Question

A 3×3 board with 9 squares on the front and 9 squares on the back is to be colored with 4 colors.

The squares can be red, green, blue or yellow. For example, one side might be a red and green checkerboard

pattern and the other side might be a blue and yellow checkerboard pattern. How many different colorings are

possible? Use the symmetry group of the square, D4, which has order 8.

This is an abstract algebra homework.

Answer 1

For One side :

	Red	Green	Blue	Yellow
Red	0	1	2	3
Green	1	0	4	5
Blue	2	4	0	6
Yellow	3	5	6	0

Here, Red+Red = Blue+Blue = Green+Green = Yellow+Yellow = 0

Red+Green = 1 = Green+Red

Red+Blue = 2 = Blue+Red

Red+Yellow = 3 = Yellow+Red

Green+Blue= 4 = Blue+Green

Green+Yellow = 5 = Yellow+Green

Blue+Yellow = 6 = Yellow+Blue

Therefore, maximum number of possibilities for each side is 6.

For both sides :

	R+G	R+B	R+Y	G+B	G+Y	B+Y
R+G	0	0	0	0	0	1
R+B	0	0	0	0	2	0
R+Y	0	0	0	3	0	0
G+B	0	0	4	0	0	0
G+Y	0	5	0	0	0	0
B+Y	6	0	0	0	0	0

Since, the front and back side of the board is to be colored with 4 different colors.

Therefore, there are only 6 ways to color the front and back side of the 3X3 board with 4 different colors.