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 ⟨r⟩.
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
⟨r⟩
. 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 = ⟨a⟩
∼
=
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
⟨b⟩
=
{e, b}
and
⟨c⟩
=
{e, c}
. So
⟨b⟩ ∩ ⟨c⟩
=
{e}
. As
G
is abelian,
⟨b⟩
and
⟨c⟩
commute. We know that
bc
=
cb
has
order 2 as well, and is the only element of
G
left. So
G
∼
=
⟨b⟩ × ⟨c⟩
∼
=
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)