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It says in my book that generators are relatively prime to $28$, so that would be a set of $\{1,3,5,9,11,13,15,17,19,23,25,27\}$, ok i get that. But why is $9$ and in there? I can express $9$ as $3^2$, so why we have to put $9$ in there as all the numbers in $Z_{28}$ could be expressed without it? Same thing goes for $25$ ($5^2$) Sorry about my english nad ty for the answers

daruma
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Nejc
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    There is a difference between additive and multiplicative generators. The book probably meant additive generators. – Somos Aug 15 '18 at 17:14

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This set isn't meant to be a generating set (i.e., a set whose members combine to generate the whole group), rather each member of the set generates the whole group. Repeated additions of $9$ to itself will generate all of $\mathbb Z_{28}$.

BallBoy
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The group $\Bbb{Z}_n$ is cyclic, so has a single generator. However, there are $\phi(n)$ different choices for this generator, namely all $1\le k\le n$ which are coprime to $n$. So we have $$ \Bbb{Z}_{28}=\langle 1\rangle=\langle 3\rangle=\langle 5\rangle=\langle 9\rangle=\cdots $$ This has been explained at this site, e.g., here:

How to find a generator of a cyclic group?

Dietrich Burde
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