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For context, this question comes from Joseph Gallian's Contemporary Abstract Algebra (9 ed.):

Let $r \in U(n)$. Prove that the mapping $\alpha: \mathbb{Z}_n → \mathbb{Z}_n$ defined by $\alpha(s) = sr\bmod n$ for all $s$ in $\mathbb{Z}_n$ is an automorphism of $\mathbb{Z}_n$.

With this, I naturally wanted to show injectivity first, so I started by supposing $\alpha(x) = \alpha(y)$ for some $x, y \in \mathbb{Z}_n$. This means $xr\text{ mod } n = yr\bmod n$, which (correct me if I'm wrong) is the same as saying $xr \equiv yr\pmod{n}$. From here, I want to show that $x \equiv y\pmod{n}$, which is equivalent to the question asked in the title since $r \in U(n)$ implies $\gcd(n, r) = 1$.

So, because $xr \equiv yr\pmod{n}$, we have that $xr - yr = nq$ for some $q \in \mathbb{Z}$, and overall $$r(x - y) = nq$$ From here, I claim that because $\gcd(n, r) = 1$, $n$ cannot divide $r$, so $n \mid x - y$ and so $x \equiv y\pmod{n} and we are done.

I am posting this question to ask whether this solution is correct, since I feel that the implication I used from $n$ and $r$ being coprime is a little hand-wavy--can't we also say that $x - y$ can divide $n$? My best argument against this is that if we divide both sides by $n$, we see that $$ \frac{r(x - y)}{n} = r\frac{x - y}{n} = q \neq \frac{r}{n}(x - y) $$ because the overall quotient must be an integer, $n$ must divide either $r$ or $x - y$, and $n$ cannot divide $r$, so $n$ must divide $x - y$. Similar logic for dividing both sides by $x - y$ wouldn't apply since we have no information about $q$, so we can't necessarily say that $x - y$ divides $n$.

However, I don't know if this is the best argument, since (1) we are working in the context of a group under addition, so an argument using division seems shaky to me, and (2) although the logic seems fine, it doesn't seem "formal" enough. Is my logic or my result correct?

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    $\gcd(n,r) =1$ implies $r$ has an inverse modulo $n$. Then right-multiply $xr \equiv yr$ by that inverse... – Eric Towers Apr 02 '24 at 15:28
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    Your proof is correct provided you remove "$n$ cannot divide $r$, so", and the argument is simply Gauss's lemma: if $\gcd(n,r)=1$ and $n\mid rz$ then $n\mid z$. – Anne Bauval Apr 02 '24 at 15:29
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    For a solution-verification question to be on topic you must specify precisely which step in the proof you question, and why so. This site is not meant to be used as a proof checking machine. – Bill Dubuque Apr 02 '24 at 15:42
  • "from here, I claim that because gcd(n,r)=1, n cannot divide r, so n∣x−y and so x≡y (mod n) and we are done." Solely as stated that's not enough and not true. $n$ is not prime so $n$ divide one or the other need not be true. Instead as $\gcd(n,r)$ then $n$ and $r$ have no factors in common so if $n|r(x-y)$ all factors of $n$ are in common with $x-y$ so $n\mid x-y$. "can't we also say that x−y can divide n? " No. We can say $(x-y)|qn$ but we know nothing about $q$. If $\gcd(x-y,q)=1$ we could say that but if $\gcd(x-y,q)\ne 1$ we can not. – fleablood Apr 02 '24 at 16:01
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    A stronger and perhaps more useful statement is that if $xr \equiv yr \pmod n$ and $\gcd(n,r)=d$ then $x\equiv y \pmod{\frac nd}$. – fleablood Apr 02 '24 at 16:04
  • You seem to be assuming that if $a\mid bc$ then we can conclude $a\mid b$ or $a \mid c$. That is ONLY true if $a$ is prime. If $a$ is not prime then it's possible that some factor of $a$ will divide $b$ while the remainder factor will divide $c$. For example $6\mid 8\times 3$ but $6\not \mid 8$ and $6\not \mid 3$. However $2$ is a factor of $6$ and $2\mid 8$ and so $\frac 62\mid 3$. – fleablood Apr 02 '24 at 16:08
  • To add to @EricTowers comment. $\gcd(n,r)=1$ implies $r$ has an inverse $\mod n$ (Because Bezout's th. says we can find $a,b$ so that $an + br=1$ so $br\equiv 1\pmod n$ and so $b$ is the mult. invers of $r$). So if $xr\equiv yr$ then $xrb\equiv yrb\implies x\cdot 1\equiv y\cdot 1$.... My addition to the comment: You don't have to actually find $b$ to do this. It's enough to know it exists. ... BTW an mult. invr to $r$ exists if and only if $\gcd(r,n)=1$. .. so it's enough to say $xr\equiv yr\implies xrr^{-1}\equiv yrr^{-1}\implies x\equiv y$. We don't need to know what $r^{-1}$ is. – fleablood Apr 02 '24 at 16:16

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