presentation of dihedral group

Dihedral Group

Renteln and Dundes (2005) give the following (bad) mathematical joke about the dihedral group:

Q: What's hot, chunky, and acts on a polygon? A: Dihedral soup.

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Math : The Dihedral Group

The general dihedral group D n is the symmetry group of the regular n-sided polygon and consists of the identity transformation, rotation about the axis through the center of the polygon, and reflection through each of the polygon's mirror planes (these planes always contain the axis of rotation and either a vertex or the center of a side). An n-sided regular polygon will always have n planes of reflection. An equilateral triangle will have the symmetry group D 3 , a square D 4 , a pentagon D 5 , etc. In each of these cases, the dihedral groups will contain the subgroups of the polygon's other symmetries. For example, the symmetry group D 3 contains the subgroup of C 3 (the rotational symmetry) and three second order subgroups (C 2 - reflections through each mirror plane).

Page author: Sasie Sealy

nLab dihedral group

Group theory.

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Dihedral groups

Binary dihedral/dicyclic groups, group cohomology, as part of the ade pattern, group presentation, quaternion group q 8 q_8 and triality.

The dihedral group, D 2 n D_{2n} , is a finite group of order 2 n 2n . It may be defined as the symmetry group of a regular n n -gon.

For instance D 6 D_6 is the symmetry group of the equilateral triangle and is isomorphic to the symmetric group , S 3 S_3 .

For n ∈ ℕ n \in \mathbb{N} , n ≥ 1 n \geq 1 , the dihedral group D 2 n D_{2n} is thus the subgroup of the orthogonal group O ( 2 ) O(2) which is generated from the finite cyclic subgroup C n ≔ ℤ / n C_n \,\coloneqq\, \mathbb{Z}/n of SO ( 2 ) SO(2) and the reflection at the x x -axis (say). It is a semi-direct product of C n C_n and a C 2 ≔ ℤ / 2 C_2 \,\coloneqq\, \mathbb{Z}/2 corresponding to that reflection, hence fitting into a short exact sequence as follows:

Under the further embedding O ( 2 ) ↪ SO ( 3 ) O(2)\hookrightarrow SO(3) the cyclic and dihedral groups are precisely those finite subgroups of SO(3) that, among their ADE classification , are not in the exceptional series.

Warning on notation

There are two different conventions for numbering the dihedral groups.

The above is the algebraic convention in which the suffix gives the order of the group: | D 2 n | = 2 n {\vert D_{2 n}\vert} = 2 n .

In the geometric convention one writes “ D n D_n ” instead of “ D 2 n D_{2n} ”, recording rather the geometric nature of the object of which it is the symmetry group.

Also beware that there is yet another group denoted D n D_n mentioned at Coxeter group .

Under the further lift through the spin group - double cover map SU ( 2 ) ≃ Spin ( 3 ) → SO ( 3 ) SU(2) \simeq Spin(3) \to SO(3) of the special orthogonal group , the dihedral group D 2 n D_{2n} is covered by the binary dihedral group , also known as the dicyclic group and denoted

Equivalently, this is the lift of the dihedral group D 2 n D_{2n} ( above ) through the pin group double cover of the orthogonal group O(2) to Pin(2)

Explicity, let ℍ ≃ ℂ ⊕ j ℂ \mathbb{H} \simeq \mathbb{C} \oplus \mathrm{j} \mathbb{C} be the quaternions realized as the Cayley-Dickson double of the complex numbers , and identify the circle group

with the unit circle in ℂ ↪ ℍ \mathbb{C} \hookrightarrow \mathbb{H} this way, with group structure given by multiplication of quaternions . Then the Pin group Pin(2) is isomorphic to the subgroup of the group of units ℍ × \mathbb{H}^\times of the quaternions which consists of this copy of SO(2) together with the multiples of the imaginary quaternion j \mathrm{j} with this copy:

The binary dihedral group 2 D 2 n 2 D_{2n} is the subgroup of that generated from

a ≔ exp ( π i 1 n ) ∈ S ( ℂ ) ⊂ Pin − ( 2 ) ⊂ Spin ( 3 ) a \coloneqq \exp\left( \pi \mathrm{i} \tfrac{1}{n} \right) \in S(\mathbb{C}) \subset Pin_-(2) \subset Spin(3)

x ≔ j ∈ Pin − ( 2 ) ⊂ Spin ( 3 ) x \coloneqq \mathrm{j} \in Pin_-(2) \subset Spin(3) .

It is manifest that these two generators satisfy the relations

and in fact these generators and relations fully determine 2 D 2 n 2 D_{2n} , up to isomorphism .

The group cohomology of the dihedral group is discussed for instance at Groupprops .

ADE classification and McKay correspondence

The dihedral group D 2 n D_{2n} has a group presentation

From this it is easy to see that it is a semi-direct product of the C n C_n generated by x x and the C 2 C_2 generated by y y . The action of y y on x x is given by y x = x − 1 \,^y x= x^{-1} .

It is a standard example considered in elementary combinatorial group theory .

The first binary dihedral group 2 D 4 2 D_4 is isomorphic to the quaternion group of order 8:

In the ADE-classification this is the entry D4 .

linear representation theory of binary dihedral group 2 D 4 2 D_4

= = dicyclic group Dic 2 Dic_2 = = quaternion group Q 8 Q_8

group order : | 2 D 4 | = 8 {\vert 2D_4\vert} = 8

character table over splitting field ℚ ( α , β ) \mathbb{Q}(\alpha,\beta) / complex numbers ℂ \mathbb{C}

character table over rational numbers ℚ \mathbb{Q} / real numbers ℝ \mathbb{R}

GroupNames , Q8 ,

Groupprops , Linear representation theory of dicyclic groups

James Montaldi , Real representations – Binary cubic – Q8

anti-cyclotomic extension

dihedral homology

group presentation

Coxeter group

Discussion in the context of the classification of finite rotation groups goes back to

• Felix Klein , chapter I.4 of Vorlesungen über das Ikosaeder und die Auflösung der Gleichungen vom fünften Grade , 1884, translated as Lectures on the Icosahedron and the Resolution of Equations of Degree Five by George Morrice 1888, online version

Wikipedia, Dihedral group

Wikipedia, Binary dihedral group

Wikipedia, Dicyclic group

Groupprops , Dicyclic group

Groupprops , Group cohomology of dihedral group:D8

GroupNames , Dicyclic groups Dic n Dic_n

Discussion as the equivariance group in equivariant cohomology theory :

• John Greenlees , Section 2 of: Rational SO ( 3 ) SO(3) -Equivariant Cohomology Theories , in Homotopy methods in algebraic topology (Boulder, CO, 1999), Contemp. Math. 271 , Amer. Math. Soc. (2001) 99 ( web , GBooks )

and specifically in equivariant K-theory and KR-theory :

• Robert Bruner , John Greenlees , Chapter 8 of: Connective Real K-Theory of Finite Groups , Mathematical Surveys and Monographs 169 AMS 2010 ( ISBN:978-0-8218-5189-0 )

Discussion of equivariant ordinary cohomology ( Bredon cohomology ) over the point but in arbitrary RO(G)-degree , for equivariance group a dihedral group of order 2 p 2p :

• Igor Kriz , Yunze Lu , On the RO ( G ) RO(G) -graded coefficients of dihedral equivariant cohomology , Mathematical Research Letters 27 4 (2020) ( arXiv:2005.01225 , doi:10.4310/MRL.2020.v27.n4.a7 )
• Yajit Jain, Bredon Equivariant Homology of Representation Spheres , 2014 ( pdf , pdf )

Last revised on December 22, 2021 at 20:51:35. See the history of this page for a list of all contributions to it.

Dihedral Group D4/Group Presentation

Group presentation of dihedral group $d_4$.

The group presentation of the dihedral group $D_4$ is given by:

We have that the group presentation of the dihedral group $D_n$ is:

Setting $n = 4, \alpha = a, \beta = b$, we get:

from which the result follows.

$\blacksquare$

• Proven Results
• Dihedral Group D4

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dihedral group properties

1 properties of dihedral groups.

Proposition 2 .

2 | n , { a 2 ⁢ i ⁢ b | i ∈ ℤ n } and { a 2 ⁢ i + 1 ⁢ b | i ∈ ℤ n }

2 ∤ n , { a i ⁢ b | i ∈ ℤ n } .

Consequently when n > 2 the center of D 2 ⁢ n is 1 when 2 ∤ n and Z ⁢ ( D 2 ⁢ n ) = ⟨ a n / 2 ⟩ when 2 | n . Furthermore C n := ⟨ a ⟩ is a characteristic subgroup of D 2 ⁢ n , provided n ≠ 2 .

For the conjugacy classes note that ( a i ) a j ⁢ b = ( a i ) b = a - i so that these conjugacy classes are established. Next

for all i ∈ ℤ . When 2 ∤ n we have ( 2 , n ) = 1 so 2 is invertible modulo n . We let j = 2 - 1 ⁢ ( k - i ) for any k ∈ ℤ and we see that a i ⁢ b is conjugate to any a k ⁢ b . However, when 2 | n we have a parity constraint that so far creates the two classes. We need to also verify conjugation by a j ⁢ b does not fuse the two classes. Indeed

Finally, the order of the elements ( a i ⁢ b ) is 2 – a fact used already. Thus the only cyclic subgroup of order n , when n > 2 , is C n and thus by its uniqueness it is characteristic. ∎

Proposition 3 .

The maximal subgroups of D 2 ⁢ n are dihedral or cyclic. In particular, the unique maximal cyclic group is C n = ⟨ a ⟩ and the maximal dihedral groups are those of the form ⟨ a n / p , a i ⁢ b ⟩ for primes p dividing n .

Corollary 4 .

If H is normal and contains an element of the form a i ⁢ b , then it contains the entire conjugacy class of a i ⁢ b . If n is odd then all reflections are conjugate to a i ⁢ b so H contains all reflections of D 2 ⁢ n and so H is D 2 ⁢ n as the relfections generate D 2 ⁢ n .

If instead n is even then H is forced only to contain one of the two conjugacy classes of reflections. If i is even then H contains b and a 2 ⁢ b so it contains a 2 . If i is odd then H contains a ⁢ b and a 3 ⁢ b so it contains a 2 = a ⁢ b ⁢ a 3 ⁢ b (note n > 3 as n > 2 and 2 | n ).

The two maximal subgroups of index 2 which can exist when n is even can be interchanged by an outer automorphism which maps a ↦ a - 1 and b ↦ a ⁢ b so these two are not characterisitic. The subgroups of a characterisitic cyclic group are necessarily characteristic. ∎

Proposition 5 .

Proposition 6 ..

D n is nilpotent if and only if n = 2 i for some i ≥ 0 .

Proposition 7 .

D 2 ⁢ n is solvable for all n ≥ 1 .

1.1 Automorphisms of D 2 ⁢ n

Theorem 8 ..

Let n > 2 . The automorphism group of D 2 ⁢ n is isomorphic to Z n × ⋉ Z n , with the canonical action of 1 : Z n × → Aut ⁡ Z n = Z n × . Explicitly,

with γ s , t defined as

Given γ ∈ Aut ⁡ D 2 ⁢ n , we know ⟨ a ⟩ is characteristic in D 2 ⁢ n so a ⁢ γ = a s for some s ∈ ℤ n . But γ is invertible so indeed ( s , n ) = 1 so that s ∈ ℤ n × . Next b ⁢ γ = a t ⁢ b as b cannot be sent to ⟨ a ⟩ .

Now we claim γ = γ s , t .

So indeed γ s , t is a homomorphism.

Finally, we show the composition of two such maps both to identify the automorphism group and to show that each γ s , t is invertible.

Hence, γ s , t ⁢ γ u , v = γ s ⁢ u , t ⁢ u + v . This agrees on a i ’s as well. This reveals the isomorphism desired: Aut ⁡ D 2 ⁢ n → ℤ n × ⋉ ℤ n by γ s , t ↦ ( s , t ) where we see the multiplications agree as

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Automorphism groups of dihedral groups

• Published: March 1977
• Volume 29 , pages 162–167, ( 1977 )

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Literature cited

N. Bourbaki, Lie Groups and Algebras, Hermann, Paris (1960).

W. Magnus, A. Karrass, and D. Solitar, Combinatorial Group Theory, Wiley (1966).

O. N. Golovin and L. E. Sadovskii, “On automorphism groups of free products,” Mat. Sb., 4 , No. 3, 505–514 (1938).

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É. Hecke, Lectures on the Theory of Algebraic Numbers, Akad. Verlagsgesellschaft, Leipzig (1923).

I. M. Vinogradov, Foundations of the Theory of Numbers [in Russian], Nauka, Moscow (1972).

M. I. Kargapolov and Yu. I. Merzlyakov, Foundations of Group Theory [in Russian], Nauka, Moscow (1972).

A. G. Kurosh, Group Theory [in Russian], Nauka, Moscow (1967).

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Translated from Ukrainskii Matematicheskii Zhurnal, Vol. 29, No. 2, pp. 216–222, March–April, 1977

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Rotma1er, F. Automorphism groups of dihedral groups. Ukr Math J 29 , 162–167 (1977). https://doi.org/10.1007/BF01089242

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DOI : https://doi.org/10.1007/BF01089242

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A presentation of a Dihedral group

I know that a dihedral group $D_{2n }$ is generated by rotation $r$ by $2\pi/n$ and reflection $s$ subject to relations $r^n = 1$ , $s^2 = 1$ , and $rs = sr^{-1}$ . So a dihedral group $D_{2n}$ has a presentation $(r, s {\,|\,} r^n, s^2, (rs)^2)$ . In other words, $$ D_{2n} \cong (r, s {\,|\,} r^n, s^2, (rs)^2) = F(r,s)/N $$ where $N$ is the smallest normal subgroup containing $\{r^n, s^2, (rs)^2\} \subset F(r, s)$ . It is not difficult to show, without using the isomorphism, that there are at most $2n$ equivalence classes in $F(r,s)/N$ , namely, $[r^is^j]$ for $i = 0, \dots, n-1$ and $j = 0, 1$ . Then the isomorphism implies that all of them are distinct. Is it possible to show that these equivalence classes are all distinct without using the isomorphism? For example, how to show that $r \nsim ()$ , or, equivalently, that $r \notin N$ ?

Edit: I don't think the link has an answer to my question. So let me rephrase it. Consider a free group $F(a,b)$ . Let $N$ be the smallest normal subgroup that contains "words" $a^n$ , $b^2$ , and $(ab)^2$ . Is it possible to show that $a \notin N$ without using any specific group "from nature"?

• group-theory
• group-presentation
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• 1 $\begingroup$ The general method would be to find a homomorphism to a group "in nature" which has $2n$ elements. The linked answer does exactly that. $\endgroup$ –  Lee Mosher Commented Jun 12, 2022 at 22:04
• $\begingroup$ Is it possible to do that without using a specific group "from nature"? @Lee Mosher $\endgroup$ –  Yerbolat Commented Jun 12, 2022 at 22:12
• 1 $\begingroup$ It might be possible, but I'm not aware of a way to do it. Here's some intuition from mathematical logic: The rewrite rules for $G$ give us a proof system which lets us verify that two words represent the same group element. Indeed, a "proof" of this fact is just a sequence of applications of the relations which converts one word into another. What you want to do is argue that two words are different , that is, you want to show that there is no proof that they are the same. $\endgroup$ –  Chris Grossack Commented Jun 12, 2022 at 23:21
• 1 $\begingroup$ But generally, there's no way to argue syntactically that no proof exists! The way we show that no proof exists is by building a model which verifies the axioms, but in which the conclusion fails. In this group theoretic setting, that means to show that two elements are different, our only recourse is to find a homomorphism from our group $G$ to some other group $H$ so that our two words get sent to different elements of $H$! Of course, this is exactly the approach that @LeeMosher is suggesting. If a purely syntactic approach exists, I would love to see it, but I suspect there isn't one. $\endgroup$ –  Chris Grossack Commented Jun 12, 2022 at 23:24
• 1 $\begingroup$ @HallaSurvivor Interesting! Thank you for sharing insights. I do understand that it is possible to do it constructing a homomorphism. Just was extremely curious if it is possible to argue "syntactically". $\endgroup$ –  Yerbolat Commented Jun 12, 2022 at 23:29

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