Homepage › Solution manuals › David S. Dummit › Abstract Algebra › Exercise 4.3.4 (Conjugates of normalizers and centralizers)

Abstract Algebra

Exercise 4.3.4 (Conjugates of normalizers and centralizers)

Prove that if S ⊆ G and g ∈ G , then g N G ( S ) g − 1 = N G ( 𝑔𝑆 g − 1 ) and g C G ( S ) g − 1 = C G ( 𝑔𝑆 g − 1 ) .

Answers

Proof. By definition of the normalizer of a subset of G , for all h ∈ G ,

h ∈ N G ( 𝑔𝑆 g − 1 ) ⟺ h ( 𝑔𝑆 g − 1 ) h − 1 = 𝑔𝑆 g − 1 ⟺ g − 1 h𝑔𝑆 ( g − 1 h𝑔 ) − 1 = S ⟺ g − 1 h𝑔 ∈ N G ( S ) ⟺ h ∈ g N G ( S ) g − 1 .

This shows that

N G ( 𝑔𝑆 g − 1 ) = g N G ( S ) g − 1 .

The centralizer of a subset S of G is

C G ( S ) = { h ∈ G ∣ ∀ ⁡ s ∈ S , h𝑠 h − 1 = s } .

Suppose that h ∈ C G ( 𝑔𝑆 g − 1 ) . For every s ∈ S , t = 𝑔𝑠 g − 1 ∈ 𝑔𝑆 g − 1 , therefore h𝑡 h − 1 = t , so h ( 𝑔𝑠 g − 1 ) h − 1 = 𝑔𝑠 g − 1 . thus ( g − 1 h𝑔 ) s ( g − 1 h𝑔 ) − 1 = s . Since this is true for every s ∈ S , g − 1 h𝑔 ∈ C G ( S ) and so h ∈ g C G ( S ) g − 1 . This proves

C G ( 𝑔𝑆 g − 1 ) ⊆ g C G ( S ) g − 1 .

Conversely, suppose that h ∈ g C G ( S ) g − 1 . Then g − 1 h𝑔 ∈ C G ( S ) . For every s ∈ S , ( g − 1 h𝑔 ) s ( g − 1 h𝑔 ) − 1 = s , so h ( 𝑔𝑠 g − 1 ) h − 1 = 𝑔𝑠 g − 1 . For every t ∈ 𝑔𝑆 g − 1 , there is some s such that t = 𝑔𝑠 g − 1 , therefore h𝑡 h − 1 = t for every t ∈ 𝑔𝑆 g − 1 . This shows h ∈ C G ( 𝑔𝑆 g − 1 ) , for every h ∈ g C G ( S ) g − 1 , so

g C G ( S ) g − 1 ⊆ C G ( 𝑔𝑆 g − 1 ) .

In conclusion,

g C G ( S ) g − 1 = C G ( 𝑔𝑆 g − 1 ) .

□

User profile picture
2026-02-03 09:24
Comments