What does a distributed lattice have to do with GCD and LCM?

Question

$\newcommand{\lcm}{\operatorname{lcm}}$I am lost while following this explanation:

Let $$A(g, i) = \gcd(F_{g}, \lcm(F_{a_1}, F_{a_2}, \ldots , F_{a_i}))$$ and $$X = \lcm(F_{a_1}, F_{a_2}, \ldots , F_{a_{i - 1}})$$

Then $A(g, i) = \gcd(F_g, \lcm(X , F_{a_i}))$

---

Because GCD distributes over LCM, and vice versa (distributive lattice), we can write:

$$A(g, i) = \lcm(\gcd(F_{g}, F_{a_i}), \gcd(F_g, X)))$$

When I looked what distributed lattice mean, I was not able to find any connection to what I see here. Can anyone explain me what is going on here?

Answer 1

As dtldarek noticed in his comment, gcd and lcm define a distributive lattice on the set of positive natural numbers. This answer is just an expanded version of this comment.

Given two positive natural numbers $a$ and $b$, denote by $a \wedge b$ their gcd and by $a \vee b$ their lcm. Then you can verify that $$ (a \wedge b) \wedge c = a \wedge (b \wedge c) \quad \text{and} \quad (a \vee b) \vee c = a \vee (b \vee c) $$

$$ (a \wedge b) \vee c = (a \vee c) \wedge (b \vee c) \quad \text{and} \quad (a \vee b) \wedge c = (a \wedge c) \vee (b \wedge c) $$ For instance, taking $a = 18$, $b = 24$ and $c = 15$ in the second line, you get

$(18 \wedge 24) \vee 15 = 6 \vee 15 = 30\ $ and $(18 \vee 15) \wedge (24 \vee 15) = 90 \wedge 120 = 30$

and

$(18 \vee 24) \wedge 15 = 72 \wedge 15 = 3\ $ and $(18 \wedge 15) \vee (24 \wedge 15) = 3 \wedge 3 = 3$.

Can you see now the connection with distributive lattices?