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I want to proof the following:

If an element $[a] \in \mathbb{Z}$/m$\mathbb{Z}$ is a unit, then $a$ and $m$ are coprime, i.e. $gcd(a,m) = 1$.

Proof:

Let $[a]$ be a unit and $[a]^{-1} = [b]$.

$\implies ab \equiv 1 \pmod m$

$\implies$ there is a $q \in \mathbb{Z}$ with $ab = qm+1$

$\implies ab-qm = 1$

$\implies \gcd(a,m)=1$

My question is: What is the reason that I am able to conclude the last step - $gcd(a,m) = 1$. Is it why I can divide the whole equation $ab-qm=1$ by a maximum of 1, because when I divide the equation by a number $> 1$, the right-hand-side would be a fraction and that is not allowed since we are in $\mathbb{Z}$. That was my guess but I am not sure at all.

Thanks for your answers.

Stefan4024
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Banause
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  • Since you proved this implication you should also try the reverse: this theorem is actually an "if and only if" statement. – Davide Trono Jun 17 '21 at 07:33
  • See the standard proof in the linked dupe (set $,b=1).,$ If is is now clear please delete the question since it is a dupe of many others. – Bill Dubuque Jun 17 '21 at 07:44

1 Answers1

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Basically, you are on the right track. More precisely, if $d$ is a common divisor of $a$ and $m$, then $d$ divides $ab$ and $mq$, as well as their difference. Therefore we get:

$$d \mid ab - mq =1$$

As the only positive divisor of $1$ is itself, we get $d=1$. Hence $gcd(a,m)=1$.

Stefan4024
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