Homepage › Solution manuals › David S. Dummit › Abstract Algebra › Problem 5.4.8 (${(xy)^n = x^n y^n [y,x]^{\frac{n(n-1)}{2}}}$ if $x$ and $y$ commute with $[x,y]$)

Abstract Algebra

Problem 5.4.8 (${(xy)^n = x^n y^n [y,x]^{\frac{n(n-1)}{2}}}$ if $x$ and $y$ commute with $[x,y]$)

Assume x , y ∈ G and both x and y commute with [ x , y ] . Prove that for all n ∈ ℤ + , ( 𝑥𝑦 ) n = x n y n [ y , x ] n ( n − 1 ) 2 .

Answers

(Text modified to obtain a more natural proof.)

Proof. Assume x , y ∈ G and both x and y commute with [ x , y ] .

First we prove by induction [ x , y ] n = [ x n , y ] . This is true for n = 1 . Assume that [ x , y ] n = [ x n , y ] for some positive integer n . Since x and y commute with [ x , y ] , we obtain

[ x n + 1 , y ] = x − n − 1 y − 1 x n + 1 y = x − 1 ( x − n y − 1 x n y ) y − 1 𝑥𝑦 = x − 1 [ x n , y ] y − 1 𝑥𝑦 = x − 1 [ x , y ] n y − 1 𝑥𝑦 = x − 1 y − 1 𝑥𝑦 [ x , y ] n = [ x , y ] n + 1 .

The induction is done, so for all positive integers n ,

[ x , y ] n = [ x n , y ] (1)

(Hence [ x , y n ] = [ y n , x ] − 1 = [ y , x ] − n = [ x , y ] n .)

Note that 𝑦𝑥 = 𝑥𝑦 [ y , x ] . Replacing y by y n , we obtain y n x = x y n [ y n , x ] , and by (1),

y n x = x y n [ y , x ] n ( n ≥ 0 ) . (2)

Then we can prove by induction ( 𝑥𝑦 ) n = x n y n [ y , x ] n ( n − 1 ) 2 . This is true for n = 1 . Assume that ( 𝑥𝑦 ) n = x n y n [ y , x ] n ( n − 1 ) 2 for some positive integer n . Then

( 𝑥𝑦 ) n + 1 = x n y n [ y , x ] n ( n − 1 ) 2 ( 𝑥𝑦 ) = x n y n 𝑥𝑦 [ y , x ] n ( n − 1 ) 2 (since  x , y  commute with  [ y , x ] ) = x n ( x y n [ y , x ] n ) y [ y , x ] n ( n − 1 ) 2 (by (2)) = x n + 1 y n + 1 [ y , x ] n ( n − 1 ) 2 + n = x n + 1 y n + 1 [ y , x ] n ( n + 1 ) 2

The induction is done, which proves, for all positive integers,

( 𝑥𝑦 ) n = x n y n [ y , x ] n ( n − 1 ) 2 ( n ≥ 1 ) .

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Note: This result is useful for groups P such that P ′ ≤ Z ( P ) , for instance the groups of order p 3 (see Exercise 7).

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2026-09-10 11:05
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