Describe generators and relations for all dihedral groups [tex]$D_{2 n}$[/tex]. (Hint: Let [tex]$x$[/tex] be reflection about a line through the center of a regular n-gon and a vertex, and let [tex]$y$[/tex] be counterclockwise rotation by [tex]$2 \pi / n$[/tex]. The group [tex]$D_{2 n}$[/tex] will be generated by [tex]$x$[/tex] and [tex]$y$[/tex], subject to three relations. To see that these relations really determine [tex]$D_{2 n}$[/tex], use them to show that any product [tex]$x^{i_1} y^{i_2} x^{i_3} y^{i_4} \cdots$[/tex] equals [tex]$x^i y^j$[/tex] for some [tex]$i, j$[/tex] with [tex]$(0 \leq i \leq 1,0 \leq j\ \textless \ n)$[/tex].
Answer (1)
The dihedral group D 2 n is generated by a reflection x and a rotation y .
The generators satisfy the relations x 2 = e , y n = e , and x y = y − 1 x .
Any element of D 2 n can be written in the form x i y j where 0 ≤ i ≤ 1 and 0 ≤ j < n .
The group D 2 n has 2 n elements: { e , y , y 2 , ... , y n − 1 , x , x y , x y 2 , ... , x y n − 1 } .
x 2 = e , y n = e , x y = y − 1 x
Explanation
Understanding the Dihedral Group We want to describe the generators and relations for the dihedral group D 2 n . The dihedral group D 2 n is the group of symmetries of a regular n -gon. We are given that x is a reflection about a line through the center of the n -gon and a vertex, and y is a counterclockwise rotation by 2 π / n .
Finding the Relations The group D 2 n is generated by x and y . We need to find the relations that x and y satisfy. Since x is a reflection, applying it twice gives the identity, so x 2 = e , where e is the identity element. Since y is a rotation by 2 π / n , rotating n times gives the identity, so y n = e . The reflection x reverses the rotation y , so x y = y − 1 x . This can also be written as x y x = y − 1 or ( x y ) 2 = e .
The Three Relations Therefore, the three relations are
x 2 = e , y n = e , x y = y − 1 x .
Rewriting Products Now we need to show that any product x i 1 y i 2 x i 3 y i 4 ⋯ can be written in the form x i y j where 0 ≤ i ≤ 1 and 0 ≤ j < n . We can use the relation x y = y − 1 x to move all the x 's to the left. For example, consider the product y 2 x y 3 x . Using the relation x y = y − 1 x , we have y 2 x y 3 x = y 2 ( y − 1 x ) y 3 x = y 1 x y 3 x = y 1 x ( y 3 ) x . We can continue moving the x to the left until all x 's are on the left.
Final Form of Elements Since x 2 = e , any x k can be reduced to either e or x , so i can be either 0 or 1. Since y n = e , any y k can be reduced to y j where 0 ≤ j < n . Therefore, any element can be written as x i y j where 0 ≤ i ≤ 1 and 0 ≤ j < n . The elements of D 2 n are { e , y , y 2 , ... , y n − 1 , x , x y , x y 2 , ... , x y n − 1 }. There are 2 n elements.
Examples
Dihedral groups are useful in understanding the symmetries of molecules in chemistry. For example, the water molecule ( H 2 O ) has C 2 v symmetry, which is isomorphic to the dihedral group D 2 . Understanding the generators and relations of dihedral groups helps chemists predict the vibrational modes and other properties of molecules. Similarly, in art and design, dihedral symmetry is often used to create visually appealing patterns and structures. Knowing the relations allows artists to manipulate these symmetries in a predictable way.
