Cancellative Abelian Monoids

Question

Is there an example of cancellative Abelian monoid $M$ in which we may find two elements $x$ and $y$ such that they have a least common multiple but not a greatest common divisor?

Answer 1

Let $a,b \in M$ and $m \in M$ be a least common multiple of $a$ and $b$. As $ab$ is a common multiple of $a$ and $b$, we must have $m \mid ab$, say $md = ab$ for some $d \in M$. Now let $t \in M$ be any common divisor of $a$ and $b$, say $a = st$, $b = rt$, define $m' = rst$. Then $m'$ is a common multiple of $a$ and $b$, so for some $d' \in M$ we have $m' = d'm$. Now we have $$ d'tm = rst^2 = ab = dm$$ As $M$ is cancellative, this implies $d't = d$. So $t$ is a divisor of $d$. As $t$ was arbitrary, $d$ is a greatest common divisor of $a$ and $b$.

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Addendum: To see that $d$ divides $a$ and $b$, note that $m$ is a multiple of $a$ and $b$, say $m = aa' = bb'$ for some $a', b' \in M$. Then we have $$ ab = md = aa'd, \quad ab = b'db $$ therefore by cancellation, $b = a'd$ and $a = b'd$.

Answer 2

Theorem $\,\ [a,b]\,$ exists $\,\Rightarrow\, (a,b) = ab/[a,b],\: $ for $\ (x,y)=\gcd(x,y),\,$ $\,[x,y] = {\rm lcm}(x,y)$

Proof $\ \ c\mid a,b \iff a,b\mid ab/c \iff [a,b]\mid ab/c\iff c\mid ab/[a,b]$