example of a non-fully invariant subgroup

example of a non-fully invariant subgroup

Every

fully invariant subgroup

is

characteristic

, but some characteristic subgroups need not be fully invariant. For example, the center of a group is characteristic but not always fully invariant. We pursue a single example.

Recall the

dihedral group

of order

2

⁢

n

, denoted

D

2

⁢

n

, can be considered as the

symmetries

of a

regular

n

-gon. If we consider a regular hexagon, so

n

=

6

, and label the vertices counterclockwise from 1 to 6 we can then encode each symmetry as a

permutation

on 6 points. So a rotation by

π

/

3

can be encoded as the permutation

ρ

=

(

123456

)

and the

reflection

fixing the axis through the vertices 1 and 4 can be encoded as

ϕ

=

(

26

)

⁢

(

35

)

. Indeed these two permutations generate a

permutation group

isomorphic

to

D

12

.

The center of a dihedral group of order

2

⁢

n

is

trivial

if

n

is odd, and of order 2 if

n

>

2

is even (if

n

=

2

it is the entire group

D

4

≅

ℤ

2

⊕

ℤ

2

, see the remark below). Specifically, if

ρ

is a rotation of order

n

, and

n

=

2

⁢

m

, then

⟨

ρ

m

⟩

is the center of

D

2

⁢

n

. (Note this is the only rotation or order 2, and in particular it is always a rotation by

π

.)
So when

n

=

6

, the center is

⟨

(

14

)

⁢

(

25

)

⁢

(

36

)

⟩

.

Now fix

n

=

6

and note the following assignment of

generators

determines an

endomorphism

f

:

D

12

→

D

12

:

(

123456

)

↦

(

26

)

⁢

(

35

)

,

(

26

)

⁢

(

35

)

↦

(

14

)

⁢

(

25

)

⁢

(

36

)

.

Note that image

K

:=

⟨

(

26

)

⁢

(

35

)

,

(

14

)

⁢

(

25

)

⁢

(

36

)

⟩

≅

ℤ

2

⊕

ℤ

2

, as

(

14

)

⁢

(

25

)

⁢

(

36

)

is

central

in

D

12

and the generators of

K

are distinct elements of order

2

. [This can be proved with the

relations

of the dihedral group.]

Remark 1

.

Geometrically we note that the kernel of the

homomorphism

is

⟨

ρ

2

⟩

– the group of rotations of order 3. So if we

quotient

by the kernel we are identifying the three inscribed (non-square)

rectangles

of the hexagon (1245, 2356 and 3461). The

symmetry group

of a non-square rectangle is none other than

Z

2

⊕

Z

2

, sometimes called

D

4

.

Now the center is mapped via

f

to the

subgroup

⟨

(

26

)

⁢

(

35

)

⟩

so

f

⁢

(

Z

⁢

(

D

12

)

)

is not

contained

in

Z

⁢

(

D

12

)

proving

Z

⁢

(

D

12

)

is not fully-invariant.

Of course the example applies without serious modification to the dihedral groups on

2

⁢

m

-gons, where

m

>

1

is odd. Here a generally offending endomorphism may be described with a

composition

of maps (the first leaves the center

invariant

, the second swaps the basis of the image of the first thus moving the image of the center):

ρ

↦

ρ

m

↦

ϕ

,

ϕ

↦

ϕ

↦

ρ

m

.

As

m

is odd and the center,

⟨

ρ

m

⟩

, has order 2, it follows

⟨

ρ

m

⟩

maps to

⟨

ρ

m

⟩

under the first map, and then can be interchanged with a reflection to violate the condition of full invariance. If

m

is even then the center lies in the kernel of the first map so no such trick can be played.

Title

example of a non-fully invariant subgroup

Canonical name

ExampleOfANonfullyInvariantSubgroup

Date of creation

2013-03-22 16:06:26

Last modified on

2013-03-22 16:06:26

Owner

Algeboy (12884)

Last modified by

Algeboy (12884)

Numerical id

7

Author

Algeboy (12884)

Entry type

Example

Classification

msc 20D99

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