Vtyjkyuk

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 $mathbbZ_n$ are the integers $r$ such that $0leq r <n$ and $gcd(r,n)=1$.

Let $mathbbZ_n=langle 1 rangle$. Since $0leq r <n$, $langle rrangle subset langle 1 rangle$. Then let $mathbbZ_n=langle r rangle$ 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 $|mathbbZ_n|=|langle 1 rangle|=n$). Then $|langle rrangle|leq s<n$. This forms a contradiction (if $|mathbbZ_n|=n$ and $mathbbZ_n=langle r rangle$ with $|langle rrangle|<n$, then $|mathbbZ_n|neq n$) by assuming that $gcd(r,n)>1$.

There are several issues. For instance, you can't say “Let $mathbbZ=langle1rangle$”, because an assertion of the type “Let $A=cdots$” should be a definition of $A$ and you are clearly not defining $mathbbZ_n$.

Your idea of proving that if $gcd(r,n)>1$, then $langle rrangleneqmathbbZ_n$ is correct. What you proved was that $bigllvertlangle rranglebigrrvert<n=lvertmathbbZ_nrvert$. 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 $langle rrangle=mathbbZ_n$. This can be done noting that, if $0<k,k'leqslant n$ and $kneq k'$, then $krneq k'r$ in $mathbbZ_n$; which means that $nnmid kr-k'r$. But that's easy: if $nmid(k-k')r$ then, since $gcd(r,n)=1$, then $nmid k-k'$, which is impossible, since$$k-k'inbiglpm1,pm2,ldotspm(n-1)bigrsetminus0.$$This proves that $bigllvertlangle rranglebigrrvert=n=lvertmathbbZ_nrvert$ and that therefore $langle rrangle=mathbbZ_n$.