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
= ha, b : a
4
= e, b
2
= a
2
, bab
1
= a
1
i
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 6
=
b G \ {e}
. By the direct product theorem,
ha, bi
=
hai × hbi
.
Then take
c 6∈ ha, bi
. By the direct product theorem, we obtain
ha, b, ci
=
hai × hbi × hci
=
C
2
× C
2
× C
2
. Since
ha, b, ci G
and
|ha, b, ci|
=
|G|
,
G = ha, b, ci
=
C
2
× C
2
× C
2
.
G
has no element of order 8 but has an order 4 element
a G
. Let
H
=
hai
. 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 6∈ H
,
bH
generates
G/H
. Then
(
bH
)
2
=
b
2
H
=
H
. So
b
2
H
. Since
b
2
hai
and
hai
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
`
. Since
a
and
b
2
commute, we know that
a
=
b
2
ab
2
=
b
(
bab
1
)
b
1
=
ba
`
b
1
=
(bab
1
)
`
= a
`
2
. So `
2
1 (mod 4). So ` ±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 = ha, bi
=
hai × hbi
=
C
4
× C
2
.
If b
2
= a
2
, then (ba
1
)
2
= e. So G = ha, ba
1
i
=
C
4
× C
2
.
If l 1 (mod 4), then bab
1
= a
1
.
If
b
2
=
e
, then
G
=
ha, b
:
a
4
=
e
=
b
2
, bab
1
=
a
1
i
. So
G
=
D
8
by definition.
If b
2
= a
2
, then we have G
=
Q
8
.