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Chapter 12: Problem 7

Show that every subgroup of rotations in a dihedral group \(D_{n}\) is normal in \(D_{n}\)

Short Answer

Expert verified
Every subgroup of rotations in a dihedral group is normal because it remains invariant under conjugation by any element of the group.

Step by step solution

01

Understanding Dihedral Group

The dihedral group, denoted as \( D_n \), consists of the symmetries of a regular \( n \)-gon, including both rotations and reflections. Specifically, \( D_n \) contains \( n \) rotations and \( n \) reflections, making a total of \( 2n \) elements.
02

Identify Normal Subgroup

A subgroup \( H \) of a group \( G \) is normal if it is invariant under conjugation by elements of \( G \). This means for every \( h \in H \) and every \( g \in G \), the element \( g h g^{-1} \) must also be in \( H \). We aim to show all rotation subgroups in \( D_n \) meet this criterion.
03

Subgroup of Rotations

The subgroup of rotations in \( D_n \) is cyclic and generated by a single rotation \( r_k \) where \( r_k = r^{k} \) and \( r \) is a basic rotation of \( 360/n \) degrees. This subgroup includes \( \{e, r, r^2, \ldots, r^{n-1}\} \).
04

Conjugation by Rotation

Consider conjugating an element \( r^{k} \) in the rotations subgroup by a rotation \( r^{m} \). \( r^{m} r^{k} (r^{m})^{-1} = r^{m} r^{k} r^{-m} = r^{k} \). Since the result is still \( r^{k} \), conjugation by a rotation leaves rotation elements unchanged.
05

Conjugation by Reflection

Now consider conjugation by a reflection \( s \). For rotation \( r^{k} \), conjugate by \( s \): \( s r^{k} s^{-1} = (sr^{k}s) = r^{-k} \). Since \( r^{-k} \) is also a rotation, it remains within the subgroup \( \{e, r, r^2, \ldots, r^{n-1}\} \).
06

Conclusion

Since both conjugations by any element \( r^{m} \) or \( s \) in \( D_n \) keep elements within the rotation subgroup, the subgroup of rotations is invariant under all conjugations by elements of \( D_n \). Hence, the subgroup is normal in \( D_n \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Normal Subgroup
Understanding what a normal subgroup is can be very helpful. Consider a group with a special subset that behaves nice when mixed with the whole group. This subset is known as a "normal subgroup". In a group, if you can conjugate any element from this subgroup with any element from the entire group, and the result stays within the subgroup, then it鈥檚 labeled a normal subgroup.
  • Conjugation means pulling a member of the subgroup to the group and back.
  • The subgroup retains its structure, despite this mixing with the entire group.
This property of being able to freely 'mix' elements is crucial. It means the subgroup is equally part of the whole, keeping its identity intact. Understanding this helps to see how elements interact in broader groups.
Subgroup of Rotations
In a dihedral group, especially one denoted as \(D_n\), rotations form their own subgroup. This subgroup is special because it includes only rotations and not reflections. Visualize this as a spinning wheel where each spoke represents a rotation. The motions of this wheel, when separated from reflections, make the rotation subgroup.
  • These rotations are characterized by finite, regular angles. Each angle is a multiple of \(360/n\) degrees.
  • Rotations include movements like \( e \, (identity), r, r^2, \ldots, r^{n-1} \).
Thus, while the dihedral group may have more symmetries, these rotations alone form a clean, orderly fashion.
Conjugation
Conjugation in mathematical groups is like a transformation or reshuffling of elements, maintaining structure. It鈥檚 an interesting way to mix elements.
  • Think of conjugation as a dance: a pair of elements performing a sequence leading them back to another position in the group.
  • In \(D_n\), conjugating a rotation with any element be it another rotation or even a reflection leads to a result that remains as a rotation within the subgroup.
The natural flow of elements in conjugation allows a deeper insight into the balanced interactions within groups, proving the stability of the subgroup of rotations.
Cyclic Subgroup
A cyclic subgroup is like repeating a simple step to cover the entire floor. Here, the step denotes one basic element that generates all others through repetition. In rotation subgroups, a single basic rotation is repeated until all rotations are expressed.
  • Formally, it takes one element, \( r \), and all subsequent elements are powers of this \( r \): \( r^1, r^2, \ldots \).
  • This structure is linear, making it easy to navigate and understand.
The magic of cyclic subgroups lies in their simplicity - one single generator, leading to a collection full of movement and symmetry.

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Most popular questions from this chapter

Let \(G\) be the group \(\\{1,-1\\}\) and let \(H\) be the group \(\\{z .-z, \bar{z},-\bar{z}\\}\) of isometries of \(\mathbb{C}\). Show that \(H\) is isomorphic to \(G \times G\).

Show that for any integers \(p_{1}, \ldots, p_{n}\) there are integers \(m_{1}, \ldots, m_{n}\) such that \(\operatorname{gcd}\left(p_{1}, \ldots, p_{n}\right)=m_{1} p_{1}+\cdots+m_{n} p_{n}\).

Show that if \(H\) is a subgroup of an abelian group \(G\), then \(H\) is a normal subgroup of \(G .\) In particular, every subspace \(U\) of a vector space \(V\) is a normal subgroup of \(V\). Find a homomorphism (a linear map) of \(V\) into itself with kernel \(U\).

Let \(G\) be \(\mathbb{Z}_{16}\) (the group \(\\{0,1, \ldots, 15\\}\) under addition modulo 16), and let \(H\) be the subgroup of \(G\) generated by the single element 4 . List all (four) cosets of \(G\) with respect to \(H\).

Show that there are exactly two homomorphisms from \(C_{6}\) to \(C_{4}\).
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