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Abstract Algebra

Exercise 4.3.16 (Actions on the left cosets of $\langle s \rangle\leq D_8$)

Find an element of S 4 which conjugates the subgroup of S 4 obtained in part (a) of Exercise 5, Section 2 to the subgroup of S 4 obtained in part (b) of the same exercise (both of these subgroups are isomorphic to D 8 ).

Answers

Proof. Let A be the set of left cosets of H = ⟨ s ⟩ ≤ D 8 .

Here the bijections f 1 : [ [ 1 , 4 ] ] → A and f 2 : [ [ 1 , 4 ] ] → A are defined by

f 1 = ( 1 2 3 4 1 H 𝑟𝐻 r 2 H r 3 H ) , f 2 = ( 1 2 3 4 1 H r 2 H 𝑟𝐻 r 3 H ) .

Let φ be the representation obtained by left multiplication on the left cosets of H , so that

φ ( g ) ( 𝑥𝐻 ) = 𝑔𝑥𝐻 ; g , x ∈ D 8 .

The permutation homomorphisms π 1 and π 2 are defined by

j = π 1 ( i ) ⟺ f 1 ( j ) = g ⋅ f 1 ( i ) , j = π 1 ( i ) ⟺ f 1 ( j ) = g ⋅ f 1 ( i ) , ( i , j ∈ [ [ 1 , 4 ] ] ) .

Therefore, for all g ∈ D 8 ,

f ∘ π 1 = φ ( g ) ∘ f 1 , f ∘ π 2 = φ ( g ) ∘ f 2 .

As in the two preceding exercises, we obtain the following commutative diagram

If τ = f 2 − 1 ∘ f 1 , then

π 2 ( g ) ∘ τ = τ ∘ π 1 ( g ) ( g ∈ D 8 ) ,

where

τ = f 2 − 1 ∘ f 1 = ( 1 2 3 4 1 3 2 4 ) = ( 2 3 ) .

Since π 2 ( g ) = τ ∘ π 1 ( g ) ∘ τ − 1 , τ = ( 2 3 ) conjugates the subgroups of S 4 obtained in Exercise 4.2.5. □

We check the equality π 2 ( g ) = τ ∘ π 1 ( g ) ∘ τ − 1 on g = r 2 , for instance:

τ ∘ π 1 ( r 2 ) ∘ τ − 1 = ( 2 3 ) ( 1 3 ) ( 2 4 ) ( 2 3 ) = ( 1 2 ) ( 3 4 ) = π 2 ( r 2 ) .
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2026-02-06 09:30
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