Chapter 29 Symmetry and Counting

Consider the task of coloring the vertices of a regular hexagon so that three are black and three are white. There are possibilities.

How many hexagonal tiles would need to be produced to obtain the 20 different designs?

Four different tiles would need to be produced. The designs produced by each of the four tiles are said to be equivalent under the group of rotations of the hexagon.

Permutation groups naturally arise in many situations involving symmetrical designs or arrangements. Some simple problems may be solved by observation. More complicated problems require a more sophisticated approach.

If G is a set of permutations on a set S and, then and

For any set X we use to denote the number of elements in X.

Definition: Elements Fixed by For any group G of permutations on a set S and any in G, we let. This set is called the elements fixed by (or more simply, “ fix of “).

An approach to solving this type of problem was provided by Georg Frobenius in Frobenius’s theorem did not become widely known until it appeared in the classic book on group theory by William Burnside in Thus Frobenius’s theorem became known as…

Burnside’s Theorem If G is a finite group of permutations on a set S, then the number of orbits of G on S is

Returning to the hexagonal ceramic tile problem… The objects being permuted are the 20 possible designs. The group of permutations is the group of six rotational symmetries of a hexagon.

The identity fixes all 20 designs. Rotations of fix none of the designs.

for the rotations of

Applying Burnside’s Theorem, the number of orbits under the group of rotations is

Applications Outside Mathematics Burnside’s Theorem enables a chemist to determine the number of benzene molecules. A benzene molecule can be viewed as a six carbon atoms arranged in a hexagon with one of the three radicals attached at each carbon atom.

Exercise #3, p. 493 Determine the number of ways the vertices of an equilateral triangle can be colored with five colors so that at least two colors are used.

There are five colors to choose from for each of the three vertices, so there are possible color schemes. Since at least two colors must be used, the five possibilities having all three vertices the same color are eliminated, leaving possibilities.

There are six permutations of the equilateral triangle (analogous to the Symmetric Group : The identity Rotation of Flip along axis 1 Flip along axis 2 Flip along axis 3

Determine for each permutation: The identity fixes all 120 possible color schemes. The two rotations fix none of the color schemes. There are 60 color schemes having two vertices the same color. Each of the flips fixes 20 of these color schemes.

Applying Burnside’s Theorem, the number of orbits under the group of permutations is

Formalizing this approach to counting the number of objects considered nonequivalent: If G is a group and S is a set of objects, we say that G acts on S if there is a homomorphism from G to sym(S), the set of all permutations on S. The homomorphism is sometimes called the group action.