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If $a$ and $b$ are elements in an integral domain with unity 1$\neq$0. Show that $a$ and $b$ have a least common multiple if $a$ and $b$ have a highest common factor.

More generally there is a problem of showing that if any finite non-empty non-zero subset of the ring has a highest common factor, then any finite non-empty non-zero subset of the ring has a least common multiple. (Actually the converse of the preceding sentence is also true.)

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    Note that your comment in the prior question has it the wrong way around.2012-04-14

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It's true that $\rm\:lcm(a,b)\:$ exists $\Rightarrow$ $\rm\:gcd(a,b)\:$ exists, but the converse fails.

THEOREM $\rm\;\; (a,b) = ab/[a,b] \;\;$ if $\;\rm\ [a,b] \;$ exists.

Proof: $\rm\quad\quad d\:|\:a,b \;\iff\; a,b\:|\:ab/d \;\iff\; [a,b]\:|\:ab/d \;\iff\; d\:|\:ab/[a,b] \quad\;\;$ QED

For further discusson see this post and see also Khurana, On GCD and LCM domains

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    @user710587 $\rm\:d\:|\:b\:$ $\Rightarrow$ $\rm\:ad\:|\:ab,\:$ $\Rightarrow$ $\rm\:d\:|\:a(b/d).\:$ By definition of LCM, $\rm\:a,b\:|\:ab\:$ $\Rightarrow$ $\rm\:[a,b]\:|\:ab.\:$ And $\rm\:d\:|\:ab/[a,b]\:$ $\Rightarrow$ $\rm\:d[a,b]\:|\:ab\:$ $(\Rightarrow$ $\rm\:d\:|\:ab)\:$ $\Rightarrow$ $\rm\:[a,b]\:|\:ab/d,\:$ etc. These are inferences one can make mentally with a little practice, analogous to scaling and reducing fractions. See the linked post for the *universal* definitions of GCD and LCM and further discussion.2012-04-14