Dihedral Groups

(Difference between revisions)
 Revision as of 15:55, 16 July 2011 (edit)← Previous diff Revision as of 15:58, 16 July 2011 (edit) (undo)Next diff → Line 326: Line 326: {{EquationRef2|Subgroup|Lemma 3.}}Given $d \mid n$, let $m = n/d$. For every $0 \leq i < n$ let $A(i,d) = \{ab^{i+km}: \ 0 \leq k < d \}$, where $a$ denotes a reflection and $b$ denotes a primitive rotation. Let $B(i,d) = A(i,d) \cup \langle b^m \rangle$. Then $B(i,d)$ is a subgroup of $D_{n}$ and $|B(i,d)|=2d$. We also have $|\{B(i,d) : \ 0 \leq i < n \}|=m=n/d.$ {{EquationRef2|Subgroup|Lemma 3.}}Given $d \mid n$, let $m = n/d$. For every $0 \leq i < n$ let $A(i,d) = \{ab^{i+km}: \ 0 \leq k < d \}$, where $a$ denotes a reflection and $b$ denotes a primitive rotation. Let $B(i,d) = A(i,d) \cup \langle b^m \rangle$. Then $B(i,d)$ is a subgroup of $D_{n}$ and $|B(i,d)|=2d$. We also have $|\{B(i,d) : \ 0 \leq i < n \}|=m=n/d.$ {{HideThis|1=The Proof|2= {{HideThis|1=The Proof|2= - '''Proof.''' If $ab^{i+km}=ab^{i+rm}$, for some $0 \leq k,r < d$, then $b^{(k-r)m}=1$ and thus $d \mid k-r$, because $o(b)=n=md$. + '''Proof.''' If $ab^{i+km}=ab^{i+rm}$, for some $0 \leq k,r < d$, then $b^{(k-r)m}=1$ and thus $d \mid k-r$, because $|b|=n=md$. Therefore $k=r$ because $0 \leq k,r < d$. Therefore $k=r$ because $0 \leq k,r < d$. Line 332: Line 332: So we have proved that $|A(i,d)|=d$. So we have proved that $|A(i,d)|=d$. - Clearly $A(i,d) \cap \langle b^m \rangle = \emptyset$ and $|\langle b^m \rangle = d$, because $o(b)=n$. + Clearly $A(i,d) \cap \langle b^m \rangle = \emptyset$ and $|\langle b^m \rangle = d$, because $|b|=n$. Thus $|B(i,d)|=|A(i,d)| + |\langle b^m \rangle|=2d$. Thus $|B(i,d)|=|A(i,d)| + |\langle b^m \rangle|=2d$.

Revision as of 15:58, 16 July 2011

Dihedral Symmetry of Order 12
Each snowflake in the main image has the dihedral symmetry of a natual regular hexagon. The group formed by these symmetries is also called the dihedral group of degree 6. Order refers to the number of elements in the group, and degree refers to the number of the sides or the number of rotations.

Basic Description

In mathematics, a dihedral group is the group of symmetries of a regular polygon, including both rotations and reflections. Dihedral groups are among the simplest examples of finite groups, and they play an important role in group theory, geometry, and chemistry.

Dihedral groups arise frequently in art and nature. Many of the decorative designs used on floor coverings, pottery, and buildings have one of the dihedral groups of symmetry. Chrysler’s logo has $D_5$ as a symmetry group, and that Mercedes-Benz has $D_3$. The ubiquitous five-pointed star has symmetry group $D_5$.

Notation

There are two different kinds of notation for a dihedral group associated to a polygon with $n$ sides.

In geometry, we usually call it $D_n$ or $Dih_n$, where $n$ indicates the number of the sides.

In algebra, we call it $D_2n$, where $2n$ indicates the number of elements in the group.

On this page, we will use the notation $D_n$ to describe a dihedral group. For $D_n$, we will call it the dihedral group of order $2n$ or the group of symmetries of a regular $n$-gon.

A More Mathematical Explanation

Note: understanding of this explanation requires: *Basic Abstract Algebra

Elements

The $n^\text{th}$ dihedral group is the symmetry group of t [...]

Elements

The $n^\text{th}$ dihedral group is the symmetry group of the regular $n$-sided polygon. The group consists of $n$ reflections, $n-1$ rotations, and the identity transformation.

Here is an example of $D_6$. This group contains 12 elements, which are all rotations and reflections. The very first one is the identity transformation.

Image 1

If $n$ is odd each axis of symmetry connects the mid-point of one side to the opposite vertex. If $n$ is even there are $\frac{n}{2}$ axes of symmetry connecting the mid-points of opposite sides and $\frac{n}{2}$ axes of symmetry connecting opposite vertices. In either case, there are n axes of symmetry altogether and $2n$ elements in the symmetry group. Reflecting in one axis of symmetry followed by reflecting in another axis of symmetry produces a rotation through twice the angle between the axes. In Image 1, through $S_0$ to $S_5$ are the axes of symmetries. All the reflections can be described as reflections of the identity through six axes of symmetries.

Definition

There are several different way to define a Dihedral Group. We will introduce three of them.

We will use $R_0$ to represent the identity, $R_k, k=1,2,\cdots,n-1$, to represent the rotations, and $S_k, k=0,1, \cdots, n-1$, to represent the reflections.

Complex Plane Presentation

For $n \geqslant 3$, the dihedral group $D_n$ is defined as the rigid motions of the plane preserving a regular $n$-gon, with the operation of composition. Our model $n$-gon will be an $n$-gon centered at the origin, with vertices at the n-th roots of unity. $1$ is always an $n$-th root of unity, but $-1$ is such a root only if $n$ is even. In general, the roots of unity form a regular polygon with $n$ sides, and each vertex lies on the unit circle.

The $n$-th roots of unity are roots $\omega = e^{\frac{2 \pi i}{n}}$ of the cyclotomic equation $x^n=1$.

Since a vector on the complex plane can be described as $p=a+bi$, a vector with an angle $\theta$ counterclockwise from x-axis can be described as $p=r\cos \theta +i r\sin \theta$, where $r=\sqrt{a^2+b^2}$ is the magnitude of the vector.

Leonhard Euler's formula says that $e^{i \theta}=\cos \theta +i \sin \theta$, so any point on complex plane is $p=re^{i \theta}$.

For a regular $n$-gon, the first angle counterclockwise from the x-axis is $\frac{2 \pi}{n}$, so the primitive root of unity is $\omega = e^{\frac{2 \pi i}{n}}$.

Letting $\omega = \frac{2 \pi i}{n}$ denote a primitive $n^ \text{th}$ root of unity, and assuming the polygon is centered at the origin, the rotations $R_k$, $k=0, 1, 2, \cdots, n-1$(Note: $R_0$ denotes the identity), are given by

$R_k:z \mapsto \omega^k z,\quad z\in\cnums.$

Notation $\mapsto$ means a function. For example: $input \mapsto output$.

For the reflections, $S_k$, $k=0, 1, 2, \cdots, n-1$, the functions are given by

$S_k: z\mapsto \omega^k \bar{z},\quad z\in\cnums.$

Group Presentation

A presentation of a group is a description of a set $I$ and a subset $R$ of the free group $F(I)$ generated by $I$, written as $\left \langle (x_i)_{i \in I}|(r)_{r \in R} \right \rangle$, where the equation $r=1$ (the identity element) is often written in place of the element $r$. A group presentation defines the quotient group of the free group $F(I)$ by the normal subgroup generated by $R$, which is the group generated by the generators $x_i$ subject to the relations $r \in R$.

We can use the presentation:

$D_n = \left \langle r,s|r^n=1,s^2=1,srs=r^{-1} \right \rangle$, or

$D_n = \left \langle x,y|x^2=1,y^n=1,(xy)^2=1 \right \rangle$ to define a group, isomorphic to the dihedral group $D_n$ of finite order $2n$, which is the group of symmetries of a regular $n$-gon.

We will use the second presentation, in which $x$ refers to a reflection, and $y$ refer to a primitive rotation.

For $x^2=1$, $x$ means an arbitrary mirror image of the $n$-gon, and $1$ means the identity. This equation means that if we reflect the $n$-gon once, you get $x$. If reflect the $n$-gon twice, the result will return to the identity.
For $y^n=1$, the equation means that $y$ is a rotation, and its $n$th power $y^n$ equals the identity. That is, if we rotate the $n$-gon $n$ times, we get back to the identity.
For $(xy)^2=1$, $xy$ means the mirror image of $y$. Reflecting the $n$-gon through the axis of symmetry of $xy$ twice, the result is the identity.

Following the group presentation, we can label all the reflections and rotations in terms of $x$ and $y$.

• Identity: $1$
• Rotations: $y, y^2, y^3, \cdots, y^{n-1}$, and $y^n=1$, which is the identity.
• Reflections: $x, xy, xy^2, \cdots, xy^{n-1}$. There is not $xy^n$, because $y^n=1$, and so $xy^n=x$ is the reflection of the identity.

Matrix Representation

If we center the regular polygon at the origin, then the elements of the dihedral group act as linear transformations of the plane. This lets us represent elements of $D_n$ as matrices, with composition being matrix multiplication. This is an example of a (2-dimensional) group representation.

For example, the elements of the group $D_4$ can be represented by the following eight matrices:

$\begin{matrix} R_0=\bigl(\begin{smallmatrix}1&0\\[0.2em]0&1\end{smallmatrix}\bigr), & R_1=\bigl(\begin{smallmatrix}0&-1\\[0.2em]1&0\end{smallmatrix}\bigr), & R_2=\bigl(\begin{smallmatrix}-1&0\\[0.2em]0&-1\end{smallmatrix}\bigr), & R_3=\bigl(\begin{smallmatrix}0&1\\[0.2em]-1&0\end{smallmatrix}\bigr), \\[1em] S_0=\bigl(\begin{smallmatrix}1&0\\[0.2em]0&-1\end{smallmatrix}\bigr), & S_1=\bigl(\begin{smallmatrix}0&1\\[0.2em]1&0\end{smallmatrix}\bigr), & S_2=\bigl(\begin{smallmatrix}-1&0\\[0.2em]0&1\end{smallmatrix}\bigr), & S_3=\bigl(\begin{smallmatrix}0&-1\\[0.2em]-1&0\end{smallmatrix}\bigr). \end{matrix}$

If we represent the columns of each matrix as basis vectors, we can observe directly all the rotations and reflections.

In general, we can write any dihedral group as:

$R_k = \begin{bmatrix} \cos \frac{2\pi k}{n} & -\sin \frac{2\pi k}{n} \\ \sin \frac{2\pi k}{n} & \cos \frac{2\pi k}{n} \end{bmatrix}$,

$S_k = \begin{bmatrix} \cos \frac{2\pi k}{n} & \sin \frac{2\pi k}{n} \\ \sin \frac{2\pi k}{n} & -\cos \frac{2\pi k}{n} \end{bmatrix}$,

where $R_k$ is a rotation matrix, expressing a counterclockwise rotation through an angle of $\frac{2 \pi k}{n}$, and $S_k$ is a reflection across a line that makes an angle of $\frac{\pi k}{n}$ with the x-axis.

Linear Transportation

Since the dihedral groups also act by linear transformations:

$(x,y)\to(\hat{x},\hat{y}),\quad (x,y)\in \reals^2$

there is a corresponding action on polynomials $p\to\hat{p}$ , defined by

$\hat{p}(\hat{x},\hat{y}) = p(x,y),\quad p\in \reals[x,y].$

Properties

Cayley Table

Cayley Table

As with any geometric object, the composition of two symmetries of a regular polygon is again a symmetry. This operation gives the symmetries of a polygon the algebraic structure of a finite group.

A Cayley table, named after the 19th century British mathematician Arthur Cayley, describes the structure of a finite group by arranging all the possible products of all the group's elements in a square table reminiscent of an addition or multiplication table.

The Cayley Table on the right shows the effect of composition in the group $D_6$ (the symmetries of a hexagon). $R_0$ denotes the identity; $R_1$ to $R_5$ denote counterclockwise rotations by 60, 120, 180, 240,and 300 degrees; and $S_0$ to $S_5$ denote reflections across the six diagonals. In general, $ab$ denotes the entry at the intersection of the row with $a$ at the left and the column with $b$ at the top.

In the table, the same or different rotations and reflections work together and result in a new rotation or reflection. For example, look first at the vertical axis to find a element, $R_4$. Then look at the horizontal axis to get the second element for our composition. We choose $S_3$. Composing two elements is just the progression of a rotation or a reflection followed by another rotation or a reflection. In this case, our elements are $R_4$ and $S_3$. First we rotate the hexagon counterclockwise 240 degrees, and then reflect it along the axis of symmetry of $S_3$. The result is the same as reflecting the identity transformation through an angle of 60 degrees, which is $S_1$. See Example 1 below.

Now, look back to Cayley Table, you will find that the intersection of $R_4$ and $S_3$ is $S_1$. If you like, you can create your own Cayley table for a dihedral group of any order and find the natual rule for it.

Explore the Cayley Table

Perhaps the most important feature of this table is that it has been completely filled in without introducing any new motions.

• Closure: Algebraically, this says that if $A$ and $B$ are in $D_6$, then so is $AB$. This property is called closure, and it is one of the requirements for a mathematical system to be a group.
• identity: Notice that if $A$ is any element of $D_6$, then $AR_0=R_0 A=A$. Thus, combining any element $A$ on either side with $R_0$ yields $A$ back again. An element $R_0$ with this property is called an identity, and every group must have one.
• Inverse: We see that for each element $B$ in $D_6$, there exists an element $A$ such that $AB=BA=R_0$. In this case, $B$ is said to be the inverse of $A$ and vise versa. The term inverse is a descriptive one, for if $A$ and $B$ are inverses of each other, then $B$ "un-does" whatever $A$ "does", in the sense that $A$ and $B$ taken together in either order produce $R_0$, representing no change.
• Non-Abelian: Another property of $D_6$ deserves special comment. Obverse that $S_1 R_3=R_3 S_1$, but $S_3 R_4 \neq R_4 S_3$. Thus in a group $ab$ may or may not be the same as $ba$. If it happens that $ab=ba$ for all choices of group elements $a$ and $b$, we say the group is commutative or --better yet-- Abelian (in honor of the great Norwegian mathematician Niel Abel). Otherwise, we say the group is non-Abelian. All dihedral groups are non-Abelian, except $D_1$ and $D_2$.
• Associativity: For all dihedral groups, it holds true that $(ab)c=a(bc)$ for all a, b, c in the group.
Multiplication Table

If we want to know what is the composition of any two elements, it is convenient to use aCayley Map, because it tells us the result directly. But when we want to know the gradual change of the compositions, we will need another tool, a multi-colored Muliplication Table.

Muptiplication Table

In this multi-colored Muliplication Table, each color represents one rotation or reflection. In the image, pink colors represent rotations, and the deepest pink represents the identity transformation. Green colors represent all the six reflections.

From the changing of color, we can observe that the gradual change of the composition of two elements in $D_6$. However, we cannot easily determine the exact result of composition by observing directly this multi-colored Muliplication Table.

The abstract group structure is given by:

$R_k R_l=R_{k+l}$

$S_k S_l=R_{k-l}$

$R_k S_l=S_{k+l}$

$S_k R_l=S_{k-l}$

Uniqueness of the Identity

In a dihedral group $D_n$, there is only one identity element.

PROOF: Suppose both $R_0$ and $R_0'$ are identities of $D_n$. Then,

Eq. 1         $aR_0=a$ for all $a$ in $D_n$, and
Eq. 2         $R_0' a=a$ for all $a$ in $D_n$.

The choices of $a=R_0'$ in Eq. 1 and $a=R_0$ in Eq. 2 yield $R_0' R_0=R_0'$ and $R_0' R_0=R_0$.

Thus, $R_0$ and $R_0'$ are both equal to $R_0' R_0$ and so are equal to each other.

Cancellation

In a dihedral group $D_n$, the right and left cancellation laws hold; that is, $ba=ca$ implies $b=c$, and $ab=ac$ implies $b=c$.

PROOF: Suppose $ba=ca$. Let $a'$ be an inverse of $a$.

Then, muliplying on the right by $a'$ yields $(ba)a'=(ca)a'$.

Associativity yields $b(aa')=c(aa')$.

Then, $bR_0=cR_0$ and, therefore, $b=c$ as desired.

Similarly, one can prove that $ab=ac$ implies $b=c$ by multiplying by $a'$ on the left.

A consequence of the cancellation property is the fact that in a Cayley table for a dihedral group, each group element occurs exactly once in each row and column. Another consequence of the cancellation property is the uniqueness of inverses.

Uniqueness of Inverses

For each element $a$ in a dihedral group $D_n$, there is a unique element $b$ in $D_n$ such that $ab=ba=R_0$.

PROOF: Suppose $b$ and $c$ are both inverses of $a$.

Then $ab=R_0$ and $ac=R_0$, so that $ab=ac$.

Canceling the $a$ on both sides gives $b=c$, as desired.

Sock-Shoes property

For dihedral group elements $a$ and $b$, $(ab)^{-1}=b^{-1} a^{-1}$.

PROOF: Since $(ab)(ab)^{-1}=R_0$ and

$(ab)(b^{-1} a^{-1})=a(bb^{-1})a^{-1}=aR_0a^{-1}=aa^{-1}=R_0$,

we have by the Uniqueness of Inverses theorem that $(ab)$ has only one inverse $x$ such that $(ab)x=R_0$.

We get $x=(ab)^{-1}=(b^{-1} a^{-1})$.

3D Rotational Symmetry

An example of abstract group $D_n$, and a common way to visualize it, is the group $D_n$ of Euclidean plane isometries which keep the origin fixed. These groups form one of the two series of discrete point groups in two dimensions. $Dn$ consists of $n$ rotations of multiples of $\frac{360^\circ}{n}$ about the origin, and reflections across $n$ lines through the origin, making angles of multiples of $\frac{180^\circ}{n}$ with each other. This is the symmetry group of a regular polygon with $n$ sides (for $n \geqslant 3$, and also for the degenerate case $n = 2$, where we have a line segment in the plane).

If we put a dihedral group in three dimensions, the reflections are also rotations of $180^\circ$

Rotation in 3D

The proper symmetry group of a regular polygon embedded in three-dimensional space (if $n \geqslant 3$). Such a figure may be considered as a degenerate regular solid with its face counted twice. Therefore it is also called a dihedron (Greek: solid with two faces), which explains the name dihedral group.

Infinite Dihedral Groups

The infinite dihedral group $Dih(C_\infty)$ is denoted by $D_\infty$. The infinite dihedral group can be described as the group of symmetries of a circle, which has infinite symmetries.

We use the group presentation:

$D_\infty = \langle r, s \mid s^2 = 1, srs = r^{-1} \rangle$, or

$D_\infty = \langle x, y \mid x^2 = y^2 = 1 \rangle$ to represente the infinite dihedral group.

In the presentation, it says that because there are infinitely many symmetries, we can never rotate back to the identity, and so there are infinitely many rotations and reflections.

Subgroups

Definition: A subgroup is a subset $H$ of group elements of a group $G$ that satisfies the four group requirements. It must therefore contain the identity element. "$H$ is a subgroup of $G$" is written as $H \subseteq G$, or sometimes $H \leq G$.[1]

Now we want to know exactly how many subgroups for $D_n$, and what they are. Fortunately, mathematician Stephan A. Cavior had already proved this for us in 1975. In the theorem, for any dihedral group in order of $2n$, there are $\tau(n) + \sigma(n)$ subgroups in total, including $\{1\}$ and $D_{n}$. $\{1\}$ is just the identity itself.

Definitions of Terms

• $\tau(n)$: the number of divisors of $n$,
e.g. $\tau(12)=6$.
• $\sigma(n)$: the sum of divisors of $n$,
e.g. $\sigma(12)=1 + 2 + 3 + 4 + 6 + 12=28$.
• $\mathbb{Z}/n \mathbb{Z}$: the notation for the cyclic group of order $n$, can be also written as $\mathbb{Z}_n$. This is a quotient group presentation.
• $d \mid n$: $d$ is a divisor of $n$.
• $N=\langle b \rangle$: group $N$ is a subgroup generated by $b$. It means $N=\{1,b,b^2,b^3,\cdots,b^{n-1}\}$.
• $|HN|$: the order of $HN$.
• $[D_{n}:N]=2$: the index. $[D_{n}:N]=2$ means $\frac{|D_n|}{|N|}=2$.
• $\{ab^{i+km}: \ 0 \leq k < d \}$: a set with elements looking like $ab^{i+km}$.
• $G(i,d)$: this means a group. $i$ is the index, which labels all the elements; $d$ is the order of the group.

This proof is complicated.

After we know what kind of group can be a subgroup of a dihedral group $D_n$, the dihedral group of order $2n$, we will start to find all subgroups of $D_n$.

Lemma 1.        The number of subgroups of a cyclic group of order $n \geq 1$ is $\tau(n)$. [2]

A cyclic group of order $n$ is the group of all the rotations including the identity of the dihedral group of order $2n$.

Proof. Let $G$ be a cyclic group of order $n$. Then $G \cong \mathbb{Z}/n \mathbb{Z}$. A subgroup of $\mathbb{Z}/n \mathbb{Z}$ is in the form $d \mathbb{Z}/n \mathbb{Z}$ where $d \mathbb{Z} \supseteq n \mathbb{Z}$. The condition $d \mathbb{Z} \supseteq n \mathbb{Z}$ is obviously equivalent to $d \mid n. \$

Lemma 2.        Let $b$ be the element of order $n$ in $D_{n}$ and let $H$ be a subgroup of $D_{n}$. Then either $H \subseteq \langle b \rangle$ or $|H \cap \langle b \rangle| =d$ and $|H|=2d$ for some $d \mid n$.

Proof. Let $N=\langle b \rangle$. Clearly $N$ is a normal subgroup of $D_{n}$ because $[D_{n}:N]=2$. Thus $HN$ is a subgroup of $D_{n}$ and hence the order of dihedral group $HN$ is a divisor of $2n$, and we use the notation:

Eq. 1         $|HN| \mid 2n.$ to represent.

On the other hand,

Eq. 2         $|HN|=\frac{|H| \cdot |N|}{|H \cap N|}=\frac{n |H|}{|H \cap N|}$.
Therefore, by Eq. 1, and Eq. 2. Hence either $|H| = |H \cap N|$ or $|H|=2|H \cap N|$. If $|H|=|H \cap N|$, then $H = H \cap N$ and thus $H \subseteq N$. If $|H|=2|H \cap N|$, then let $|H \cap N|=d$ and so $|H|=2d$. Clearly $d \mid n$ because $H \cap N$ is a subgroup of $N$ and $|N|=n. \$

Lemma 3.        Given $d \mid n$, let $m = n/d$. For every $0 \leq i < n$ let $A(i,d) = \{ab^{i+km}: \ 0 \leq k < d \}$, where $a$ denotes a reflection and $b$ denotes a primitive rotation. Let $B(i,d) = A(i,d) \cup \langle b^m \rangle$. Then $B(i,d)$ is a subgroup of $D_{n}$ and $|B(i,d)|=2d$. We also have $|\{B(i,d) : \ 0 \leq i < n \}|=m=n/d.$

Proof. If $ab^{i+km}=ab^{i+rm}$, for some $0 \leq k,r < d$, then $b^{(k-r)m}=1$ and thus $d \mid k-r$, because $|b|=n=md$.

Therefore $k=r$ because $0 \leq k,r < d$.

So we have proved that $|A(i,d)|=d$.

Clearly $A(i,d) \cap \langle b^m \rangle = \emptyset$ and $|\langle b^m \rangle = d$, because $|b|=n$.

Thus $|B(i,d)|=|A(i,d)| + |\langle b^m \rangle|=2d$.

Proving that $B(i,d)$ is a subgroup of $D_{n}$ is very easy.

Just note that every element of $A(i,d)$ is the inverse of itself (because they all have order two) and also note that $ab^s = b^{-s}a$, for all $s$, because $ab=b^{-1}a$.

Finally, the set $\{B(i,d) : \ 0 \leq i < n \}$ has $m$ elements because clearly $B(i,d)=B(j,d)$ if and only if $A(i,d)=A(j,d)$ if and only if $i \equiv j \mod m. \$

Suppose that $H$ is a subgroup of $D_{n}$. There are two disjoint cases to consider.

Case 1. $H \subseteq \langle b \rangle$.

By Lemma 1. the number of these subgroups is $\tau(n)$.

Case 2. $H \nsubseteq \langle b \rangle$.

In this case, by Lemma 2. we have $|H|=2d$ and $|H \cap \langle b \rangle|=d$, for some $d \mid n$.
Let $n = md$. Since $H \cap \langle b \rangle$ is a subgroup of $\langle b \rangle$, which is a cyclic group of order $n$, we have
Eq. 1        $H \cap \langle b \rangle = \langle b^m \rangle.$
Let $A(i,d)$ and $B(i,d)$ be as were defined in Lemma 3.
Now, since $H$ is not contained in $\langle b \rangle$, there exists some $0 \leq i < n$ such that $ab^i \in H$.
Then, since $H$ is a subgroup, we must have $ab^ib^{km} \in H$, for all $k$.
Thus $ab^{i + km} \in H$ and so $A(i,d) \subseteq H$ and therefore, by Eq. 1, we have $B(i,d) \subseteq H$.
Thus, since $|H|=|B(i,d)|=2d$, we must have $H=B(i,d)$.
The converse is obvously true, i.e. given $d \mid n$ and $0 \leq i < n$, $B(i,d)$ is a subgroup of $D_{n}$, by Lemma 3., and $B(i,d) \nsubseteq \langle b \rangle$ because it contains $A(i,d)$.
So the subgroups in this case are exactly the ones in the form $B(i,d)$, where $0 \leq i < n$ and $d \mid n$.
Thus, by Lemma 3. the number of subgroups in this case is
$\sum_{d \mid n} | \{B(i,d) : \ 0 \leq i < n \} | = \sum_{d \mid n} n/d = \sum_{d \mid n} d = \sigma(n).$

So, by Case 1. and Case 2. the number of subgroups of $D_{n}$ is $\tau(n) + \sigma(n)\$.

Why It's Interesting

In Music

The sequence of pitches which form a musical melody can be transposed or inverted. Since the 1970s, music theorists have modeled musical transposition and inversion in terms of an action of the dihedral group of order 24. More recently music theorists have found an intriguing second way that the dihedral group of order 24 acts on the set of major and minor chords.[3]

Dihedral groups as a kind of special symmetric groups are studied in music. In music, we use operations Transportation and Inversion, which are denoted as $T$ and $I$, to represente rotations and reflections in dihedral groups.

Musicians usually study about $D_{12}$, because 12 is a normal cycle in music.C C D E E F F G G A B B, and then C again.

Based on this 12 elements cycle, $D_{12}$ is important in music theory. Musicians use Transportations and Inversions(rotations and reflections) of a simple note to create other notes to complete a final composition.

A transposition of a sequence $x$ of pitch classes by $n$ semitones is the sequence $T^n(x)$ in which each of the pitch classes in $x$ has been increased by $n$ semitones.

So for example if

$x=3 0 8$, which the numbers denote to notes in numbers,

then

$T^4(x)=T^4(3 0 8) = 7 4 0$.

When doing the operation $T^n(x)$, add $n$ to each digit of $x$, and use Modular arithmetic of 12, when the resulting digit is over 12. For instance, add 4 to 8, the result is 12, then $12 \equiv 0 \pmod 12.\,$

Turning to the next operation, inversion $I(x)$ of a sequence $x$ just re-places each pitch class by its negative (in clock arithmetic).

So in the first example above with $x = 3 0 8$, we have

$I(x) = 9 0 4$.

To do the operation $I(x)$, we need to do the opposite way of $T^n(x)$. For instance, if we want to get 12 from 3, we need to add 9. 0 is already 12, so we need to add 0.

References

[2] de Cornulier, Yves. (n.d). Group Presentation. From MathWorld--A Wolfram Web Resource, created by Eric W. Weisstein. http://mathworld.wolfram.com/GroupPresentation.html

[4] Milson, Robert and Foregger, Thomas. dihedral group. From PlanetMath.org. June, 12. 2007. Retrieved from http://planetmath.org/encyclopedia/DihedralGroup.html

[5] Gallian, Joseph A. Contemporary Abstract Algebra Seventh Edition. Belmont: Brooks/Cole, Cengage Learning. 2010.

[6] Dahlke, Karl. (n.d). Groups, Dihedral and General Linear Groups. Retrieved from http://www.mathreference.com/grp,dih.html

[7] Sharifi, Yaghoub. Subgroups of dihedral groups (1)&(2). Feb, 17, 2011. Retrieved from http://ysharifi.wordpress.com/2011/02/17/subgroups-of-dihedral-groups-1/

[8] [1]Scott, W. R. Group Theory. New York: Dover, 1987.

[9] [2]Hungerford, Thomas W. Graduate Texts in Mathematics - Algebra. New York: Springer, 1974.

[10] [3]Crans, Alissa S., Fiore, Thomas M. and Satyendra, Ramon. Musical Actions of Dihedral Groups. University of South Florida. Nov 3, 2007. Retrieved from http://myweb.lmu.edu/acrans/MusicalActions.PDF

[11] Benson, Dave J. Music: A Mathematical Offering. Cambridge University Press. Nov 2006. Retrieved from http://www.maths.abdn.ac.uk/~bensondj/html/music.pdf

[12] Rowland, Todd and Weisstein, Eric W. (n.d). Root of Unity. From MathWorld--A Wolfram Web Resource. Retrieved from http://mathworld.wolfram.com/RootofUnity.html

2. Add more about Dihedral Groups in 3D. I only talk one property in 3D, but there must be some more.
3. In subgroup part, it is hard to explain only in words, so I use lots of notations, but they are still not very clear, I hope can find a better way to illustrate it.