example of non-permutable subgroup

example of non-permutable subgroup

Example 1

.

There are groups (even

finitely generated

) with

subnormal subgroups

which are
not

permutable

.

Proof.

Let

D

8

be the

dihedral group

of order 8. The classic

presentation

is

D

8

=

⟨

a

,

b

:

a

4

=

b

2

=

1

,

b

a

b

=

a

-

1

⟩

.

As the group is

nilpotent

we know every

subgroup

is subnormal; however, not every subgroup is permutable. In particular, observe that for two general subgroups

H

and

K

of

D

8

, it may be possible that

H

⁢

K

is not a subgroup. In this situation we find our counterexample.

⟨

b

⟩

⁢

⟨

a

⁢

b

⟩

=

{

1

,

b

,

a

⁢

b

,

b

⁢

a

⁢

b

}

=

{

1

,

b

,

a

⁢

b

,

a

-

1

}

.

Yet

⟨

a

⁢

b

⟩

⁢

⟨

b

⟩

=

{

1

,

a

⁢

b

,

b

,

a

⁢

b

⁢

b

}

=

{

1

,

b

,

a

⁢

b

,

a

}

.

More generally, in any dihedral group

D

2

⁢

n

=

⟨

a

,

b

:

a

n

=

b

2

=

1

,

b

a

b

=

a

-

1

⟩

,

for

n

>

2

, then

⟨

b

⟩

⁢

⟨

a

⁢

b

⟩

≠

⟨

a

⁢

b

⟩

⁢

⟨

b

⟩

and both are subnormal whenever

n

=

2

i

.
∎

However, we do observe the competing observation that the group generated
by

⟨

b

⟩

and

⟨

a

⁢

b

⟩

is the same as the group
generated by

⟨

a

⁢

b

⟩

and

⟨

b

⟩

, namely

D

8

.
Indeed in any

group with

subgroups

H

and

K

,

⟨

H

,

K

⟩

=

⟨

K

,

H

⟩

so the condition of permutability is one which must be tested as complexes (sets

H

⁢

K

), not as subgroups. This is a

consequence

of the following general result:

Claim 1

.

H

⁢

K

=

K

⁢

H

if and only if

H

⁢

K

=

⟨

H

,

K

⟩

.

Proof.

We will show that every

element

in

⟨

H

,

K

⟩

can be written in
the form

h

⁢

k

for some

h

∈

H

and

k

∈

K

. To see this first note
every element in

⟨

H

,

K

⟩

is a word over elements in

H

and in

K

. If the word involves only elements in

H

or only elements in

K

then
we are done. Now for

induction

suppose all words of length

m

in

⟨

H

,

K

⟩

can be expressed in the form

h

⁢

k

for some

h

∈

H

and

k

∈

K

.
Then given a word of length

m

+

1

we have either

h

′

⁢

w

for

h

′

∈

H

and

w

a word of length

m

, in which case we are done, or

k

′

⁢

w

for some

k

′

∈

K

and

w

a word of length

m

. Then by induction

w

=

h

⁢

k

form some

h

∈

H

and

k

∈

K

. Hence

k

′

⁢

w

=

k

′

⁢

h

⁢

k

. Then

k

′

⁢

h

∈

K

⁢

H

=

H

⁢

K

so there exists

h

′

∈

H

and

k

′′

∈

K

such that

k

′

⁢

h

=

h

′

⁢

k

′′

. Thus

k

′

⁢

w

=

h

′

⁢

k

′′

⁢

k

=

h

′

⁢

(

k

′′

⁢

k

)

is of the
desired format. Hence

H

⁢

K

⊆

⟨

H

,

K

⟩

⊆

H

⁢

K

so

H

⁢

K

=

⟨

H

,

K

⟩

.

For the

converse

suppose

H

⁢

K

=

⟨

H

,

K

⟩

. Then

K

⁢

H

⊆

H

⁢

K

.
This means for every

k

∈

K

and

h

∈

H

there exists

h

′

∈

H

and

k

′

∈

K

such that

k

-

1

⁢

h

-

1

=

h

′

⁢

k

′

. Thus

h

⁢

k

=

(

k

-

1

⁢

h

-

1

)

-

1

=

(

h

′

⁢

k

′

)

-

1

=

(

k

′

)

-

1

⁢

(

h

′

)

-

1

∈

K

⁢

H

.

Thus

H

⁢

K

⊆

K

⁢

H

and

K

⁢

H

=

H

⁢

K

.
∎

This helps illustrate how permutability is such a useful condition in the study of subgroup lattices (one of Ore’s main research interests). For these are the subgroups whose complexes are also subgroups. Thus we can relate the order of

⟨

H

,

K

⟩

to the order of

H

and

K

and many other combinatorial

relations

.

Title

example of non-permutable subgroup

Canonical name

ExampleOfNonpermutableSubgroup

Date of creation

2013-03-22 16:15:56

Last modified on

2013-03-22 16:15:56

Owner

Algeboy (12884)

Last modified by

Algeboy (12884)

Numerical id

10

Author

Algeboy (12884)

Entry type

Example

Classification

msc 20E07

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