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

.