Question

If $\mathit{a}\mathbf{,}\mathit{b}\mathbf{\in }\mathit{G}$ and $\mathit{a}\mathit{b}\mathbf{=}\mathit{b}\mathit{a}$, prove that${{\mathbf{\left(}\mathbf{a}\mathbf{b}\mathbf{\right)}}^{\mathbf{\left|}\mathbf{a}\mathbf{\right|}}}^{\mathbf{\left|}\mathbf{b}\mathbf{\right|}}\mathbf{=}\mathit{e}$ .

Short answer

It is proved that,${{\left(ab\right)}^{\left|a\right|}}^{\left|b\right|}=e$

Step-by-step solution

Basic Properties of Group Elements

• $a^{mn}={\left(a^m\right)}^n$
  
• $e^n=1$
  
• role="math" localid="1654260068736" $a^{\left|a\right|}=e$
  

Proving that (ab)|a||b| =  e

As given in the question, $a,b\in G$ and $ab=ba$.

Multiplying both sides by a from left on $ab=ba$as:

$\begin{array}{c}ab=ba\\ aab=aba\end{array}$

Multiplying both sides by b from right as follows.

$\begin{array}{l}aabb=abab\\ a^2{b}^2={\left(ab\right)}^2\end{array}$

Again, multiplying with a from left and b from right as follows:

$a^3{b}^3={\left(ab\right)}^3$

Similarly, by repeating the steps we get:

${\left(ab\right)}^n=a^n{b}^n$

Taking $n=\text{\hspace{0.17em}}\left|a\right|\left|b\right|$.

Find ${\left(ab\right)}^{\text{\hspace{0.17em}}\left|a\right|\left|b\right|}$ as:

role="math" localid="1654260444972" $\begin{array}{c}{\left(ab\right)}^{\text{\hspace{0.17em}}\left|a\right|\left|b\right|}=a^{\text{\hspace{0.17em}}\left|a\right|\left|b\right|}{b}^{\text{\hspace{0.17em}}\left|a\right|\left|b\right|}\\ ={\left(a^{\left|a\right|}\right)}^{\left|b\right|}{\left(b^{\left|b\right|}\right)}^{\left|a\right|}\\ =e^{\left|b\right|}{e}^{\left|a\right|}\\ =e.e\end{array}$

Further simplify as:

${{\left(ab\right)}^{\left|a\right|}}^{\left|b\right|}=e$

Hence, it is proved that ${{\left(ab\right)}^{\left|a\right|}}^{\left|b\right|}=e$.