7Quaternions
IA Groups
7 Quaternions
In the remaining of the course, we will look at different important groups. Here,
we will have a brief look at
Definition (Quaternions). The quaternions is the set of matrices
1 0
0 1
,
i 0
0 −i
,
0 1
−1 0
,
0 i
i 0
−1 0
0 −1
,
−i 0
0 i
,
0 −1
1 0
,
0 −i
−i 0
which is a subgroup of GL
2
(C).
Notation. We can also write the quaternions as
Q
8
= ⟨a, b : a
4
= e, b
2
= a
2
, bab
−1
= a
−1
⟩
Even better, we can write
Q
8
= {1, −1, i, −i, j, −j, k, −k}
with
(i) (−1)
2
= 1
(ii) i
2
= j
2
= k
2
= −1
(iii) (−1)i = −i etc.
(iv) ij = k, jk = i, ki = j
(v) ji = −k, kj = −i, ik = −j
We have
1 =
1 0
0 1
, i =
i 0
0 −i
, j =
0 1
−1 0
, k =
0 i
i 0
−1 =
−1 0
0 −1
, −i =
−i 0
0 i
, −j =
0 −1
1 0
, −k =
0 −i
−i 0
Lemma. If
G
has order 8, then either
G
is abelian (i.e.
∼
=
C
8
, C
4
× C
2
or
C
2
× C
2
× C
2
), or
G
is not abelian and isomorphic to
D
8
or
Q
8
(dihedral or
quaternion).
Proof. Consider the different possible cases:
– If G contains an element of order 8, then G
∼
=
C
8
.
– If all non-identity elements have order 2, then G is abelian (Sheet 1, Q8).
Let
a
=
b ∈ G \ {e}
. By the direct product theorem,
⟨a, b⟩
=
⟨a⟩ × ⟨b⟩
.
Then take
c ∈ ⟨a, b⟩
. By the direct product theorem, we obtain
⟨a, b, c⟩
=
⟨a⟩ × ⟨b⟩ × ⟨c⟩
=
C
2
× C
2
× C
2
. Since
⟨a, b, c⟩ ⊆ G
and
|⟨a, b, c⟩|
=
|G|
,
G = ⟨a, b, c⟩
∼
=
C
2
× C
2
× C
2
.
– G
has no element of order 8 but has an order 4 element
a ∈ G
. Let
H
=
⟨a⟩
. Since
H
has index 2, it is normal in
G
. So
G/H
∼
=
C
2
since
|G/H|
= 2. This means that for any
b ∈ H
,
bH
generates
G/H
. Then
(
bH
)
2
=
b
2
H
=
H
. So
b
2
∈ H
. Since
b
2
∈ ⟨a⟩
and
⟨a⟩
is a cyclic group,
b
2
commutes with a.
If b
2
= a or a
3
, then b has order 8. Contradiction. So b
2
= e or a
2
.
We also know that
H
is normal, so
bab
−1
∈ H
. Let
bab
−1
=
a
ℓ
. Since
a
and
b
2
commute, we know that
a
=
b
2
ab
−2
=
b
(
bab
−1
)
b
−1
=
ba
ℓ
b
−1
=
(bab
−1
)
ℓ
= a
ℓ
2
. So ℓ
2
≡ 1 (mod 4). So ℓ ≡ ±1 (mod 4).
◦ When l ≡ 1 (mod 4), bab
−1
= a, i.e. ba = ab. So G is abelian.
∗ If b
2
= e, then G = ⟨a, b⟩
∼
=
⟨a⟩ × ⟨b⟩
∼
=
C
4
× C
2
.
∗ If b
2
= a
2
, then (ba
−1
)
2
= e. So G = ⟨a, ba
−1
⟩
∼
=
C
4
× C
2
.
◦ If l ≡ −1 (mod 4), then bab
−1
= a
−1
.
∗
If
b
2
=
e
, then
G
=
⟨a, b
:
a
4
=
e
=
b
2
, bab
−1
=
a
−1
⟩
. So
G
∼
=
D
8
by definition.
∗ If b
2
= a
2
, then we have G
∼
=
Q
8
.