Do you still remember that I am still only fourteen?” I said, starting to lose my temper and grasping how Mum should feel like when Dad says things like, “Such a sophisticated perfume honey, how many litres and for how long did you need to soak in it? Come to my localhost, I will turn off my firewall, disable my antivirus, grant you sudo access, and show you my source code,” Apocalypse, Anawim, #justothepoint

A polygon is a two-dimensional or plane figure that has a finite number of sides. It is described by a finite number of straight line segments connected to form a closed polygonal chain. A **regular polygon is a polygon that is both equiangular** (all angles are equal in measure) and **equilateral** (all sides have the same length).

The dihedral group D_{n} is the group of symmetries of a regular polygon with n vertices, which includes rotations and reflections or, in other words, the right motions taking a regular n-gon back to itself, or the arrangement of the figure preserving sides, vertices, distances, and angles.

A regular polygon can be rotated around its center in n different ways to come back to itself. More specifically, we can rotate it around the center by $\frac{2kπ}{n}$ radians where 0 ≤ k ≤ n − 1 (Figure 1.a).

Any regular n-gon can be inscribed into a circle whose center coincides with the center of the n-gon. As the side lengths of the polygon are irrelevant in terms of symmetry, we could **assume that our polygon is inscribed in the unit circle**.

Let be r the counterclockwise rotation by θ = ^{2π}⁄_{n}, then the n rotations in D_{n} are e, r, r^{2},…, r^{n-1}, namely, 1, e^{2π/n}, e^{2(2π/n)}, e^{3(2π/n)},···, e^{(n-1)(2π/n)}, Figure 01.

Theorem. Let s be a reflection across the real axis. Let D_{n} be the dihedral group. Then, sr^{k}s^{-1} = r^{-k}. In particular, sr^{k} = r^{-k}s (by multiplying the previous equation by s).

Proof.

Consider sr^{k}s^{-1} = [s^{2} = 1 ⇒ s = s^{-1}] sr^{k}s = [sr^{k}s means to **reflect across the x-axis, then rotate by θ = ^{2πk}⁄_{n} counterclockwise, that is, k rotations by r = ^{2π}⁄_{n} counterclockwise, and finally reflect across the x-axis back**, but this is just a rotation by θ but clockwise, that is, k rotations by r

Theorem. D_{n} is generated by the symmetries r and s, D_{n} = ⟨r, s⟩. In particular, every symmetry σ can be expressed in the form σ = r^{k}s^{l} for k, l ∈ ℤ.

Proof.

We already know that the only symmetries of the unit circle are rotations and reflections. These are the only types of symmetries for the regular n-gon which is a subset of the unit circle.

- Rotations. All counterclockwise rotational symmetries of the n-gon are power of r because they
**fix the center of the circle and every vertex of the n-gon needs to land on another vertex**. Besides, all clockwise rotational symmetries of the n-gon are powers of r^{-1}. - Let ρ be a reflection across the line L. If ρ is a symmetry of the n-gon, then
**L must pass through both by the origin and a by vertex or a midpoint of an edge of the polygon**. In either case, the angle θ that L makes with the positive real axis equals $k\frac{π}{n}$ for some k (Figure 11).It is not $k\frac{2π}{n}$ because L could pass thought the origin and a midpoint of an edge of the polygon.

Let r_{θ} be the rotation counterclockwise by θ. ρ = $r_θsr_θ^{-1}$ (Figure 100, r_{θ}^{-1} is the rotation clockwise by θ) = [By the previous lemma, **sr ^{k}s^{-1} = r^{-k}**, sr

Each product of r’s and s’s is called a **word**, and we can simplify words by collapsing down adjacent r’s and s’s together, and we can consider that we don’t need inverses since s^{2}=1 (s = s^{-1}) and sr^{k} = r^{-k}s. Therefore, a generic word in D_{n} has the following form, $σ=r^{k_0}sr^{k_1}sr^{k_2}···sr^{k_m}$ (k_{0} and k_{m} could be zero, that is, a generic word can start and/or end with a rotation or a reflection).

Claim: σ = r^{k}s^{l} for k, l ∈ ℤ.

Let’s induct on the number of s’s in σ.

Case base. The number of s’s is 0 ⇒ σ = r^{k0} = r^{k0}s^{0}.

Suppose that if σ involves m-1 s’s, then σ = r^{k}s^{l} for some k, l ∈ ℤ.

$σ=r^{k_0}sr^{k_1}sr^{k_2}···sr^{k_m}=(r^{k_0}sr^{k_1}sr^{k_2}···sr^{k_{m-1}})sr^{k_m}$ = [By our induction hypothesis] $(r^ks^l)sr^{k_m}=r^ks^{l+1}r^{k_m}$ =

Notice: s···_{l+1 times}···sr···_{km times}···r = … [By the previous lema, sr^{k}=r^{-k}s each of the k_{m} r’s can be flipped by an s and needs to be moved l+1 times] = s···_{l times lefts to go}···(r^{-1}s)·r···_{km-1 times}···r = r^{(-1)l+1}s···_{l+1 times}···sr···_{km-1 times}···r

= $r^kr^{(-1)^{l+1}}··_{k_m}··r^{(-1)^{l+1}}s^{l+1} = r^kr^{(-1)^{l+1}k_m}s^{l+1}$ ∎

Lemma. Every symmetry σ can be expressed in the form σ = r^{k}s^{l} for k, l ∈ ℤ. It is called the normal form of σ and is unique up to modulo n on k and modulo 2 on l.

Proof.

Let’s suppose by reduction to the absurd that σ = r^{k}s^{l} = r^{k’}s^{l’} ⇒ r^{-k’}r^{k} = s^{l’}s^{-l}, the product of rotations is itself a rotation, and the product of reflections is also a reflection, that’s impossible, therefore the only option left is the identity.

r^{-k’}r^{k} = s^{l’}s^{-l} ⇒ r^{-k’}r^{k} = s^{l’}s^{-l} = id ⇒ r^{k} = r^{k’} and s^{l’} = s^{l} ⇒ **k ≡ k’ (mod n) and l ≡ l’ (mod 2)** since s^{2} = r^{n} = e∎

Theorem. The size of D_{n} is 2n.

Proof.

We have already demonstrated in the previous lemma that ∀σ ∈ D_{n}, σ = r^{k}s^{l} and this expression is unique up to modulo n on k and module 2 on l since s^{2} = r^{n} = e, so the options are 0 ≤ k ≤ n-1 and 0 ≤ l ≤ 2 ⇒ |D_{n}| = 2n ∎

Theorem. Let s be reflection across a line through a vertex. The n reflections in D_{n} are s, rs, r^{2}s,..., r^{n-1}s. Figure 1.b and 1.c.

Proof. The rigid motions s, rs, r^{2}s,…, r^{n-1}s are different since e, r, r^{2},…, r^{n-1} are different. Futhermore, s is not a rotation. Then, we claim that **r ^{k}s is not a rotation either**.

Suppose for the sake of contradiction, r^{k}s is a rotation ⇒ r^{k}s = r^{l} ⇒ s = r^{l-k}, but s is not a rotation, but a reflection ⊥ (contradiction).

Therefore, D_{n} = {e, r, r^{2},…, r^{n-1}, s, rs, r^{2}s,…, r^{n-1}s}

A reflection has order 2, so s^{2} = 1 and s^{-1} = s. The elements of D_{n} are rotations or reflections and the product of a rotation and a reflection is always a reflection. The geometric interpretation of s, rs, r^{2}s,…, r^{n-1}s is as follows: **starting from a vertex fixed by s, we draw all lines of reflections for a regular n-gon by moving clockwise $\frac{2kπ}{n}$ radians** where 0 ≤ k ≤ n − 1.

The rigid motions r and s do not commute, but sr^{n-1} = rs, and therefore D_{n} = {e, r, r^{2},…, r^{n-1}, s, rs, r^{2}s,…, r^{n-1}s} = ⟨r, s | r^{n} = s^{2} = e, rs = sr^{n-1}⟩.

A reflection is a transformation that acts like a mirror. A line that reflects a figure onto itself is called a line of symmetry.

Exercise. Let n = 6. Compute rsr^{4}sr^{3} = [∴ Associativity] (rs)r^{4}sr^{3} = [∴ n=6, rs=sr^{n-1}=sr^{5}] sr^{5}r^{4}sr^{3} = sr^{9}sr^{3} = [∴ r^{9} = r^{3}, 9 ≡_{6} 3] sr^{3}sr^{3} = (sr^{3})^{2} = [A rotation compose with a reflection is always a reflection and a reflection compose with itself is the identity] e.

Theorem. r^{k}s = sr^{n-k}, 1 ≤ k ≤ n-1.

Proof.
Induction. Base case (k = 1) is already based on the definition of the dihedral group, D_{n} = ⟨r, s | r^{n} = s^{2} = e, rs = sr^{n-1}⟩.

Let’s assume that the hypothesis holds for k.

r^{k+1}s = r (r^{k}s) = [k induction hypothesis] r(sr^{n-k}) = [Associativity] (rs)r^{n-k} = sr^{n-1}r^{n-k} = sr^{n+n-(k+1)} = [r^{n} = e] sr^{n-(k+1)} ∎

D_{3} is the symmetry group of the equilateral triangle. An equilateral triangle is a triangle in which all three sides are the same length, that is, **a regular polygon with three equal sides with angles of the same measure** (60°).

D_{3} = {id (e), rotations (ρ_{1} -120° around the center anticlockwise-, ρ_{2} -240°-), reflections or mirror images on all its vertices (μ_{1}, μ_{2}, and μ_{3})} = {e, r, r^{2}, s, rs, r^{2}s} = ⟨r, s | r^{3} = s^{2} = e, rs = sr^{2}⟩.

- 1 = e = ().
- 120° around the center anticlockwise, r = ρ
_{1}= $(\begin{smallmatrix}A & B & C\\ B & C & A\end{smallmatrix}) = (\begin{smallmatrix}1 & 2 & 3\\ 2 & 3 & 1\end{smallmatrix}) = $ (123). - r
^{2}= ρ_{2}= $(\begin{smallmatrix}A & B & C\\ C & A & B\end{smallmatrix}) = (\begin{smallmatrix}1 & 2 & 3\\ 3 & 1 & 2\end{smallmatrix}) = $(132). - s = μ
_{1}= $(\begin{smallmatrix}A & B & C\\ A & C & B\end{smallmatrix}) = (\begin{smallmatrix}1 & 2 & 3\\ 1 & 3 & 2\end{smallmatrix})$ = (23). - rs = μ
_{3}= $(\begin{smallmatrix}A & B & C\\ B & A & C\end{smallmatrix}) = (\begin{smallmatrix}1 & 2 & 3\\ 2 & 1 & 3\end{smallmatrix})$ = (12). - r
^{2}s = μ_{2}= $(\begin{smallmatrix}A & B & C\\ C & B & A\end{smallmatrix}) = (\begin{smallmatrix}1 & 2 & 3\\ 3 & 2 & 1\end{smallmatrix})$ = (13). Figure 101.

D_{3} **is not Abelian**, e.g., ρ_{1}μ_{1} = (123)(23) = (12) = μ_{3} ≠ μ_{1}ρ_{1} = (23)(123) = (13) = μ_{2}

**A square is a regular quadrilateral** (four-sided polygon), which means that it has four equal sides and four equal right angles (90° angles, π/2 radian angles). The dihedral group D4 is the symmetry group of the square.

D4 = {id (e), rotations (ρ_{1} -90° around the center anticlockwise-, ρ_{2} -180°-, and ρ_{3} -270°-), reflections or mirror images in the x -horizontal- and y -diagonal- axes (μ_{1}, μ_{2}), and reflections in the diagonals δ_{1} and δ_{2} respectively} = {e, r, r^{2}, r^{3}, s, rs, r^{2}s, r^{3}s} = ⟨r, s | r^{4} = s^{2} = e, rs = sr^{3}⟩.

- 1 = e = R
_{0}= (). - r = ρ
_{1}= R_{90}-90° around the center anticlockwise- = $(\begin{smallmatrix}1 & 2 & 3 & 4\\ 2 & 3 & 4 & 1\end{smallmatrix})$ = (1234). - r
^{2}= ρ_{2}= R_{180}-180° around the center anticlockwise- = $(\begin{smallmatrix}1 & 2 & 3 & 4\\ 3 & 4 & 1 & 2\end{smallmatrix})$ = (13)(24). - r
^{3}= ρ_{3}= R_{270}-270° around the center anticlockwise- = $(\begin{smallmatrix}1 & 2 & 3 & 4\\ 4 & 1 & 2 & 3\end{smallmatrix})$ = (1432). - s = μ
_{1}= H = $(\begin{smallmatrix}1 & 2 & 3 & 4\\ 2 & 1 & 4 & 3\end{smallmatrix})$ = (12)(34). - rs = δ
_{1}= D = $(\begin{smallmatrix}1 & 2 & 3 & 4\\ 3 & 2 & 1 & 4\end{smallmatrix})$ = (1234)(12)(34) = (13). - r
^{2}s = μ_{2}= V = $(\begin{smallmatrix}1 & 2 & 3 & 4\\ 4 & 3 & 2 & 1\end{smallmatrix})$ = (13)(24)(12)(34) = (14)(23). - r
^{3}s = δ_{2}= D’ = $(\begin{smallmatrix}1 & 2 & 3 & 4\\ 1 & 4 & 3 & 2\end{smallmatrix})$ = (1432)(12)(34) = (24)

D4 is not Abelian, e.g. μ_{1}ρ_{1} = (12)(34)(1234) = (24) = δ_{2} ≠ ρ_{1}μ_{1} = (1234)(12)(34) = (13) = δ_{1}.

$K$ = {R_{0}, R_{180}} = ⟨R_{180}⟩. Let $K$ be the subgroup of D_{4} generated by R_{180}, this subgroup is normal, $K$ ◁ D_{4}.

$D_4/K$ = {$K, R_{90}K, HK, DK$} where $K$ = {R_{0}, R_{180}}, $R_{90}K$ = {R_{90}, R_{270}}, $HK$ = {HR_{0}, HR_{180}} = {H, V}, $DK$ = {DR_{0}, DR_{180}} = {D, D’}.

Its multiplication table is given in Table 2.3. Observe that the multiplication table for D_{4} can be simplified into small boxes that are cosets of $K$. We can represent an entire box by a single element of the box!

Definition. The center of a group, G, is the set of elements that commute with all the elements of G. Z(G) = {g ∈ G | gx = xg, ∀x ∈ G}.

Center of D_{n} ℤ(D_{n}) = $
\begin{cases}
e, n~is~ odd\\\\
e~ and~ α^{\frac{n}{2}}, n~is~ even
\end{cases}$

The dihedral group D_{n} is the group of symmetries of a regular polygon with n vertices, which includes rotations and reflections or, alternatively, the rigid motions taking a regular n-gon back to itself. D_{n} = {e, r, r^{2},…, r^{n-1}, s, rs, r^{2}s,…, r^{n-1}s} = ⟨r, s | r^{n} = s^{2} = e, rs = sr^{n-1}⟩.

- 1, r, r
^{2},…, r^{n-1}are all distinct, r^{n}= 1, |r| = n. - |s| = 2.
- s ≠ r
^{i}∀i, sr^{i}≠ sr^{j}, ∀i, j: 0 ≤ i, j ≤ n-1, i ≠ j. - rs = sr
^{-1} - r
^{i}s = sr^{-i}∀i, j: 0 ≤ i ≤ n

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. This post relies heavily on the following resources, specially on NPTEL-NOC IITM, Introduction to Galois Theory, Michael Penn, and Contemporary Abstract Algebra, Joseph, A. Gallian.

- NPTEL-NOC IITM, Introduction to Galois Theory.
- Algebra, Second Edition, by Michael Artin.
- LibreTexts, Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
- Field and Galois Theory, by Patrick Morandi. Springer.
- Michael Penn (Abstract Algebra), and MathMajor.
- Contemporary Abstract Algebra, Joseph, A. Gallian.
- Andrew Misseldine: College Algebra and Abstract Algebra.