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Let $n\geq 1$ be an integer and let $d|n$ be a divisor. I want to prove the following statement:

For any $1\leq k<d$ with $\gcd(k,d)=1$, there exists a number $1\leq u<n$ with $\gcd(u,n)=1$ such that $u=k$ mod $d$.

Can someone provide a proof or a reference?

Drew Armstrong
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    In other words, the canonical map $U(n) \to U(d)$ is surjective. See also https://mathoverflow.net/questions/31495/when-does-a-ring-surjection-imply-a-surjection-of-the-group-of-units – lhf Feb 26 '19 at 23:03
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    See also https://math.stackexchange.com/questions/885216/why-f-colon-mathbbz-n-times-to-mathbbz-m-times-is-surjective – lhf Feb 26 '19 at 23:07

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Hint $ $ We seek $u = k\!+\!jd\,$ with $\,\gcd(k\!+\!jd,n)=1.\,$ Such a $j$ exists by here and $\,\gcd(k,d)=1$

Bill Dubuque
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