Homepage › Solution manuals › David S. Dummit › Abstract Algebra › Exercise 4.3.14 (Comparison of two permutation representations of $V_4$)

Abstract Algebra

Exercise 4.3.14 (Comparison of two permutation representations of $V_4$)

In Exercise 1 of Section 2 two labellings of the elements { 1 , a , b , c } of the Klein’s 4 -group V were chosen to give two versions of the left regular representation of V into S 4 . Let π 1 be the version of regular representation obtained in part (a) of that exercise and let π 2 be the version obtained via the labelling in part (b). Let τ = ( 2 4 ) . Show that τ ∘ π 1 ( g ) ∘ τ − 1 = π 2 ( g ) for each g ∈ V (i.e., conjugation by τ sends the image of π 1 to the image of π 2 elementwise).

Answers

Proof. To define π 1 , we label 1 , a , b , c by the bijection f 1 , where

k 1 2 3 4
f 1 ( k ) 1 a c b

so that g i = f 1 ( i ) for all i ∈ [ [ 1 , 4 ] ] .

By definition of the permutation representation π 1 , for all i , j ∈ [ [ 1 , 4 ] ] , and all g ∈ G ,

π 1 ( g ) ( i ) = j ⟺ x ⋅ g i = g j ⟺ x f 1 ( i ) = f 1 ( j ) .

Hence, if φ is the homomorphism afforded by the action of G on itself, i.e., φ ( g ) ( x ) = 𝑔𝑥 for all g , x ∈ G , then φ ( g ) ( f 1 ( i ) ) = f 1 ( j ) = f 1 ( π 1 ( g ) ( i ) ) for all i ∈ [ [ 1 , 4 ] ] , thus

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

so the following diagram is commutative:

ByExercise 4.2.1,

π 1 ( 1 ) = ( ) , π 1 ( a ) = ( 1 2 ) ( 3 4 ) , π 1 ( b ) = ( 1 4 ) ( 2 3 ) , π 1 ( c ) = ( 1 3 ) ( 2 4 ) .

Now we relabel 1 , a , b , c by the bijection f 2 , where

k 1 2 3 4
f 2 ( k ) 1 b c a

As in the first part, π 2 is defined by f 2 ∘ π 2 ( g ) = φ ( g ) ∘ f 2 , so π 2 ( g ) = f 2 − 1 ∘ φ ( g ) ∘ f 2 , for all g ∈ G .

ByExercise 4.2.1,

π 2 ( 1 ) = ( ) , π 2 ( a ) = ( 1 4 ) ( 2 3 ) , π 2 ( b ) = ( 1 2 ) ( 3 4 ) , π 2 ( c ) = ( 1 3 ) ( 2 4 ) .

If we put together the two preceding diagrams, we obtain the following commutative diagram

Put τ = f 2 − 1 ∘ f 1 . Since

f 1 = ( 1 2 3 4 1 a c b ) , f 2 = ( 1 2 3 4 1 b c a ) ,

we obtain

τ = ( 1 2 3 4 1 4 3 2 ) = ( 2 4 ) ,

So τ is the transposition ( 2 4 ) of the statement.

By omitting the intermediate nodes, we obtain the commutative diagram

so π 2 ( g ) ∘ τ = τ ∘ π 1 ( g ) .

Explicitly, since f 2 ∘ π 2 ( g ) = φ ( g ) ∘ f 2 and f 1 ∘ π 1 ( g ) = φ ( g ) ∘ f 1 ,

π 2 ( g ) ∘ τ = π 2 ( g ) ∘ f 2 − 1 ∘ f 1 = f 2 − 1 ∘ φ ( g ) ∘ f 1 = f 2 − 1 ∘ f 1 ∘ φ ( g ) = τ ∘ φ ( g )

This shows that

∀ ⁡ g ∈ G , π 2 ( g ) = τ ∘ π 1 ( g ) ∘ τ − 1 .

□

We can check this general result for g = a :

τ ∘ π 1 ( a ) ∘ τ − 1 = ( 2 4 ) ( 1 2 ) ( 3 4 ) ( 2 4 ) = ( 1 4 2 ) ( 2 3 4 ) = ( 1 4 ) ( 2 3 ) = π 2 ( a )
User profile picture
2026-02-05 10:53
Comments