For what values of n the set of divisors of n under partial order relation divides is a complemented lattice

Question

For what values of n the set of divisors of n under partial order relation divides is a complemented lattice?

Complemented lattice: If every element of a lattice has at least one complement, it is called complemented lattice.

Answer 1

If $n$ is a product of distinct primes, then this lattice is complemented. The top element is $n$, the bottom element is $1$. Any divisor $j$ has $n/j$ for a complement, since $n/j$ and $j$ are relatively prime (i.e., have meet $1$) and have join $n$.

On the other hand, if $\gcd(p, n) > p$ for some prime $p$, i.e., if $n$ has a squared-prime factor $p^2$, then $n/p$ has no complement in the lattice. So we have characterized all $n$ which yield a complemented lattice.

Answer 2

The lattice in question is a product over all primes$~p$ (or just those that divide$~n$) of partial orders (lattices) that are chains of length the multiplicity of$~p$ as factor of$~n$. A product lattice is complemented if and only if all its factors are. So it boils down to finding which chains are complemented lattices; it is obvious that that chains of length $l\leq1$ are and those of length $l>1$ are not. So the necessary and sufficient condition is that every prime have multiplicity $0$ or $1$ in$~n$, in other words that $n$ be square-free.