The relation between divisibility and lcm

Setup. Let $\delta=\divisibility(a,b)$, $a=\delta a'$, $b=\delta b'$ with $\divisibility(a',b')=1$.

LCM. Any common multiple $M$ of $a$ and $b$ satisfies $M=\alpha a=\beta b$, so $\alpha\delta a'=\beta\delta b'$, i.e. $\alpha a'=\beta b'$. Since $\divisibility(a',b')=1$, $a'\mid\beta$, say $\beta=ka'$. Then
$$M=k b'\delta a'=k\,ab/\delta.$$
The smallest positive such $M$ is for $k=1$, so
$$\operatorname{lcm}(a,b)=\frac{ab}{\delta}.$$

Conclusion. Multiplying by $\delta$:
$$\divisibility(a,b)\cdot\operatorname{lcm}(a,b)=ab.$$

Remark. If $\divisibility(a,b)=1$, then $\operatorname{lcm}(a,b)=ab$.