Greatest Common Divisor implies Least Common Multiple? 
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.)

It's true that $\rm\,lcm(a,b)\:$ exists $\Rightarrow$ $\rm\:gcd(a,b)\:$ exists - see the Theorem below.  But the converse fails, e.g. as here, in $\Bbb Q[x^2,x^3]$ we have $\,\gcd(x^2,x^3)=1\,$ but ${\rm lcm}(x^2,x^3)$ does not exist. Or, for simple well-known counterexamples in quadratic number fields see the paper linked below.
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 discussion see this post and  see also Khurana, On GCD and LCM domains, and for basic properties of gcd and lcm (in cancellative commutative monoids) see also Section 1.6, Factorization in Commutative Monoids,
in Robert Gilmer, Commutative Semigroup Rings, 1984.
