Question

If $\mathit{a}\mathbf{,}\mathit{b}\mathbf{\in }\mathit{G}$and $\mathit{n}$is any integer, show that ${\left(ab{a}^{-1}\right)}^{\mathbf{n}}\mathbf{=}\mathit{a}{\mathit{b}}^{\mathbf{n}}{\mathit{a}}^{\mathbf{-}\mathbf{1}}$.

Short answer

It is shown that ${\left(ab{a}^{-1}\right)}^n=a{b}^n{a}^{-1}$ by using induction.

Step-by-step solution

Determine induction propert 

Consider the function $a,b\in G$and $n$is any integer.

Now, using induction on integer $n$, if $\left|n\right|=0$then, prove ${\left(ab{a}^{-1}\right)}^n=a{b}^n{a}^{-1}$as:

role="math" localid="1653669344256" $$\begin{gathered} {\left(aba\right)}^{-1}=e \\ =a{b}^0{a}^{-1} \end{gathered}$$

Assume that $\left|n\right|>0$and the equation holds for all integer $k$so that $\left|k\right|<\left|n\right|$.

Determine (aba-1)n=abna-1

Assume that $n>0$.

Then, prove ${\left(ab{a}^{-1}\right)}^n=a{b}^n{a}^{-1}$as:

role="math" localid="1653670172027" $$\begin{gathered} {\left(ab{a}^{-1}\right)}^n={\left(ab{a}^{-1}\right)}^{n-1}\left(ab{a}^{-1}\right) \\ =a{b}^{n-1}{a}^{-1}ab{a}^{-1} \\ =a{b}^n{a}^{-1} \end{gathered}$$

And if $n<0$then, prove ${\left(ab{a}^{-1}\right)}^n=a{b}^n{a}^{-1}$as:

$$\begin{gathered} {\left(ab{a}^{-1}\right)}^n={\left(ab{a}^{-1}\right)}^{n+1}\left(ab{a}^{-1}\right) \\ =a{b}^{n+1}{a}^{-1}ab{a}^{-1} \\ =a{b}^n{a}^{-1} \end{gathered}$$

Here, Corollary 7.6 is used because $n<0!$as observed the above equations,$n+1$instead of $n-1$.

Hence, ${\left(ab{a}^{-1}\right)}^n=a{b}^n{a}^{-1}$.