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[/tex]
Answer (2)
The dihedral group D 2 n is generated by a reflection x and a rotation y .
The reflection x satisfies the relation x 2 = e .
The rotation y satisfies the relation y n = e .
The interaction between x and y is given by x y = y − 1 x .
Any element in D 2 n can be written in the form x i y j for some 0 ≤ i ≤ 1 and 0 ≤ j < n .
The relations x 2 = e , y n = e , and x y = y − 1 x completely determine D 2 n .
The generators and relations for the dihedral group D 2 n are ⟨ x , y ∣ x 2 = e , y n = e , x y = y − 1 x ⟩ .
Explanation
Understanding the Problem We want to describe the generators and relations for the dihedral group D 2 n . This group represents the 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 . We need to find the relations that x and y satisfy and show that these relations completely determine the group D 2 n .
Relation for Reflection Since x is a reflection, applying it twice results in the identity. Thus, x 2 = e , where e is the identity element.
Relation for Rotation Since y is a rotation by 2 π / n , applying it n times results in a full rotation, which is the identity. Thus, y n = e .
Interaction between Reflection and Rotation The reflection x and rotation y interact in a specific way. Reflecting, rotating, and then reflecting again is the same as rotating in the opposite direction. This can be expressed as x y = y − 1 x , or equivalently x y x = y − 1 .
Summary of 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 for some i , j with 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 x y 2 x y − 1 x . Using the relation x y = y − 1 x , we can rewrite this as y − 1 x y 2 y − 1 x = y − 1 y − 2 x y − 1 x = y − 3 x y − 1 x . Continuing this process, we can move all x 's to the left.
Reducing Exponents Since x 2 = e , we can reduce the exponent of x to be either 0 or 1. Since y n = e , we can reduce the exponent of y to be between 0 and n − 1 . Therefore, any element in D 2 n can be written in the form x i y j for some 0 ≤ i ≤ 1 and 0 ≤ j < n .
Conclusion The order of D 2 n is 2 n . The elements are { e , y , y 2 , ... , y n − 1 , x , x y , x y 2 , ... , x y n − 1 }. Thus, the relations x 2 = e , y n = e , and x y = y − 1 x completely determine D 2 n .
Examples
Dihedral groups are useful in understanding the symmetries of molecules in chemistry. For example, the water molecule (H2O) has C2v symmetry, which is isomorphic to the dihedral group D2. Understanding the generators and relations of dihedral groups helps chemists predict the vibrational modes and other properties of molecules. Similarly, in physics, dihedral groups can be used to describe the symmetries of crystals and other physical systems.
The dihedral group D 2 n is generated by a reflection x and a rotation y , characterized by the relations x 2 = e , y n = e , and x y = y − 1 x . This group captures the symmetries of a regular polygon with n sides, allowing any element to be expressed in the form x i y j . In summary, the group can be presented as ( D_{2n} = \langle x, y \mid x^2 = e, y^n = e, xy = y^{-1}x \rangle.
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