dihedral group properties

dihedral group properties

1

Properties of Dihedral Groups

A group generated by two

involutions

is a

dihedral group

. When
the group is finite it is possible to show that the group has
order

2

⁢

n

for some

n

>

0

and takes the

presentation

D

2

⁢

n

=

⟨

a

,

b

|

a

n

=

1

,

b

2

=

1

,

a

b

=

a

-

1

⟩

.

Remark 1

.

Contemporary group theorists prefer

D

2

⁢

n

over

D

n

as the notation for the
dihedral group of order

2

⁢

n

. Although this notation is overly explicit, it does help to resolve the ambiguity with the Lie type

D

l

which corresponds to the

orthogonal group

Ω

+

⁢

(

2

⁢

l

,

q

)

. However, introductory texts in

algebra

still make use of the more appropriate

D

n

notation to emphasize the

connection

to the

symmetries

of a

regular

n

-gon (

n

-sided

polygon

).

D

2

≅

ℤ

2

,

D

4

≅

ℤ

2

×

ℤ

2

are the two

abelian

examples of dihedral groups. They can be considered as dihedral groups of the respective order because they

satisfy

the

relations

, though the geometric

interpretations

are slightly modified. Often

D

2

can be termed the symmetries of a

line segment

, and the

D

4

the symmetries of a non-square

rectangle

, as the

symmetry groups

of each of these object is

isomorphic

to

D

2

and

D

4

respectively. These exceptions cause problems for most

theorems

on dihedral groups so it is convenient to insist that

n

>

2

for theorems.

Proposition 2

.

D

2

⁢

n

=

{

a

i

|

i

∈

ℤ

n

}

⊔

{

a

i

⁢

b

|

i

∈

ℤ

n

}

is a

irredundant

list of the elements of

D

2

⁢

n

. Moreover the

conjugacy
classes

of

D

2

⁢

n

are

{

a

i

,

a

-

i

}

for all

i

∈

Z

n

and

•

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

.

Proof.

The

conjugation

relation

a

b

=

a

-

1

allows us to place every

element

in the normal form

a

i

⁢

b

j

. If

a

i

⁢

b

j

=

a

k

⁢

b

l

then

a

i

-

k

=

b

l

-

j

. Yet

⟨

a

⟩

∩

⟨

b

⟩

=

1

so

i

-

k

≡

0

(

mod

n

)

and

l

-

j

≡

0

(

mod

2

)

. Thus we have and
irredundant list as required.

For the conjugacy classes note that

(

a

i

)

a

j

⁢

b

=

(

a

i

)

b

=

a

-

i

so
that these conjugacy classes are established. Next

(

a

i

⁢

b

)

a

j

=

a

-

j

+

i

⁢

b

⁢

a

j

=

a

-

j

+

i

⁢

a

-

j

⁢

b

=

a

-

2

⁢

j

+

i

⁢

b

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

(

a

i

⁢

b

)

a

j

⁢

b

=

(

a

-

2

⁢

j

+

i

⁢

b

)

b

=

b

⁢

a

-

2

⁢

j

+

i

⁢

b

=

a

2

⁢

j

-

i

⁢

b

,

thus we retain two conjugacy classes amongst the

reflections

.

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

.

We will prove this with a more general claim. First we pause to note that as a corollary to these two

propositions

we can determine the entire

lattice

of normal and characteristic subgroups of a dihedral group.

Corollary 4

.

A

proper subgroup

H

of

D

2

⁢

n

is normal in

D

2

⁢

n

if and only if

H

≤

⟨

a

⟩

or

2

|

n

, and

H

is one the following two maximal subgroups
of index 2:

M

1

=

⟨

a

2

,

b

⟩

,

M

2

=

⟨

a

2

,

a

⁢

b

⟩

.

The proper characteristic subgroups of

D

2

⁢

n

are all the

subgroups

of

⟨

a

⟩

.

Proof.

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

.

Quotient groups

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

Proof.

The

homomorphic image

of a dihedral group has two

generators

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

induction

. When

n

=

1

the result is clear.
Now suppose that

D

2

⁢

n

has some proper subgroup

H

that is not dihedral
or cyclic.

H

is

contained

in some maximal subgroup

M

of

D

2

⁢

n

. However
the maximal subgroups of

D

2

⁢

n

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

contradictions

.
∎

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

.

Proof.

When

n

=

1

,

D

2

⁢

n

≅

ℤ

2

which is nilpotent and so also solvable.
Now let

n

>

1

. Then

D

2

⁢

n

/

⟨

a

⟩

≅

ℤ

2

and

⟨

a

⟩

≅

ℤ

n

. Both

ℤ

n

and

ℤ

2

are nilpotent and so they are both solvable.
As

extensions

of

solvable groups

are solvable,

D

2

⁢

n

is solvable for all

n

>

0

.
∎

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,

Aut

⁡

D

2

⁢

n

=

{

γ

s

,

t

|

s

∈

ℤ

n

×

,

t

∈

ℤ

n

}

with

γ

s

,

t

defined as

(

a

i

)

⁢

γ

s

,

t

=

α

i

⁢

s

,

(

a

i

⁢

b

)

⁢

γ

s

,

t

=

a

i

⁢

s

+

t

⁢

b

.

Proof.

We apply the needle-in-the-haystack heuristic and search first to
explain why these are the only possible forms for the

automorphisms

.
We will then prove all such are indeed automorphisms.

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

.

(

a

i

)

⁢

γ

=

a

i

⁢

s

=

(

a

i

)

⁢

γ

s

,

t

and

(

a

i

⁢

b

)

⁢

γ

=

a

i

⁢

s

⁢

a

t

⁢

b

=

a

i

⁢

s

+

t

⁢

b

=

(

a

i

⁢

b

)

⁢

γ

s

,

t

.

Now we must show all

γ

s

,

t

are indeed

homomorphisms

when

s

∈

ℤ

n

×

and

t

∈

ℤ

n

. First we note that

γ

is well-defined as we have an irredundant listing of the elements. Next we verify the homomorphism cases.

(

a

i

⁢

a

j

)

⁢

γ

s

,

t

=

a

(

i

+

j

)

⁢

s

=

a

i

⁢

s

⁢

a

j

⁢

s

=

(

a

i

)

⁢

γ

s

,

t

⁢

(

a

j

)

⁢

γ

s

,

t

.

(

a

i

⁢

a

j

⁢

b

)

⁢

γ

s

,

t

=

a

(

i

+

j

)

⁢

s

+

t

⁢

b

=

a

i

⁢

s

⁢

(

a

j

⁢

s

+

t

⁢

b

)

=

(

a

i

)

⁢

γ

s

,

t

⁢

(

a

i

⁢

b

)

⁢

γ

s

,

t

.

(

a

i

⁢

b

⁢

a

j

)

⁢

γ

s

,

t

=

(

a

i

-

j

⁢

b

)

⁢

γ

s

,

t

=

a

(

i

-

j

)

⁢

s

+

t

⁢

b

=

a

i

⁢

s

+

t

⁢

b

⁢

a

j

⁢

s

=

(

a

i

⁢

b

)

⁢

γ

s

,

t

⁢

(

a

j

)

⁢

γ

s

,

t

.

(

a

i

⁢

b

⁢

a

j

⁢

b

)

⁢

γ

s

,

t

=

(

a

i

-

j

)

⁢

γ

s

,

t

=

a

i

⁢

s

-

j

⁢

s

=

a

i

⁢

s

+

t

-

t

-

i

⁢

s

=

(

a

i

⁢

s

+

t

⁢

b

)

⁢

(

b

⁢

a

-

t

-

i

⁢

s

)

=

(

a

i

⁢

s

+

t

⁢

b

)

⁢

(

a

j

⁢

s

+

t

⁢

b

)

=

(

a

i

⁢

b

)

⁢

γ

s

,

t

⁢

(

a

j

⁢

b

)

⁢

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.

(

a

i

⁢

b

)

⁢

γ

s

,

t

⁢

γ

u

,

v

=

(

a

i

⁢

s

+

t

⁢

b

)

⁢

γ

u

,

v

=

a

i

⁢

s

⁢

u

+

t

⁢

u

+

v

⁢

b

.

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

(

s

,

t

)

⁢

(

u

,

v

)

=

(

s

⁢

u

,

t

⁢

u

+

v

)

.

In fact this demonstrates that the

inverse

of

γ

s

,

t

is simply

γ

s

-

1

,

-

t

⁢

s

-

and the

identity map

is

γ

1

,

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

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