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Abstract Algebra

Problem 5.1.8 (Action of $S_n$ on $H^n$)

Let G 1 = G 2 = ⋯ = G n and let G = G 1 × ⋯ × G n . Under the notations of the preceding exercise show that φ π ∈ Aut ( G ) . Show also that the map π ↦ φ π is an injective homomorphism of S n into Aut ( G ) .

Answers

Proof. If G 1 = G 2 = ⋯ = G n , then the isomorphism φ π : G → G is an automorphism, so

φ π ∈ Aut ( G ) .

Consider now

χ { S n → Aut ( G ) π ↦ φ π

For every g = ( g 1 , g 2 , … , g n ) ∈ G , and for all permutations π 1 , π 2 ∈ S n ,

( φ π 1 ∘ φ π 2 ) ( g ) = φ π 1 ( g π 2 − 1 ( 1 ) , g π 2 − 1 ( 2 ) , … , g π 2 − 1 ( n ) ) = φ π 1 ( h 1 , h 2 , … , h n ) ( where  h i = g π 2 − 1 ( i )  for all  i ) = ( h π 1 − 1 ( 1 ) , h π 1 − 1 ( 2 ) , … , h π 1 − 1 ( n ) ) = ( g π 2 − 1 ( π 1 − 1 ( 1 ) ) , g π 2 − 1 ( π 1 − 1 ( 2 ) ) , … , g π 2 − 1 ( π 1 − 1 ( n ) ) ) = ( g ( π 1 π 2 ) − 1 ( 1 ) , g ( π 1 π 2 ) − 1 ( 2 ) , … , g ( π 1 π 2 ) − 1 ( n ) ) ( since  ( π 1 π 2 ) − 1 = π 2 − 1 π 1 − 1 ) = φ π 1 π 2 ( g ) .

This shows that for all permutations π 1 , π 2 ∈ S n ,

φ π 1 ∘ φ π 2 = φ π 1 π 2 .

So

χ ( π 1 π 2 ) = χ ( π 1 ) χ ( π 2 ) ,

and χ : S n → Aut ( G ) is a homomorphism, associate to the action of S n on G defined by

π ⋅ ( g 1 , g 2 , … , g n ) = ( g π − 1 ( 1 ) , … , g π − 1 ( n ) ) .

This explains the use of π − 1 in the definition of φ π . □

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2026-09-05 12:05
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