subgroup

subgroup

Definition:

Let

(

G

,

*

)

be a group and let

K

be subset of

G

. Then

K

is a

subgroup

of

G

defined under the same

operation

if

K

is a group by itself (with respect to

*

), that is:

•

K

is

closed under

the

*

operation.

•

There exists an

identity element

e

∈

K

such that for all

k

∈

K

,

k

*

e

=

k

=

e

*

k

.

•

Let

k

∈

K

then there exists an

inverse

k

-

1

∈

K

such that

k

-

1

*

k

=

e

=

k

*

k

-

1

.

The subgroup is denoted likewise

(

K

,

*

)

. We denote

K

being a subgroup of

G

by writing

K

≤

G

.

In

addition

the notion of a subgroup of a

semigroup

can be defined in the following manner. Let

(

S

,

*

)

be a semigroup and

H

be a subset of

S

. Then

H

is a subgroup of

S

if

H

is a subsemigroup of

S

and

H

is a group.

Properties:

•

The set

{

e

}

whose only

element

is the

identity

is a subgroup of any group. It is called the trivial subgroup.

•

Every group is a subgroup of itself.

•

The

null set

{

}

is never a subgroup (since the definition of group states that the set must be non-empty).

There is a very useful

theorem

that allows proving a given subset is a subgroup.

Theorem:

If

K

is a nonempty subset of the group

G

. Then

K

is a

subgroup

of

G

if and only if

s

,

t

∈

K

implies that

s

⁢

t

-

1

∈

K

.

Proof:

First we need to show if

K

is a subgroup of

G

then

s

⁢

t

-

1

∈

K

. Since

s

,

t

∈

K

then

s

⁢

t

-

1

∈

K

, because

K

is a group by itself.

Now, suppose that if for any

s

,

t

∈

K

⊆

G

we have

s

⁢

t

-

1

∈

K

. We want to show that

K

is a subgroup, which we will accomplish by proving it holds the group axioms.

Since

t

⁢

t

-

1

∈

K

by

hypothesis

, we conclude that the identity element is in

K

:

e

∈

K

. (

Existence

of identity)

Now that we know

e

∈

K

, for all

t

in

K

we have that

e

⁢

t

-

1

=

t

-

1

∈

K

so the inverses of elements in

K

are also in

K

. (Existence of inverses).

Let

s

,

t

∈

K

. Then we know that

t

-

1

∈

K

by last step. Applying hypothesis shows that

s

⁢

(

t

-

1

)

-

1

=

s

⁢

t

∈

K

so

K

is closed under the operation.

Q

⁢

E

⁢

D

Example:

•

Consider the group

(

ℤ

,

+

)

. Show that

(

2

⁢

ℤ

,

+

)

is a subgroup.

The subgroup is closed under addition since the sum of

even integers

is even.

The identity

0

of

ℤ

is also on

2

⁢

ℤ

since

2

divides

0

.
For every

k

∈

2

⁢

ℤ

there is an

-

k

∈

2

⁢

ℤ

which is the inverse under addition and

satisfies

-

k

+

k

=

0

=

k

+

(

-

k

)

. Therefore

(

2

⁢

ℤ

,

+

)

is a subgroup of

(

ℤ

,

+

)

.

Another way to show

(

2

⁢

ℤ

,

+

)

is a subgroup
is by using the

proposition

stated above. If

s

,

t

∈

2

⁢

ℤ

then

s

,

t

are even numbers and

s

-

t

∈

2

⁢

ℤ

since the

difference

of even numbers is always an even number.

See also:

•

Wikipedia,

http://www.wikipedia.org/wiki/Subgroup

subgroup

Title

subgroup

Canonical name

Subgroup

Date of creation

2013-03-22 12:02:10

Last modified on

2013-03-22 12:02:10

Owner

Daume (40)

Last modified by

Daume (40)

Numerical id

18

Author

Daume (40)

Entry type

Definition

Classification

msc 20A05

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Defines

trivial subgroup

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