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File 13701346985.png - (13.60KB , 927x879 , webgl.png )
323 No. 323
Background: I'm a CS major, I've had a handful of calculus classes and only one discrete math class. I loved complex analysis and hated real analysis. I've never had an abstract algebra class. So my question is as follows:

I want a closed field (ring? I don’t know the difference) of six elements, each corresponding to the face of a cube. Call them S := {a, b, c, d, e, f}. I want to have two functions (“operators” I guess) defined, one to travel vertically, one horizontally. Let’s call them v:S→S and h:S→S. Obviously they’ll need to be invertible, so that v-1(v(x)) = v(v-1(x)) = x, and they’ll have to be sort-of almost-involutions so that v4(x) = (v-1)4(x) = x. Also, v(v(x)) = h(h(x)).

Apologies, the superscript notation was lost there. Pretend all those -1s and 4s are superscript.

So how can I do this? Is it possible to have S ⊂ N, or would I have to let S ⊂ Z or even S ⊂ Z × Z?

Is there a way to solve this algebraically without knowing the nature of S besides its size?

(It’s for a puzzle game. I hate brute force solutions and huge switches but if I'll use them if I have to.)
>> No. 325
Essentially what you would have is called a group.
http://en.wikipedia.org/wiki/Group_(mathematics)

The elements of your group would be all sequences of h and v (taking into account the relations you described), together with the empty sequence to serve as the identity.
h and v are said to act on the points which make up the cube(or equivalently the vertices).
Since the cube has 8 vertices and h and v permute them, the group you get would be a subgroup of the set of all permutations on the set of vertices. This is usually denoted S8. S8 isn't a subgroup of N, Z or ZxZ.
http://en.wikipedia.org/wiki/Permutation_group


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