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Let $M$ be a commutative monoid for which the cancellation laws hold. Given $a,b\in M$, show that if $a$ and $b$ are relatively prime then $\gcd(a,b)=\mathcal{U}(M)$, where $\mathcal{U}(M)$ is the group of units of the monoid $M$.

I find the analogous proof for integers and integral domains everywhere, but I need to prove this for monoids. Thanks in advance.

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  • $\begingroup$ What is your definition of "relatively prime"? $\endgroup$ Oct 28 '19 at 19:55
  • $\begingroup$ For $a,b\in M$, $a$ is relatively prime to $b$ if for every $x\in M$ such that $a\mid bx$ then $a\mid x$. In the statement that I gave, $a$ is relatively prime to $b$ and $b$ to $a$. $\endgroup$ Oct 28 '19 at 20:03
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Hint $\,\ c\mid a,b\,\Rightarrow\, a\mid b(a/c)\,\Rightarrow\, a\mid a/c\,\Rightarrow\, c\mid 1$

Remark $ $ There are various notions of $\,a,b\,$ are "relatively prime / coprime" in rings, e.g. below excerpted from the linked post. You might find it instructive to investigate their relationships also.

[0] $\ \ \ x\mid a,b\,\Rightarrow\, x\mid 1$

[a] $\ \ \ a\mid bx \,\Rightarrow\, a\mid x $

[b] $\ \ \ a,b\mid x \,\Rightarrow\, ab\mid x $

[c] $\ \ \ (a)\cap (b) = (ab)$

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  • $\begingroup$ What does the notation $(a)\cap (b)=(ab)$ mean? sorry I'm sorta new to abstract algebra. Also, what did you mean when you wrote "[0], [a],[b], and [c]"? $\endgroup$ Oct 28 '19 at 23:11
  • $\begingroup$ @SimonFraser $\ (a)\,$ denotes the ideal $aR$ in a commutative ring $R$, i.e. all multiples of $a$, The brackets are just labels (consistent with the linked post). S $\endgroup$ Oct 28 '19 at 23:15
  • $\begingroup$ @SimonFraser The definition of gcd doesn't vary but "relatively prime" may. $\endgroup$ Oct 28 '19 at 23:22
  • $\begingroup$ @Simon No, it is correct as written, see here. $\endgroup$ Oct 29 '19 at 17:28
  • $\begingroup$ @BillDubuqueI completely misunderstood. I was thinking of something completely illogical. You are correct. $\endgroup$ Oct 29 '19 at 18:06

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