Find the order of 5 in $\mathbb Z_{12}$ The textbook answer is 12. I am not sure how they got that. My work is:
$$\langle5\rangle=\{0,5,10\}$$ Therefore, the order is 3.
 A: Since we are working modulo $12$, we have $$\left \langle 5 \right \rangle=\{5, 10, \overbrace3^{15\pmod{12}}, 8, \overbrace1^{13\pmod{12}}, 6, 11, \overbrace4^{16\pmod{12}}, 9, \overbrace2^{14\pmod{12}}, 7, \overbrace0^{12\pmod{12}}\}$$ Hence the order is $12$ - in fact, $5$ generates the entire group.
A: Don't stop at 12, keep going around!
$$\langle5\rangle=\{0,5,10,3,8,1,6,11,4,9,2,7\};\ |\langle5\rangle|=12$$
As a check, the order of any element in a group must divide into the order of the whole group.
A: $$5\times 12 =60 \equiv 0\pmod {12}$$ and no multiple of $5$  less than 60 has this property. 
A: The order is not $3$. You are right that $0, 5, 10$ can all be generated from $5$. But what about $3$? Note that
$$3\equiv 5+5+5+5\text{ (mod 12)}$$
In that way you can generate all elements of $\mathbb{Z}_{12}$.
The more general result follows from the Bezout's identity:

Lemma. Let $a\in\mathbb{Z}_n$. If $\gcd(a,n)=1$ then $a$ generates $\mathbb{Z}_n$. In particular $|a|=n$.

Proof. By the Bezout's identity there are integers $x, y\in\mathbb{Z}$ such that
$$ax+ny=\gcd(a,n)=1$$
Apply $\text{(mod n)}$ to both sides:
$$ax\equiv 1\text{ (mod n)}$$
Now pick any $k\in\mathbb{Z}$. It follows that
$$a(xk)\equiv (ax)k\equiv 1k\equiv k\text{ (mod n)}$$
and since $xk\in\mathbb{Z}$ then the first multiplication can be written as a sum 
$$a(xk)=\pm \sum_{i=0}^{|xk|} a$$
and thus we've generated $k$ from $a$ in $\mathbb{Z}_{n}$. Since $k$ was chosen arbitrarly then $a$ generates $\mathbb{Z}_n$. $\Box$
Now since $5$ and $12$ are relatively prime then (by the Lemma) $5$ generates $\mathbb{Z}_{12}$ hence $|5|=12$.
A: For $m\in\mathbb Z_n$, $\langle\overline m\rangle=\mathbb Z_n$ if (and only if, which can also be verified easily) $(m,n)=1$. Indeed, if so, then $$um+vn=1$$ for some $u,v\in\mathbb Z$ and therefore $$u\overline m=\overline u\overline m=\overline 1,$$ while $\overline 1$ generates the whole $\mathbb Z_n$. For your question, $(5,12)=1$ and thus the order of $\overline 5$ in $\mathbb Z_{12}$ is $12$.
