dihedral group properties

1 Properties of Dihedral Groups

A group generated by two involutions is a dihedral groupMathworldPlanetmath. When the group is finite it is possible to show that the group has order 2n for some n>0 and takes the presentationMathworldPlanetmathPlanetmathPlanetmath

Remark 1.

Contemporary group theorists prefer D2n over Dn as the notation for the dihedral group of order 2n. Although this notation is overly explicit, it does help to resolve the ambiguity with the Lie type Dl which corresponds to the orthogonal groupMathworldPlanetmath Ω+(2l,q). However, introductory texts in algebraMathworldPlanetmathPlanetmath still make use of the more appropriate Dn notation to emphasize the connection to the symmetriesMathworldPlanetmathPlanetmathPlanetmath of a regularPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath n-gon (n-sided polygonMathworldPlanetmathPlanetmath).

D22, D42×2 are the two abelianMathworldPlanetmath examples of dihedral groups. They can be considered as dihedral groups of the respective order because they satisfy the relationsMathworldPlanetmathPlanetmath, though the geometric interpretationsMathworldPlanetmathPlanetmath are slightly modified. Often D2 can be termed the symmetries of a line segmentMathworldPlanetmath, and the D4 the symmetries of a non-square rectangleMathworldPlanetmathPlanetmath, as the symmetry groups of each of these object is isomorphicPlanetmathPlanetmathPlanetmath to D2 and D4 respectively. These exceptions cause problems for most theorems on dihedral groups so it is convenient to insist that n>2 for theorems.

Proposition 2.

is a irredundant list of the elements of D2n. Moreover the conjugacy classesMathworldPlanetmathPlanetmath of D2n are {ai,a-i} for all iZn and

  • 2|n, {a2ib|in} and {a2i+1b|in}

  • 2n, {aib|in}.

Consequently when n>2 the center of D2n is 1 when 2n and Z(D2n)=an/2 when 2|n. Furthermore Cn:=a is a characteristic subgroup of D2n, provided n2.


The conjugationMathworldPlanetmath relation ab=a-1 allows us to place every element in the normal form aibj. If aibj=akbl then ai-k=bl-j. Yet ab=1 so i-k0(modn) and l-j0(mod2). Thus we have and irredundant list as required.

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


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


thus we retain two conjugacy classes amongst the reflectionsPlanetmathPlanetmath.

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

Proposition 3.

The maximal subgroups of D2n are dihedral or cyclic. In particular, the unique maximal cyclic group is Cn=a and the maximal dihedral groups are those of the form an/p,aib for primes p dividing n.

We will prove this with a more general claim. First we pause to note that as a corollary to these two propositionsPlanetmathPlanetmath we can determine the entire latticeMathworldPlanetmath of normal and characteristic subgroups of a dihedral group.

Corollary 4.

A proper subgroupMathworldPlanetmath H of D2n is normal in D2n if and only if Ha or 2|n, and H is one the following two maximal subgroups of index 2:


The proper characteristic subgroups of D2n are all the subgroupsMathworldPlanetmathPlanetmath of a.


If H is normal and contains an element of the form aib, then it contains the entire conjugacy class of aib. If n is odd then all reflections are conjugate to aib so H contains all reflections of D2n and so H is D2n as the relfections generate D2n.

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 a2b so it contains a2. If i is odd then H contains ab and a3b so it contains a2=aba3b (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 aa-1 and bab so these two are not characterisitic. The subgroups of a characterisitic cyclic group are necessarily characteristic. ∎

Proposition 5.

Quotient groupsMathworldPlanetmath of dihedral groups are dihedral, and subgroups of dihedral groups are dihedral or cyclic.


The homomorphic imagePlanetmathPlanetmathPlanetmath of a dihedral group has two generatorsPlanetmathPlanetmathPlanetmath a^ and b^ which satisfy the conditions a^b^=a^-1 and a^n=1 and b^2=1, therefore the image is a dihedral group.

For subgroups we proceed by inductionMathworldPlanetmath. When n=1 the result is clear. Now suppose that D2n has some proper subgroup H that is not dihedral or cyclic. H is contained in some maximal subgroup M of D2n. However the maximal subgroups of D2n are cyclic or dihedral so H falls to the induction step for M – together with the fact that subgroups of cyclic groups are cyclic. Thus H must actually be dihedral or cyclic to avoid contradictionsMathworldPlanetmathPlanetmath. ∎

Proposition 6.

Dn is nilpotent if and only if n=2i for some i0.

Proposition 7.

D2n is solvable for all n1.


When n=1, D2n2 which is nilpotent and so also solvable. Now let n>1. Then D2n/a2 and an. Both n and 2 are nilpotent and so they are both solvable. As extensionsPlanetmathPlanetmathPlanetmath of solvable groups are solvable, D2n is solvable for all n>0. ∎

1.1 Automorphisms of D2n

Theorem 8.

Let n>2. The automorphism group of D2n is isomorphic to Zn×Zn, with the canonical action of 1:Zn×AutZn=Zn×. Explicitly,


with γs,t defined as


We apply the needle-in-the-haystack heuristic and search first to explain why these are the only possible forms for the automorphismsMathworldPlanetmathPlanetmathPlanetmathPlanetmath. We will then prove all such are indeed automorphisms.

Given γAutD2n, we know a is characteristic in D2n so aγ=as for some sn. But γ is invertible so indeed (s,n)=1 so that sn×. Next bγ=atb as b cannot be sent to a.

Now we claim γ=γs,t.




Now we must show all γs,t are indeed homomorphismsMathworldPlanetmathPlanetmath when sn× and tn. First we note that γ is well-defined as we have an irredundant listing of the elements. Next we verify the homomorphism cases.

(aiaj)γs,t = a(i+j)s=aisajs=(ai)γs,t(aj)γs,t.
(aiajb)γs,t = a(i+j)s+tb=ais(ajs+tb)=(ai)γs,t(aib)γs,t.
(aibaj)γs,t = (ai-jb)γs,t=a(i-j)s+tb=ais+tbajs=(aib)γs,t(aj)γs,t.
(aibajb)γs,t = (ai-j)γs,t=ais-js=ais+t-t-is
= (ais+tb)(ba-t-is)=(ais+tb)(ajs+tb)=(aib)γs,t(ajb)γs,t.

So indeed γs,t is a homomorphism.

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


Hence, γs,tγu,v=γsu,tu+v. This agrees on ai’s as well. This reveals the isomorphism desired: AutD2nn×n by γs,t(s,t) where we see the multiplications agree as


In fact this demonstrates that the inverseMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath of γs,t is simply γs-1,-ts- and the identity mapMathworldPlanetmath is γ1,0. ∎

Title dihedral group properties
Canonical name DihedralGroupProperties
Date of creation 2013-03-22 16:06:35
Last modified on 2013-03-22 16:06:35
Owner Algeboy (12884)
Last modified by Algeboy (12884)
Numerical id 10
Author Algeboy (12884)
Entry type Topic
Classification msc 20F55
Related topic GeneralizedQuaternionGroup