Group of sphere transformations, impressing friends

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?

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I love the nature of this question. Group Theory and Being Popular –  000 Nov 17 '12 at 0:11
"Commute the operations however you like "... Sorry that doesn't always work. –  user641 Nov 17 '12 at 0:13
@TaraB My mathematician friends were moderately impressed, the others weren't. Really, I'm aiming at the mathematician friends. (The heathens are no fun.) –  user1296727 Nov 17 '12 at 0:36
Ah, I would have thought you'd have more hope in general of impressing non-mathematicians, since the mathematicians might just see that you are taking cyclic permutations of their sequences, and realise that this would produce an element of the same order, and hence not be all that impressed. –  Tara B Nov 17 '12 at 0:38
By the way, I would recommend changing the bit about 'commute these operations however you like' in your question, otherwise people reading it are likely to think you think something false is true (not everyone reads the comments). –  Tara B Nov 17 '12 at 0:44

The magic for the mathematicians fades away if they know the basic

$$ba=a^{-1}(ab)a$$

which show that in any group, the elements $\,ab\,,\,ba\,$ are conjugate and thus have the very same order.

The non-mathematicians most probably will not only not understand the above, but in fact they won't even be interested at all.

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Well, for starters I might use a different object. Since the nonabelian group you are describing is the group of symmetries of a cube (or an octahedron) why not use one of those? Maybe use a six sided die so you can track rotations fairly easily?

There are a some other interesting thought puzzles about group order. One of my favorite is identifying elements of $S_{52}$ with ways to shuffle a deck of cards. Finding the largest order of an element in that group yields an interesting fact: though there are about $10^{68}$ ways to shuffle a deck, any deterministic shuffle will yield the original deck after $\leq 360360$ iterations!

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I think the suggestion of using a die rather than a ball is good. –  Tara B Nov 17 '12 at 0:39
The excitement you've displayed for a wholly impractical method of shuffling is the mark of a true mathematics enthusiast. I'd have to say I'm right there with you; it impresses me that any amount of iterations can yield the original deck. –  000 Nov 21 '12 at 15:05
Talking about $52$-card decks, try computing the order of the most obvious permutation to do by hand: ditribute the cards one by one in fout piles, then stack the four piles in order. (In mathematical terms, write the cards to a $4\times13$ matrix by columns, and then collect them by rows.) This permutation has amazingly small order, which I can neither remember nor explain. –  Marc van Leeuwen Nov 21 '12 at 15:42
I just computed, the above order is $4$. –  Marc van Leeuwen Nov 21 '12 at 15:52

I'm going out on a limb here and thinking entirely outside the box (or, in your shape, sphere).

Why not do something a tad more simple, but much more intriguing (in the sense that it is more imaginative)? Being 16 and enjoying mathematics is difficult, so I have had to think outside of the sphere. Here's a scenario I thought about:

There's this classic rice problem involving someone (a mathematician in hiding?) and a king. The someone does something awesome for the king and the king is just like, "I will give you any wish you want." So the someone responds, "I want you to fill a chess board with rice in the following process: Put one grain on the first space, double that and put it on the next space, and so forth."

As you may know, the total amount of rice is a geometric series. Ultimately, in the story, the someone bankrupted the king's kingdom after the first few spaces (the 18th, perhaps?). Ignoring the terrible ambiguity of my story, here's what I propose: Find a nonmathematical friend (or possibly a mathematical one) and simply ask them, "Can rice bankrupt a kingdom?"

Presuming you have good social skills and are a good story teller, there's a fun and mathematical conversation right there!

P.S. If they say yes, just ask them how. If they provide a good explanation or no explanation, tell them you have a clever one and tell them the story.

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