Question

If ${\mathbf{\left(}\mathbf{a}\mathbf{b}\mathbf{\right)}}^{\mathbf{3}}\mathbf{=}{\mathit{a}}^{\mathbf{3}}{\mathit{b}}^{\mathbf{3}}$and${\mathbf{\left(}\mathbf{a}\mathbf{b}\mathbf{\right)}}^{\mathbf{5}}\mathbf{=}{\mathit{a}}^{\mathbf{5}}{\mathit{b}}^{\mathbf{5}}$ for all $\mathit{a}\mathit{b}\mathbf{\in }\mathit{G}$, prove that G is abelian.

Short answer

It is proved that, G is abelian.

Step-by-step solution

Commutativity Property of Abelian Group

If group Ais an abelian group, then for all $a,b\in A$

role="math" localid="1654268585596" $\mathit{a}\mathbf{\ast }\mathit{b}\mathbf{=}\mathit{b}\mathbf{\ast }\mathit{a}$

It is given that ${\left(ab\right)}^3=a^3{b}^3$and ${\left(ab\right)}^5=a^5{b}^5$.

Proving that G is abelian

Multiplying both sides of ${\left(ab\right)}^3=a^3{b}^3$with $a^{-1}$and $b^{-1}$as:

$$\begin{gathered} a^{-1}{\left(ab\right)}^3{b}^{-1}=a^{-1}{a}^3{b}^3{b}^{-1} \\ {\left(ab\right)}^2=a^2{b}^2\mathrm{...........}\left(1\right) \end{gathered}$$

Now, multiplying both sides of ${\left(ab\right)}^5=a^5{b}^5$with $a^{-1}$and $b^{-1}$as:

$\begin{array}{c}{a}^{-1}{\left(ab\right)}^5{b}^{-1}=a^{-1}{a}^5{b}^5{b}^{-1}\\ {\left(ab\right)}^4=a^4{b}^4\\ {\left({\left(ab\right)}^2\right)}^2=a^4{b}^4...........\left(2\right)\end{array}$

From equation and , we get${\left(a^2{b}^2\right)}^2=a^4{b}^4$

Multiplying both sides of ${\left(a^2{b}^2\right)}^2=a^4{b}^4$with $a^{-2}$and $b^{-2}$as:

$b^2{a}^2=a^2{b}^2\mathrm{..........}\left(3\right)$

Again considering ${\left(ab\right)}^3=a^3{b}^3$as:

$$\begin{gathered} {\left(ab\right)}^3=a^3{b}^3 \\ {\left(ab\right)}^3=a\left(a^2{b}^2\right)b \end{gathered}$$

From equation (3), we know$b^2{a}^2=a^2{b}^2$and simplify as:

$$\begin{gathered} {\left(ab\right)}^3=a\left(b^2{a}^2\right)b \\ ababab=abbaab \end{gathered}$$

Multiplying both sides of equation with role="math" localid="1654269359102" $b^{-1}{a}^{-1}$as:

role="math" localid="1654269341771" $$\begin{gathered} b^{-1}{a}^{-1}ababab{b}^{-1}{a}^{-1}=b^{-1}{a}^{-1}abbaab{b}^{-1}{a}^{-1} \\ ab=ba \end{gathered}$$

This is the property of an abelian group.

Hence, it is proved that G is abelian.

${\left(a^2{b}^2\right)}^2=a^4{b}^4$