Homework #7
Due: October 18, 2010
Do the following exercises from Lax: Page 124: 9.l, 9.3, 9.5.
Do the following exercises from the Burnside Lemma supplement: Page 235: 6, 7, 12.
This supplement has many worked out examples that you may study for how to apply
Burnside’s Lemma.
9.1.
a) Find the number of different squares with vertices colored red, white, or blue.
b) Find the number of different 𝑚-colored squares for any 𝑚.
▶ Solution. The symmetry group of a square is
{
}
𝐷4 = 𝜖, 𝛼, 𝛼2 , 𝛼3 , 𝛽, 𝛼𝛽, 𝛼2 𝛽, 𝛼3 𝛽
and the set 𝑋 on which we want 𝐷4 = 𝐺 to act is the set of all functions from
the vertices of the square to the set of 𝑚 colors. Identifying the elements of 𝐷4
with permutations of the vertices of the square, as described on page 114, we
can compute the cycle type of each permutation determined by 𝑔 ∈ 𝐷4 and then
compute ∣𝑋𝑔 ∣ using Theorem 5.5.5, page 231 of the supplement (or corollary 10.3,
page 127 of Lax). The results are, for 𝑚-colors:
𝜌
Cycle Rep. 𝑙(𝜌) ∣𝑋𝜌 ∣
𝜖 (1)(2)(3)(4) 4
𝑚4
𝛼
(1 2 3 4)
1
𝑚
2
𝛼
(1 3)(2 4)
2
𝑚2
𝛼3
(1 4 3 2)
1
𝑚
𝛽
(1 2)(3 4)
2
𝑚2
𝛼𝛽 (1 3)(2)(4)
3
𝑚3
2
𝛼 𝛽 (1, 4)(2, 3)
2
𝑚2
𝛼3 𝛽 (1)(2 4)(3)
3
𝑚3
Hence, Burnside’s Lemma give the number of orbits as
1
𝑁 = (𝑚4 + 2𝑚3 + 3𝑚2 + 2𝑚).
8
This is the answer to part b). To get part a), let 𝑚 = 3 to get
1
168
𝑁 = (81 + 54 + 27 + 6) =
= 21.
8
8
◀
9.3. Find the number of different regular hexagons with vertices colored red or blue.
▶ Solution. This is a coloring problem with 𝑚 = 2 colors. The symmetry group is
the symmetry group of a regular hexagon, that is, the dihedral group 𝐷6 of degree 6,
which consists of 6 rotations and 6 reflections. If the vertices are labeled from 1 to 6
counterclockwise, then the elements of 𝐷6 can be represented by permutations in 𝑆6 .
Math 4023
1
Homework #7
Due: October 18, 2010
The set 𝑋 on which 𝐺 = 𝐷6 acts is the set of all functions from the set of vertices of
the hexagon to the set {red, blue}. The table of sizes of fixed point sets is
𝜌
𝑙(𝜌) ∣𝑋𝜌 ∣
(1)(2)(3)(4)(5)(6) 6
26
(1 2 3 4 5 6)
1
21
(1 3 5)(2 4 6)
2
22
(1 4)(2 5)(3 6)
3
23
(1 5 3)(2 6 4)
2
22
(1 6 5 4 3 2)
1
21
(1 6)(2 5)(3 4)
3
23
(1 2)(3 6)(4 5)
3
23
(1 4)(2 3)(5 6)
3
23
(1)(4)(2 6)(3 5)
4
24
4
24
(2)(5)(1 3)(4 6)
(3)(6)(1 5)(2 4)
4
24
The number of different color patterns is the number of orbits of 𝐺 acting on 𝑋, which
by Burnside’s Theorem is 𝑁 = (26 + 3 ⋅ 24 + 4 ⋅ 23 + 2 ⋅ 22 + 2 ⋅ 2)/12 = 13.
◀
9.5. A wheel is divided evenly into six different compartments. Each compartment can be
painted red or white. The back of the wheel is black. How many different color wheels
are there?
▶ Solution. This is similar to the previous problem except that instead of vertices,
you are asked to color six equal wedges of the circle, that we shall label 1 to 6 in a
counterclockwise direction. Since the back of the wheel is black, it follows that the
only symmetries are rotations. A reflection is equivalent to turning the wheel over
across a diagonal, but this would turn the back to the front and the back is different.
Thus, the symmetry group is the cyclic group 𝐺 = 𝐶6 = (𝛼) of rotations by multiples
of 60 degrees, so that ∣𝐺∣ = 6. This corresponds to the first six rows of the table in
problem 9.3, when viewed as permutations. Hence, the number of distinct patterns is,
according to Burnside’s Lemma:
1
𝑁 = (26 + 23 + 2 ⋅ 22 + 2 ⋅ 2) = 14.
6
◀
6. A rectangular design consists of 11 parallel stripes of equal width. If each stripe can
be painted red, blue, or green, find the number of possible patterns.
▶ Solution. This shape is like the Polya’s necktie problem in Example 5.5.3, Page
228. In our case, 𝑛 = 11. The group of symmetries is the group 𝐺 = {𝜖, 𝛼} where 𝛼
is the rotation about the center of the rectangle by 180 degrees. Labeling the stripes
from the left to the right as 1 to 11, the rotation 𝛼 corresponds to the permutation
𝛼 = (1 11)(2 10)(3 9)(4 8)(5 7)(6)
Math 4023
2
Homework #7
Due: October 18, 2010
so that 𝑙(𝛼) = 6. The identity is of course a product of 11 1-cycles. Since there are 3
possible colors, we conclude that 𝑋 is the set of all functions from the set of 11 stripes
to the set of 3 colors. Hence, ∣𝑋𝜖 ∣ = 311 and ∣𝑋𝛼 ∣ = 36 . Thus, the number of distinct
patterns is
1
177876
𝑁 = (311 + 36 ) =
= 88938.
2
2
◀
7. Each side of an equilateral triangle is divided into two equal parts, and each part is
colored red or green. Find the number of patterns.
▶ Solution. If each side of the triangle is divided into two parts, label all the parts
sequentially, starting at the left hand half of the bottom and proceeding counterclockwise. Thus, the bottom has parts 1 and 2, the right hand side has 3 and 4, and the left
hand side has 5 and 6. The symmetry group of this figure is the same as the symmetry
group 𝐷3 of the equilateral triangle, namely 3 rotations, 𝜖, 𝛼, and 𝛼2 of multiples of
120 degrees and three reflections across the line joining a vertex to the middle of the
opposite side. Since we are coloring each half of each side, the set 𝑋 on which 𝐷3 acts
is the set of all functions from the set 𝐶 of 6 half-lines to the set of 2 colors red or
green. Each element of 𝐺 corresponds to a permutation of 𝐶. The size of the fixed
point sets ∣𝑋𝜌 ∣ are recorded in the following table:
Symmetry Cycle Structure 𝑙(𝜌) ∣𝑋𝜌 ∣
𝜖
(1)(2)(3)(4)(5)(6) 6
26
𝛼
(1 3 5)(2 4 6)
2
22
2
𝛼
(1 5 3)(2 6 4)
2
22
𝛽
(4 5)(3 6)(1 2)
3
23
𝛼𝛽
(1 4)(2 3)(5 6)
3
23
2
𝛼 𝛽
(1 6)(2 5)(3 4)
3
23
By Burnside’s Lemma the number of distinct patterns is
1
96
𝑁 = (26 + 3 ⋅ 23 + 2 ⋅ 22 ) =
= 16.
6
6
◀
12. The interior of an equilateral triangle is divided into six parts by the medians. Each
part is painted with one of 𝑚 colors. Find the pattern.
▶ Solution. This problem is essentially the same as the previous problem, in that the
six half-lines on the edge of the triangle are replaced with the six interior subdivisions,
which we can again label from 1 to 6 starting in the lower left hand corner and proceeding counterclockwise. The symmetry group is still 𝐷3 and the set 𝑋 is the set of
Math 4023
3
Homework #7
Due: October 18, 2010
all functions from the set of 6 interior parts of the triangle to the set of 𝑚 colors. The
size of the fixed point sets ∣𝑋𝜌 ∣ are recorded in the following table:
Symmetry Cycle Structure 𝑙(𝜌) ∣𝑋𝜌 ∣
𝜖
(1)(2)(3)(4)(5)(6) 6
𝑚6
𝛼
(1 3 5)(2 4 6)
2
𝑚2
𝛼2
(1 5 3)(2 6 4)
2
𝑚2
𝛽
(4 5)(3 6)(1 2)
3
𝑚3
𝛼𝛽
(1 4)(2 3)(5 6)
3
𝑚3
2
(1 6)(2 5)(3 4)
3
𝑚3
𝛼 𝛽
By Burnside’s Lemma the number of distinct patterns is
1
𝑁 = (𝑚6 + 3 ⋅ 𝑚3 + 2 ⋅ 𝑚2 ).
6
◀
Math 4023
4