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Abstract Algebra

Exercise 4.3.15 (Two permutation representations of $D_8$)

Find an element of S 8 which conjugates the subgroup of S 8 obtained in part (a) of Exercise 3, Section 2 to the subgroup of S 8 obtained in part (b) of the same exercise (both of these subgroups are isomorphic to D 8 ).

Answers

Proof. To label the elements of G = D 8 , we use the following bijection f 1 :

k 1 2 3 4 5 6 7 8
f 1 ( k ) 1 r r 2 r 3 s 𝑠𝑟 s r 2 s r 3

so that g i = f 1 ( i ) for all i ∈ [ [ 1 , 8 ] ] .

By definition of the corresponding permutation representation π 1 , as in Exercise 14,

f 1 ∘ π 1 ( g ) = φ ( g ) ∘ f 1  for all  g ∈ G ,

where φ is the regular representation of G , defined by φ ( g ) ( x ) = 𝑔𝑥 for all g , x in G .

So the following diagram is commutative:

Now we relabel the elements of D 8 with the bijection f 2 :

k 1 2 3 4 5 6 7 8
f 2 ( k ) 1 s r 𝑠𝑟 r 2 s r 2 r 3 s r 3

The corresponding permutation representation π 2 satisfies

f 2 ∘ π 2 ( g ) = φ ( g ) ∘ f 2  for all  g ∈ G .

We obtain as in Exercise 14 the commutative diagram

Put τ = f 2 − 1 ∘ f 1 . Then

π 2 ( g ) ∘ τ = τ ∘ π 1 ( g ) . (1)

Since

f 1 = ( 1 2 3 4 5 6 7 8 1 r r 2 r 3 s 𝑠𝑟 s r 2 s r 3 ) , f 2 = ( 1 2 3 4 5 6 7 8 1 s r 𝑠𝑟 r 2 s r 2 r 3 s r 3 )

we obtain

τ = ( 1 2 3 4 5 6 7 8 1 3 5 7 2 4 6 8 ) = ( 2 3 5 ) ( 4 7 6 ) .

By (1), π 2 ( g ) = τ ∘ π 1 ( g ) ∘ τ − 1 , so τ = ( 2 3 5 ) ( 4 7 6 ) conjugates the two permutation groups obtained in Exercise 4.2.3. □

We check this result on π 1 ( s r 2 ) = ( 1 7 ) ( 2 8 ) ( 3 5 ) ( 4 6 ) :

τ ∘ π 1 ( s r 2 ) ∘ τ − 1 = ( 2 3 5 ) ( 4 7 6 ) ( 1 7 ) ( 2 8 ) ( 3 5 ) ( 4 6 ) ( 4 6 7 ) ( 2 5 3 ) = ( 1 6 ) ( 2 5 ) ( 3 8 ) ( 4 7 ) = π 2 ( s r 2 ) .

(see Exercise 4.2.3).

With Gap

gap> t := (2, 3,5)(4,7,6);
(2,3,5)(4,7,6)
gap> s := (1,7)(2,8)(3,5)(4,6);
(1,7)(2,8)(3,5)(4,6)
gap> t^-1 * s *t;
(1,6)(2,5)(3,8)(4,7)

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2026-02-05 11:47
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