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This question already has an answer here:

Can someone tell me whether my solution is okay? I based it off a proof for a more general theorem about cyclic groups and generators.

Prove that the generators of $\mathbb{Z}_n$ are the integers $r$ such that $0\leq r <n$ and $\gcd(r,n)=1$.

Let $\mathbb{Z}_{n}=\langle 1 \rangle$. Since $0\leq r <n$, $\langle r\rangle \subset \langle 1 \rangle$. Then let $\mathbb{Z}_n=\langle r \rangle$ and assume that $d=\gcd(r,n)>1$. Then there exist integers $t$ and $s$ such that $r=td$ and $n=sd$. Then $rs=tds=tn=1$ (since $|\mathbb{Z}_n|=|\langle 1 \rangle|=n$). Then $|\langle r\rangle|\leq s<n$. This forms a contradiction (if $|\mathbb{Z}_n|=n$ and $\mathbb{Z}_{n}=\langle r \rangle$ with $|\langle r\rangle|<n$, then $|\mathbb{Z}_n|\neq n$) by assuming that $\gcd(r,n)>1$.

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marked as duplicate by Jyrki Lahtonen group-theory Aug 17 '18 at 6:42

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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There are several issues. For instance, you can't say “Let $\mathbb{Z}=\langle1\rangle$”, because an assertion of the type “Let $A=\cdots$” should be a definition of $A$ and you are clearly not defining $\mathbb{Z}_n$.

Your idea of proving that if $\gcd(r,n)>1$, then $\langle r\rangle\neq\mathbb{Z}_n$ is correct. What you proved was that $\bigl\lvert\langle r\rangle\bigr\rvert<n=\lvert\mathbb{Z}_n\rvert$. But this is only half of what you were supposed to have proved. Indeed, it remains to be proved that if $\gcd(r,n)=1$, then $\langle r\rangle=\mathbb{Z}_n$. This can be done noting that, if $0<k,k'\leqslant n$ and $k\neq k'$, then $kr\neq k'r$ in $\mathbb{Z}_n$; which means that $n\nmid kr-k'r$. But that's easy: if $n\mid(k-k')r$ then, since $\gcd(r,n)=1$, then $n\mid k-k'$, which is impossible, since$$k-k'\in\bigl\{\pm1,\pm2,\ldots\pm(n-1)\bigr\}\setminus\{0\}.$$This proves that $\bigl\lvert\langle r\rangle\bigr\rvert=n=\lvert\mathbb{Z}_n\rvert$ and that therefore $\langle r\rangle=\mathbb{Z}_n$.

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