Problem 5.4.3 (Commutators)

Proof. Let $a,b,c\in G$.

(a)

$$\begin{array}{llll}\hfill [a,\mathit{𝑏𝑐}]& =a^{-1}{(\mathit{𝑏𝑐})}^{-1}a(\mathit{𝑏𝑐})& \hfill & \\ \hfill & =a^{-1}{c}^{-1}{b}^{-1}\mathit{𝑎𝑏𝑐}& \hfill & \\ \hfill & =(a^{-1}{c}^{-1}\mathit{𝑎𝑐})(c^{-1}(a^{-1}{b}^{-1}\mathit{𝑎𝑏})c)& \hfill & \\ \hfill & =[a,c](c^{-1}[a,b]c).& \hfill & \end{array}$$

(b)

By Exercise 1, $[\mathit{𝑎𝑏},c]={[c,\mathit{𝑎𝑏}]}^{-1}$. Using part (a), we obtain $$\begin{array}{llll}\hfill [\mathit{𝑎𝑏},c]& ={[c,\mathit{𝑎𝑏}]}^{-1}& \hfill & \\ \hfill & ={([c,b](b^{-1}[c,a]b))}^{-1}& \hfill & \\ \hfill & =(b^{-1}{[c,a]}^{-1}b){[c,b]}^{-1}& \hfill & \\ \hfill & =(b^{-1}[a,c]b)[b,c]& \hfill & \end{array}$$