Since D₁ is isomorphic to Z/2Z and D₂ is isomorphic to Z/2Z×Z/2Z, we shall assume n>2.

Then the following permutation representation on n points is transitive and injective:

c -> (1,2, ...,n),  and
a -> (1,n)(2,n - 1) ··· ((n - 1)/2,(n + 1)/2) if n is odd,
a -> (1,n)(2,n - 1) ··· (n/2,n/2 + 1) if n is even.

An injective morphism D_n -> GL₂(C) is determined by

c ->

cos(2/n) sin(2/n) -sin(2/n) cos(2/n)

,

a ->

0 1 1 0

,

Check that the matrices satisfy the relations imposed on c and a.

Left multiplication gives the transitive permutation representation determined by the following assignments:

-1 -> (1,-1)(i,-i)(j,-j)(k,-k)
i -> (1,i,-1,-i)(j,k,-j,-k)
j -> (1,j,-1,-j)(i,-k,-i,k)
k -> (1,k,-1,-k)(i,j,-i,-j)

Replacing the elements by numbers 1, ..., 8, a more usual description is obtained.

An injective morphism Q -> GL₂(C) is determined by

i ->

i 0 0 -i

,

j ->

0 1 -1 0

,

Verify that this forces

k ->

0 i i 0

,

-1 ->

-1 0 0 -1

.