My task is to draw Hasse diagram. I have "$|$" relation, which means divide. The relation is defined by partial order $D = \{ 1, 2, 3, 5, 6, 10, 15, 30 \}$ including divisors to number $30$.
My Logic:
I would draw a straight chain line from $1$ to $30$, because all of numbers are divisors to $30$. Does my logic correct? If not why, and how must looks like Hasse diagram?
Best Answer
There is a similar Hasse diagram in the example here.
In this case, it's the lattice of divisors of $60$, under the same order relation (notice this posets are always lattices, under the operations $\gcd$—greatest common divisor—and $\mathrm{lcm}$—least common multiple).
In the case of $30$, it's a square-free number (not divisible by $n^2$, for any $n$); in particular, it's not divisible by $4$, and that's the difference between the diagram of $D_{30}$ and the one of $D_{60}$.
More to the point, $D_{30}$ is the diagram given by all the elements below $30$, in the diagram of $D_{60}$ in the link above.
As an aside, the diagram of $D_n$ is a chain, as you were implying, iff $n = m^k$, for some integer $m$ and positive integer $k$.
As another aside, in the case of $D_{30}$, it is a Boolean lattice, because $30$ is square-free. It can be made into a Boolean algebra by defining the complement $n' = 30/n$.