Ok, so here's the story: I am reading a book on algebra and, via some exercises, discovered that in any group $G$, the order of $x \cdot y$, written $o(x \cdot y)$, equals $o(y \cdot x)$. Now, this is trivial in an abelian group, but I was looking for examples of a non-abelian group (simply because the result was interesting) to see this happen.
Of course, I knew $GL(2, \mathbb{R})$ and the permutation groups. However, literally by chance (I had a ball in my hand), I realized that $m(90)$ degree rotations of a sphere - $m \in \mathbb{N}$ - are also a non-abelian group. (That is, let $G$ be the set of transformations of some particular distinguished point on a sphere through right angles, like the transformation forward, or clockwise, or right. The group operation is composition, and the identity is the "don't do anything" transformation.)
This discovery lends itself to a trick I find neat: take a sequence of operations, and find their order. (Like $o(\text{fwd cwise left left})$, which is $3$.) Then, take any cyclic permutation of that sequence, and you have a sequence of the same order.
If you actually have a sphere on you (I took a ball and drew a little arrow on it) and you ask someone for an eight-term sequence, and then instantly give them back a (well-mixed, unrecognizable) sequence of the same order, and show them that you're right, on the spot - this is impressive.
Well, okay... actually, that's the thing. I find it impressive; my friends don't. This bums me out.
So, my question: how can I amp this trick up? I thought about memorizing a "basis" for all sequences of a certain length, i.e., knowing enough sequences that any of them are commutation-equivalent with the ones I know; but, unfortunately, this is impractical. Does anybody know how to make this cooler?