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APA
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Engineering
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Math Problem
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Decomposition (Math Problem Sample)

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Abstract Algebra

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Content:


Decomposition
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The symmetry group of a cube is isomorphic to S4 the permutation group on 4 elements. If we number the vertices of the cube from 1 to 4, and where opposite vertices are given the same number, the permutation corresponding to asymmetry can be read out from one of the faces. If we e.g. let the top face be numbered 1, 2, 3, 4 clockwise, then perform a rotation around the 1-diagonal, we see that the numbers of the top face have changed to 1, 3, 4, 2, so this symmetry corresponds to the permutation (234). Proceeding in this way we can classify the 24 elements of the symmetry group by their order.
The set X of all paintings of the cube by up to n different colors has 6^n6n elements. It is reasonable to say that two paintings are equivalent if they only differ by an asymmetry of the cube. Thus the number of different paintings is the number of orbits in X under the action of the symmetry group of the cube. To tool for computing this number is the Burnside Formula: 
Symmetry

Permutation

Order

Number

Xg

Identity

e

1

1

n6

Edge-midpoint rotation

(12), (13), (14), (23), (24), (34)

2

6

n3

Face-midpoint rotation

(1234), (1432), (1324), (1423), (1243), (1342)

4

6

n3

Square of Face-midpoint rotation

(12), (34), (13), (24), (14), (23)

2

3

n4

Diagonal rotation

(123), (132), (124), (142), (134), (143), (234), (243)

3

8

n2

Inserting this into the Burnside Formula we get
r=rn=124n6+6n3+6n3+3n3+8n2
r=rn=n2n4+12n+3n2+824
r=rn=n2(n+1)n3-n2+4n+824
For some small values of n we get, r (1) =1, r (2) =10, r (3) =57, r (4) =240
Faces: Call this permutation representation X. The identity permutation does not move the cube at all, leaving the 6 faces fixed. The transposition (12) rotates the cube 180 degrees about the axis connecting the midpoints of edges a and b. This motion moves every face to a new location and therefore fixes no faces. The 3-cycle (123) rotates the cube 120 degrees about long diagonal number 4 and also fixes no faces. The 4-cycle (1234) rotates the cube 90 degrees about the axis z, leaving the top and bottom faces fixed but moving the other four around. Finally, (12) (34) rotates the cube 180 degrees about z, again leaving only the top and bottom faces fixed. Therefore, by the Fixed Point Formula, Xx has the values 6, 0, 0, 2, 2 on the conjugacy classes of r, (12), (123), (1234), and (12)(34), respectively. This gives us 
=1S462.1+02.6+02.8+22.6+22.3=12436+24+12=124.72=3
Since the only way to express 3 as the sum of squares is 3 = 12 + 12+ 12, we know that X is the direct sum of exactly three irreducible representations. See that 
=1S46.1.1+0.1.6+0.1.8+2.1.6+2.1.3=1246+12+6=124.24=1
=1S46.1.1+0.-1.6+0.1.8+2.-1.6+2.1.3=124.0=0
=1S46.3.1+0.1.6+0.0.8+2.(1.6+2.-1.3=124.0=0
=1S46.3.1+0.-1.6+0.0.8+2.1.6+2.-1.3=124.24=1
At this point, we know that X must contain a copy of W. Therefore, we have the decomposition X=U⊕V′⊕W
Vertices: Call this permutation representation Y. The identity permutation fixes all 8 vertices. Transpositions fix no vertices. Each 3-cycle fixes 2 vertices. Each 4-cycle fixes no vertices. Finally, elements in the

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