Question

If $\mathit{a}\mathbf{,}\mathit{b}\mathbf{\in }\mathit{G}$, prove that $\mathbf{\left|}\mathbf{b}\mathbf{a}{\mathbf{b}}^{\mathbf{-}\mathbf{1}}\mathbf{\right|}\mathbf{=}\mathbf{\left|}\mathbf{a}\mathbf{\right|}$.

Short answer

It is proved that,$\left|ba{b}^{-1}\right|=\left|a\right|$.

Step-by-step solution

Determine |bab−1|=|a|

Consider the given function as,$a,b\in G$.

Now, find ${\left(ba{b}^{-1}\right)}^{\left|a\right|}$as:

$$\begin{gathered} {\left(ba{b}^{-1}\right)}^{\left|a\right|}=b{a}^{\left|a\right|}{b}^{-1} \\ =be{b}^{-1} \\ =e \end{gathered}$$

Thus,$\left|ba{b}^{-1}\right|$ divides $\left|a\right|$.

Determine further solution of |bab−1|=|a|

On other side, find $\mathit{e}$as:

$\begin{array}{c}e={\left(ba{b}^{-1}\right)}^{\left|ba{b}^{-1}\right|}\\ =b{a}^{\left|ba{b}^{-1}\right|}{b}^{-1}\end{array}$

Multiply by $b^{-1}$on left hand side and multiply by $b$on right hand side as follows:

$a^{\left|ba{b}^{-1}\right|}=e$

So that, role="math" localid="1654254894062" $\left|a\right|$divides $\left|ba{b}^{-1}\right|$.

Therefore,$\left|ba{b}^{-1}\right|=\left|a\right|$ .