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The set of all divisors of a square-free number, partially ordered by divisibility, is trivially isomorphic to the set of all subsets of the set of prime factors, partially ordered by inclusion.

Are there examples that are almost as simple as that one where the isomorphism would be somewhat unexpected, and that would be quickly understood by students seeing the concept of partial ordering and the concept of isomorphism for the first time?

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The sets of divisors of two different numbers whose prime factorization happen to have the same structure? E.g. divisors of 60 versus divisors of 90 or 525. –  Henning Makholm Apr 19 '12 at 15:58
    
Subsets by inclusion and sequences of $0$'s and $1$'s with partial order given by $x \le y$ if $x_i \le y_i$ for all $i$. Too simple? –  André Nicolas Apr 19 '12 at 16:01
    
Every poset $(A, \leq)$ is isomorphic to a subposet of $(\mathbb P(A), \subseteq)$. If they already seen groups and fields: subfields of $Q(\sqrt 2, \sqrt 3)$ and subgroups of $\mathbb Z_2 \times \mathbb Z_2$. –  sdcvvc Apr 19 '12 at 16:27
    
@HenningMakholm : I decided to set an exercise using your example, so if you want to post that as an answer, I'll accept it. –  Michael Hardy Apr 20 '12 at 22:29

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up vote 2 down vote accepted

By transitivity of isomorphism one can extend your initial idea to something that looks slightly more unexpected:

Consider the sets of divisors of two different numbers, each partially ordered by divisibility. If the two given numbers happen to have prime factorizations with the same structure (i.e., the same multiset of exponents), then the two sets of divisors are isomorphic partial orders.

For example the numbers could be $60=2^2\cdot 3^1\cdot 5^1$ versus $90=2^1\cdot3^2\cdot 5^1$ or $525=3^1\cdot5^2\cdot 7^1$.

This also provides an opportunity to make the additional point that some divisors such as $10$ (for the 60/90 example) are points in both partial orders yet there is no isomorphism that maps $10$ to $10$..

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