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I have the following exercise:

Having the lattice $D_{165}$ of the divisors of 165, ordered by divisibility

1) Draw Hasse Diagram;
2) Find all complements;
3) Check if it's a distributive lattice;
4) Check if it's a Boolean lattice.

My development:

The divisors of 165 are $D_{165} = \left \{ 1,3,5,11,15,33,55,165 \right \}$

1)
this is the diagram I have done, look at the figure on the left:
Hasse Diagram for $D_{165}$

2)
in this table I have put an asterisk when an element has its complement.
i.e. every time is true this condition $\forall a \in D_{165}$
$a \land a' = 0 \quad \mbox{ and } \quad a \lor a' = 1$ \begin{array}{c|c} \, & 1 & 3 & 5 & 11 & 15 & 33 & 55 & 165 \\ \hline 1 & - & - & - & - & - & - & - & * \\ \hline 3 & - & - & - & - & - & - & * & - \\ \hline 5 & - & - & - & - & - & * & - & - \\ \hline 11 & - & - & - & - & * & - & - & - \\ \hline 15 & - & - & - & * & - & - & - & - \\ \hline 33 & - & - & * & - & - & - & - & - \\ \hline 55 & - & * & - & - & - & - & - & - \\ \hline 165 & * & - & - & - & - & - & - & - \\ \end{array} so this lattice is NOT a Complemented lattice.

3)
it is NOT possible to get a sublattice like one of the two nondistributive lattices on the right in the previous image.
So the lattice $D_{165}$ is distributive.

4)
Since NOT all elements in the lattice have a complement, i.e. it is not a Complemented lattice. The lattice $D_{165}$ is not a Boolean lattice.

Please, can you tell me if the exercise is correct? Can you give me any suggestion?
Many thanks!

1 Answers 1

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By re-arranging the elements of the Hasse diagram, you can check that it's isomorphic to the 8 elements Boolean lattice (aka, the cube).
So it is indeed complemented, and I suppose it answers all of your queries.

By the way, in the same page I put a link above, in the section examples, you can see that whenever $n$ is a square-free number, as it is the case of $165$, the lattice of divisors is Boolean.
Clearly this is a necessary and sufficient condition.

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    Still I'm not convinced that my lattice $D_{164}$ is complemented. For example, if you look to my Hasse diagram, see 33 and 165: $33 \land 165 = 165 \land 33 = 33 \ne 0, \quad \mbox{ and } \quad 33 \lor 165 = 165 \lor 33 = 165 = 1$. or is it wrong? – 2017-02-11
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    @JB-Franco: here's the complements. $1'=165$, $3'=55$, $5'=33$, $11'=15$, and the other follow by symmetry. More generally, for the lattice of divisors of a square-free integer $n$, we have $m' = n/m$. I don't understand why you picked those meets and joins... – 2017-02-11
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    Ok! So I have calculated the right complements for each element, and my table is right! And therefore my lattice is complemented and is a Boolean Algebra! – 2017-02-11
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    @JB-Franco By the way, in the link I send in my answer, the cube (of which your lattice is an instance) is in the beginning of the page, on the right, under the name "Boolean lattice of subsets" – 2017-02-11
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    @JB-Franco. Yes, that's what I said: it is a Boolean lattice. Therefore, with the complementation it forms a Boolean algebra. – 2017-02-11
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    Yes! In my head I was wrongly convinced that every element must be comlement with every element! Your answer has been useful! Are the other points of the exercise right? – 2017-02-11
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    @JB-Franco, once it is a Boolean lattice, it is distributive (as you had already concluded) and complemented. You have the complements and the Hasse diagram (consider the re-arranged form in the link I gave, but the one in yours it's ok). It seems like you have everything. By the way you can forget about that table. – 2017-02-11
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    Rearranging the elements in my Hasse diagram in the way as in the cube, is it enough to prove that there is an isomorphism? – 2017-02-14
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    @JB-Franco, yes, because the cube is a Boolean lattice, as it is said in that page. If a lattice has the same Hasse diagram as another lattice, then they are isomorphic (the isomorphism is just a re-labelling of the elements). – 2017-02-15