When I was at the JMM in San Antonio in January, I was happy to catch a workshop by George Hart:

I went to some great talks at the JMM, but a hands-on, interactive workshop was a nice change of pace in the schedule. Having seen some of George’s artwork before, I couldn’t resist. In the workshop, he taught us to build his sculpture/puzzle which he calls the 12 Card Star. Here’s what mine looked like:

He supplied the necessary materials: 13 decks of cards, all pre-cut (presumably with a band saw), like this:

We each took 13 card from these decks—the 13th, he said, was “just in case something terrible happens.”

He showed us how to put three cards together:

Then he gave us a clue for assembling the star: *the symmetry group is the same as that of a … *

*Wait!* Let’s not give that away just yet. Better, let’s have some fun figuring out the symmetry group.

Let’s start by counting how many symmetries there are. There are twelve cards in the star, all identically situated in relation to their neighbors, so that’s already 12 symmetries: given any card, I can move it to the position of my favorite card, which is the one I’ll mark here with a blue line along its fold:

But my favorite card also has one symmetry: I can rotate it 180$^\circ$, flipping that blue line from end to end around its midpoint, and this gives a symmetry of the whole star. (Actually, this symmetry is slightly spoiled since I drew the five of hearts: that heart right in the middle of the card isn’t symmetric under a 180$^\circ$ rotation, but never mind that. This would be better if I had drawn a better card, say the *two* of hearts, or the five of *diamonds*.)

So there are $12\times 2 = 24$ symmetries in total, and we’re looking for a group of order 24. Since $24 = %4 \times 3 \times 2 \times 1 =

4!$, the most obvious guess is the group of permutations of a 4-element set. Is it that? If so, then it would be nice to have a concrete isomorphism.

By a *concrete* isomorphism, I mean a specific 4-element set such that a permutation of that set corresponds uniquely to a symmetry of the 12-card star. Where do we get such a 4-element set? Well, since there are conveniently *four* card suits, let’s get a specific isomorphism between the symmetry group of Hart’s star and the group of permutations of the set

At the workshop, each participant got all identical cards, as you can see in the picture of mine. But if I could choose, I’d use a deck with three 2’s of each suit:

From this deck, there is an essentially unique way to build a 12-Card Star so that the isomorphism between the symmetry group and the group of permutations of suits becomes obvious! The proof is `constructive,’ in that to really convince yourself you might need to *construct* such a 12-card star. You can cut cards using the template on George’s website. He’s also got some instructions there. But here I’ll tell you some stuff about the case with the deck of twelve 2’s. % and from these it will be clear that (if you succeed) your star will have the desired properties.

First notice that there are places where four cards come together, like this:

In fact, there are *six* such places—call these the six **4-fold rotation points**—and it’s no coincidence that six is also the number of *cyclic orderings* of card suits:

Now, out of the deck of twelve 2’s, I claim you can build a 12-card star so that each of these cyclic orderings appears at one 4-fold rotation point, and that you can do it in an essentially unique way.

This should be enough information to build such a 12-card star. If you do, then you can check the isomorphism. Think up an permutation of the set of suits, like this one:

and check that you can rotate your 12-card star in such a way that all of the suit symbols on all of the cards in the 12-card star are permuted in that way. The rest follows by counting.

Sometime I should get hold of the right cards to actually build one like this.

Of course, there are other ways to figure out the symmetry group. What George Hart actually told us during the workshop was not that the symmetry group was the permutation group on 4 elements, but rather that the symmetry group was the same as that of the *cube*. One way to see this is by figuring out what the `convex hull’ of the 12-card star is. The convex hull of an object in Euclidean space is just the smallest convex shape that the object can fit in. Here it is:

This convex polyhedron has eight hexagonal faces and six square faces. You might recognize as a truncated octahedron, which is so named because you can get it by starting with an octahedron and cutting off its corners:

The truncated octahedron has the same symmetry group as the octahedron, which is the same as the symmetry group of the cube, since the cube and octahedron are dual.

Thanks to Chris Aguilar for the Vectorized Playing Cards, which I used in one of the pictures here.