Need Help With Using Cycle Permutation to Rotate Squares I'm a high school math teacher and I'm working with some calculus students who are exploring other areas of math for an end-of-year project. I thought I would introduce them to some basic group theory by exploring the symmetries of a square. However, I'm a little rusty, and I'm getting some results that don't make (physical) sense.
This is how I'm laying out the square, and I've calculated the elements of the dihedral group as such in cycle notation ($R$ represents a $90$ degree rotation counter-clockwise):
$R=(1234)$
$RR=(13)(24)$
$RRR=(1432)$
$V=(12)(34)$
$H=(14)(23)$
$D_1=(13)$
$D_2=(24)$
I can put the Cayley table together just fine, but I run into a problem when I try and actually compose two permutations that involve either $D_1$ or $D_2$. For example, flipping across the $D_1$ axis and then rotating counter-clockwise by $90$ degrees should be equivalent to $R$ composed with $D_1$. And this is equivalent to $(1234)(13)=(14)(32)=H$. However, when I do those two actions physically with an actual square, I get $V$ - that is, the square is in a position equivalent to merely flipping around the vertical axis - not $H$. I've tried compositions with all sorts of other situations, and they all work, so it's only compositions with $D_1$ or $D_2$ involved that give results not matching their physical representation.
I'm really trying to figure out what's going on here, but it just isn't clear. Can anyone point me in the right direction? Thank you so much!
 A: There are two ways to think about composing permutations with this algebraic notation. I will illustrate using your "first $(13)$, then $(1234)$" example:

*

*The corners of the square are labeled, and the labels move when you move the square. This way we do get $(14)(23)$ as the result.

*The spots on the table where the square lies are labeled, and the labels stay put when you move the square. This way we get $(12)(34)$ when we compose the two given permutations in the prescribed order.

When you move the square in front of you, I suspect you're using the latter approach, thinking of $R$ as a counterclockwise rotation to you, no matter how the square has previously been moved. But your calculations (assuming you calculate permutation products from right to left) use the former notion, where $R$ is a counterclockwise rotation relative to the square, which after flipping it over looks clockwise to you.
Fortunately, the remedy is relatively simple. Merely evaluate the product in the opposite order of application. I.e. use $(1234)$ first to see that, for instance, $2$ goes to $3$, then use $(13)$ to see that that $3$ then becomes $1$. In total $2\mapsto 1$, which is on the way to building the end result of $(12)(34)$. (Whether you write this product as $(1234)(13)$ or $(13)(1234)$ is up to you; either way, it will have to either be constructed "backwards" or calculated "backwards".) This may confuse students, but after demonstrating with exactly an example like yours, they may have to admit that that's the way that works.
Or you could try the other way around, manipulating the square relative to the square's local labels, but that also has some downsides. Maybe it's easier for the students to swallow conceptually, but I suspect it's more difficult for them to actually perform correctly with on square in front of them. Actually putting labels on the square, on both sides, may help in this endeavor.
It is, in fact, a very cool piece of trivia that swapping between "the reference labels move with the transformation" and "the reference labels stay put" exactly corresponds to reversing which order you apply the factors in the product during calculation. Some clever use of conjugation proves this generally.
A: Remember: You are composing (multiplying) cycles right-to-left. That means you should do D1 before R on the physical square to correspond to $(1234)(13)$.
