Exercise 1.1.24 (If $ab = ba$ then $(ab)^n = a^n b^n$)

Proof. Let $a$ and $b$ be commuting elements of $G$, i.e., $\mathit{𝑎𝑏}=\mathit{𝑏𝑎}$. We prove first by induction on $b$ that $a{b}^n=b^{n}a$ for all integers $n\ge 0$.

• $a{b}^0=a=b^{0}a$.

• If $a{b}^n=b^{n}a$ for some integer $n\ge 0$, then

  $$a{b}^{n+1}=a{b}^{n}b=b^n\mathit{𝑎𝑏}=b^n\mathit{𝑏𝑎}=b^{n+1}a.$$

• The induction is done, which proves

  $$\begin{array}{lll}\hfill a{b}^n=b^{n}a(n\in \mathbb{Z},n\ge 0).& & \hfill \text{(1)}\end{array}$$

Now

• ${(\mathit{𝑎𝑏})}^0=1=a^0{b}^0$.

• Assume that ${(\mathit{𝑎𝑏})}^n=a^n{b}^n$ for some integer $n\ge 0$. Then

  $$\begin{array}{llllll}\hfill {(\mathit{𝑎𝑏})}^{n+1}& ={(\mathit{𝑎𝑏})}^n(\mathit{𝑎𝑏})& \hfill & & \hfill & \\ \hfill & =a^n{b}^n\mathit{𝑎𝑏}& \hfill (\text{by the induction hypothesis})& & \hfill & & \hfill \\ \hfill & =a^{n}a{b}^{n}b& \hfill (\text{by (1)})& & \hfill & & \hfill \\ \hfill & =a^{n+1}{b}^{n+1}.& \hfill & & \hfill & \end{array}$$

• The induction is done, which proves that if $\mathit{𝑎𝑏}=\mathit{𝑏𝑎}$,

  $${(\mathit{𝑎𝑏})}^n=a^n{b}^n(n\in \mathbb{Z},n\ge 0).$$

If $n<0$, then $m=-n>0$. By the solution of Exercise 19 (equality (5)),

So if $\mathit{𝑎𝑏}=\mathit{𝑏𝑎}$, then

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