $k$ is if $j = 66$ and indeed $7^{66} = 66\times 7 \equiv 0 \pmod{66}$ and we know that no smaller such number exists because.... well, $7$ and $66$ are relatively prime....
In general the formula if if $|a| = k$ in $\mathbb Z_m$ so that $a^k = 0 \implies k\times a = m\times N$ for some $N$ the smallest such $k$ that can do this is if $k\times a = m\times N$ is the least common multiple of $a$ and $m$.  ANd that would be when $k = \frac {m}{\gcd(a,m)}$.
Hence the formula you saw.  $|a| = \frac {|\mathbb Z_m|}{\gcd(a, |\mathbb Z_m|)}$.  (because $|\mathbb Z_m| = m$).
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Actually John Nash's answer has a good insight for this.
$\mathbb Z_n$ is a cyclic group generated by $1$.  $a= 1+1+1....= 1^a$ and $a^k = 1^{ak}$ and $0 = 1^{nm}$ fr some $m$.
So the order $k$ of $a$ will be occur with $ak=nm$ is the least common multiple of $n$ and $a$.  I.E. if $k = \frac{n}{\gcd(n,a)}$. 
(If you are curious what rule $m = \frac {a}{\gcd(n,a)}$ plays... well it doesn't really play any role.  It's how many times $a$ has to "double back" because its remainder doesn't divide "nicely" into $n$.  It's  ... only as significant as we want it to be.)
A: $\mathbb{Z}_{12}$ is a cyclic group under addition. So the generator $'a'$ of the form $<a>={na |~ for ~some n\in \mathbb{Z}}$.
In order to find order  of $3$ we need to do as follows:
$<3>=\{3,6,9,0\}=\{0,3,6,9\}$
here $n=4$ for which $<3> $is zero. Hence order of $3$ is $4$.
