What are the generators of $\Bbb{Z}_6, \Bbb{Z}_8,$ and $\Bbb{Z}_{20}$?

[karush]

nmh{707}
$\textit{Find all generators of $\Bbb{Z}_6, \Bbb{Z}_8,$ and $\Bbb{Z}_{20}$}$
$$\begin707{align*}
\Bbb{Z}_6&\quad=6, \textit{ all generators of } \Bbb{Z}_6 \textit{ are of the form } k\cdot1=k.
where gcd(6,k)=1\\
&\quad \textit{ So } k=1,5 \textit{ and there are two generators of } \Bbb{Z}_6 1 \textit{ and }5 \\
\Bbb{Z}_8&\quad \textit{ For } k \in \Bbb{Z}_8, \gcd(8; k)=1 \textit{ iff } k=1,3,5,7. \textit{So there are four
generators.}\\
\Bbb{Z}_{20}&\quad \textit{ For } k \in \Z_{20}, \gcd(20;k)=1 \textit{ iff } k=1,3,7,9,11,13,17,19.
\textit{ They are generators of } \Bbb{Z}_{20}
\end{align*}$$

ok this is c/p answer
but I don't think I understand still what a generarator is and how it is used

 

[Olinguito]

A generator of a cyclic group $G$ is an element $a\in G$ such that all the elements of $G$ are of the form $a^n=\underbrace{a\cdot a\cdot\cdots\cdot a}_{n\ \text{times}}$ for some non-negative integer $n$ (where $a^0$ is defined to be the identity element of $G$). In the case when $G$ is Abelian and additive notion (as in your case) the notation becomes $na=\underbrace{a+a+\cdots+a}_{n\ \text{times}}$ (with 0a being the identity).

Take for example the element $5\in\mathbb Z_6$. It is a generator because
$$0\cdot5\ =\ 0\ \equiv\ 0\pmod5 \\ 1\cdot5\ =\ 5\ \equiv\ 1\pmod5 \\ 2\cdot5\ =\ 10\ \equiv\ 4\pmod6 \\ 3\cdot5\ =\ 15\ \equiv\ 3\pmod6 \\ 4\cdot5\ =\ 20\ \equiv\ 2\pmod6 \\ 5\cdot5\ =\ 25\ \equiv\ 1\pmod6$$
which gives us all the elements of $\mathbb Z_6$. Similarly, 5 is a generator of $\mathbb Z_8$ because going through $0\cdot5,\,1\cdot5,\,2\cdot5,\,\ldots\pmod8$ gives all the elements of $\mathbb Z_8$.

But $5$ is not a generator of $\mathbb Z_{10}$ because
$$0\cdot5\ =\ 0 \\ 1\cdot5\ =\ 5\ \\ 2\cdot5\ =\ 10\ \equiv\ 0\pmod{10} \\ 3\cdot5\ =\ 15\ \equiv\ 5\pmod{10} \\ 4\cdot5\ =\ 20\ \equiv\ 0\pmod{10} \\ \qquad\vdots$$
so we don’t get all the elements of $\mathbb Z_{10}$.

In general, $a\in\mathbb Z_n$ is a generator of $\mathbb Z_n$ if and only if $\gcd(a,n)=1$. As an exercise, try and find the generators of $\mathbb Z_9$. (Hint: There are 6 of them.)

 

[karush]

In general, $a\in\mathbb Z_n$ is a generator of $\mathbb Z_n$ if and only if $\gcd(a,n)=1$. As an exercise, try and find the generators of $\mathbb Z_9$. (Hint: There are 6 of them.)

$0\cdot2\ =\ 0\ \equiv\ 0\pmod2 \\
1\cdot2\ =\ 2\ \equiv\ 0\pmod2 \\
2\cdot2\ =\ 4\ \equiv\ 1\pmod3 \\
3\cdot2\ =\ 6 \equiv\ 2\pmod4\\
4\cdot2\ =\ 8\ \equiv\ 3\pmod5 \\
5\cdot2\ =\ 10\ \equiv\ 4\pmod6 \\
6\cdot2\ =\ 12\ \equiv\ 5\pmod7\\
7\cdot2\ =\ 14\equiv\ 6\pmod8\\
8\cdot2\ =\ 16\ \equiv\ 7\pmod9 \\
9\cdot2\ =\ 18\ \equiv\ 8\pmod10
$at least one pass
sorta or is it a derail?

 

[Olinguito]

Sorry, I made a few typos in my post above. I meant the following:

Take for example the element $5\in\mathbb Z_6$. It is a generator because
$$\color{black}0\cdot5\ =\ 0\ \equiv\ 0\pmod{\color{red}6\color{black}} \\ \color{black}1\cdot5\ =\ 5\ \equiv\ \color{red}5\color{black}\pmod{\color{red}6\color{black}} \\ 2\cdot5\ =\ 10\ \equiv\ 4\pmod6 \\ 3\cdot5\ =\ 15\ \equiv\ 3\pmod6 \\ 4\cdot5\ =\ 20\ \equiv\ 2\pmod6 \\ 5\cdot5\ =\ 25\ \equiv\ 1\pmod6$$
which gives us all the elements of $\mathbb Z_6$.

 

[karush]

thusly?

$$
0\cdot8\ =\ 0\ \equiv\ 0\pmod9 \\
1\cdot8\ =\ 8\ \equiv\ 8\pmod 9 \\
2\cdot8\ =\ 16\ \equiv\ 7\pmod9 \\
3\cdot8\ =\ 24 \equiv\ 6\pmod9\\
4\cdot8\ =\ 32\ \equiv\ 5\pmod9\\
5\cdot8\ =\ 40\ \equiv\ 4\pmod9 \\
6\cdot8\ =\ 48\ \equiv\ 3\pmod9\\
7\cdot8\ =\ 56\equiv\ 2\pmod9\\
8\cdot8\ =\ 64\ \equiv\ 1\pmod9 \\
$$

 

[karush]

https://dl.orangedox.com/GXEVNm73NxaGC9F7Cy