3Lagrange's Theorem

IA Groups

3.1 Small groups

We will study the structures of certain small groups.

Example

(Using Lagrange theorem to find subgroups)

.

To find subgroups of

D

10

, we know that the subgroups must have size 1, 2, 5 or 10:

1: {e}

2: The groups generated by the 5 reflections of order 2

5:

The group must be cyclic since it has prime order 5. It is then generated

by an element of order 5, i.e.

r, r

2

, r

3

and

r

4

. They generate the same

group hri.

10: D

10

As for D

8

, subgroups must have order 1, 2, 4 or 8.

1: {e}

2: 5 elements of order 2, namely 4 reflections and r

2

.

4:

First consider the subgroup isomorphic to

C

4

, which is

hri

. There are two

other non-cyclic group.

8: D

8

Proposition. Any group of order 4 is either isomorphic to C

4

or C

2

× C

2

.

Proof.

Let

|G|

= 4. By Lagrange theorem, possible element orders are 1 (

e

only),

2 and 4. If there is an element a ∈ G of order 4, then G = hai

∼

=

C

4

.

Otherwise all non-identity elements have order 2. Then

G

must be abelian

(For any

a, b

, (

ab

)

2

= 1

⇒ ab

= (

ab

)

−1

⇒ ab

=

b

−1

a

−1

⇒ ab

=

ba

). Pick

2 elements of order 2, say

b, c ∈ G

, then

hbi

=

{e, b}

and

hci

=

{e, c}

. So

hbi ∩ hci

=

{e}

. As

G

is abelian,

hbi

and

hci

commute. We know that

bc

=

cb

has

order 2 as well, and is the only element of

G

left. So

G

∼

=

hbi × hci

∼

=

C

2

× C

2

by the direct product theorem.

Proposition.

A group of order 6 is either cyclic or dihedral (i.e. is isomorphic

to C

6

or D

6

). (See proof in next section)