Dihedral Group Explorer — Leandro Junes

Polygon and group

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Right to left. A product $AB = A\circ B$ means: apply $B$ first, then $A$ . Here $r$ is the rotation by one step counterclockwise and $s$ is the reflection in the axis through $v_0$ .

Number of sides

n

Examples

Animation speed

The polygon

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Kind of symmetry

Which one

Exponent

k

r

Show all reflection axes

The selected symmetry

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Where the vertices go

vertex	image

Two symmetries

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$A\circ B$ means: apply $B$ first, then $A$ . The symmetry written on the right acts first, as with functions: $(A\circ B)(v)=A(B(v))$ .

A

(applied second)

B

(applied first)

How the product is worked out from the relations

Step by step on the polygon

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Where the vertices go

v	B(v)	A(B(v))	(A ∘ B)(v)	(B ∘ A)(v)

An element

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Kind of element

Element

All elements

Center and commutativity

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Conjugacy classes

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Two elements $x$ and $y$ are conjugate when $y=gxg^{-1}$ for some $g$ : they are the same symmetry seen after moving the polygon by $g$ .

class	representative	size	members

The selected class

Generating sets

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The chosen elements

Add an element

Subgroups

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Subgroup lattice

List of subgroups

subgroup	order	index	structure	normal	generators

The selected subgroup

Cayley graph

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The Cayley graph is a picture of the abstract group, not of the polygon: it has one vertex for every element, and an edge from $g$ to $gr$ and to $gs$ (multiplication on the right by the two generators).

About dihedral groups

Notation. Here $D_n=\langle r,s\mid r^n=e,\ s^2=e,\ srs=r^{-1}\rangle$ is the group of the $2n$ symmetries of a regular $n$ -gon, for $n\ge 3$ . Some textbooks name the same group after its order and write $D_{2n}$ . Every element is written in exactly one way as $r^i s^j$ with $0\le i<n$ and $j\in\{0,1\}$ , and $(r^is^j)(r^ks^l)=r^{\,i+(-1)^jk}\,s^{\,j+l}$ , the exponent of $r$ taken modulo $n$ and that of $s$ modulo 2.

What a dihedral group is

A symmetry of a regular polygon is a rigid motion of the plane that puts the polygon back onto itself. Doing one symmetry after another is again a symmetry, doing nothing is a symmetry, and every symmetry can be undone: the symmetries form a group under composition. That group is the dihedral group.

Why there are 2n elements

A symmetry is known once you know where it sends the vertex $v_0$ and its neighbour $v_1$ . There are $n$ places for $v_0$ , and then $v_1$ must go to one of the two neighbours of that place: $n\cdot 2=2n$ symmetries, the $n$ rotations (the neighbour stays on the counterclockwise side) and the $n$ reflections (it changes sides).

How the group acts on the vertices

The vertices are $v_0,\dots,v_{n-1}$ , counterclockwise, with $v_0$ at the top. The generators act by $r(v_k)=v_{k+1}$ and $s(v_k)=v_{-k}$ , so $(r^is^j)(v_k)=v_{\,i+(-1)^jk}$ , indices modulo $n$ . Every table of vertices on this page comes from that formula, and every product from the multiplication rule above; the tests of this page check that the two always agree.

Rotations and reflections

The rotation $r^k$ turns the polygon by $360k/n$ degrees counterclockwise about its centre and has order $n/\gcd(n,k)$ . The reflection $r^ks$ turns the polygon over about an axis through the centre: the axis of $s$ turned counterclockwise by $180k/n$ degrees. For odd $n$ every axis joins a vertex to the midpoint of the opposite edge. For even $n$ the axes with even $k$ join two opposite vertices and those with odd $k$ join the midpoints of two opposite edges.

Why a reflection reverses orientation

Walk round the polygon counterclockwise: $v_0,v_1,v_2,\dots$ After a rotation the images still run counterclockwise. After a reflection they run clockwise, because $k\mapsto i-k$ counts backwards. No motion that stays inside the plane can do that: the polygon has to be lifted and turned over. That is why the page shows a reflection as a flip through the third dimension, and why the shapes in between are not symmetries.

Why reflections have order 2

Turning the polygon over twice about the same axis puts every vertex back. In symbols, $(r^ks)(r^ks)=r^k(sr^k)s=r^kr^{-k}ss=e$ .

Why r and s do not commute, and the relation srs = r⁻¹

Reflect, rotate one step counterclockwise, reflect back: seen from the other side, the counterclockwise step was a clockwise one. So $srs=r^{-1}$ , or $sr=r^{-1}s$ : an $s$ may move past an $r$ , but the $r$ becomes $r^{-1}$ . Since $r^{-1}\ne r$ for $n\ge 3$ , $sr\ne rs$ , and the group is not abelian.

Why the product of two reflections is a rotation

Two flips bring the same face up again, so the result preserves orientation: it is a rotation. Precisely, $(r^is)(r^ks)=r^{\,i-k}$ . The two axes meet at the angle $180(i-k)/n$ degrees, and the product turns by twice that angle, from the axis of the reflection applied first towards the other.

Why the center depends on the parity of n

A rotation $r^k$ commutes with $s$ exactly when $r^k=r^{-k}$ , that is when $2k$ is a multiple of $n$ . For odd $n$ only $k=0$ does it; for even $n$ so does the half turn $r^{n/2}$ . No reflection commutes with $r$ . Hence $Z(D_n)=\{e\}$ for odd $n$ and $\{e,r^{n/2}\}$ for even $n$ .

How the subgroups are classified

A subgroup either has no reflection or has as many reflections as rotations. Its rotations form a subgroup of $\langle r\rangle$ , so they are $\langle r^d\rangle$ for a divisor $d$ of $n$ . That gives the rotation subgroups $H_d=\langle r^d\rangle$ , of order $n/d$ , and the subgroups $K_{d,k}=\langle r^d,\,r^ks\rangle$ with $0\le k<d$ , of order $2n/d$ : one for each divisor, and $d$ more for each divisor, $\tau(n)+\sigma(n)$ in all. Every $H_d$ is normal; $K_{d,k}$ is normal exactly when $d$ divides 2.

Limits. Everything is computed in your browser, for $3\le n\le 1000$ . The polygon is drawn with every vertex labelled for $n\le 24$ , without individual labels for $n\le 60$ , and described in symbols beyond that. The subgroup lattice is drawn for up to 60 subgroups and the Cayley graph for groups of at most 48 elements. Long lists are shortened and say so; no result is left out because a picture is not drawn.