Question

If ${\mathbf{\left(}\mathbf{a}\mathbf{b}\mathbf{\right)}}^{\mathbf{i}}\mathbf{=}{\mathit{a}}^{\mathbf{i}}{\mathit{b}}^{\mathbf{i}}$for three consecutive integers i and 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 $a,b\in A$

$\mathit{a}\mathbf{\ast }\mathit{b}\mathbf{=}\mathit{b}\mathbf{\ast }\mathit{a}$

It is given that ${\left(ab\right)}^i=a^i{b}^i$ for three consecutive integersi.

Proving that G is abelian 

write ${\left(ab\right)}^i=a^i{b}^i$ for three consecutive integers i as:

$$\begin{gathered} {\left(ab\right)}^i=a^i{b}^i \\ {\left(ab\right)}^{i+1}=a^{i+1}{b}^{i+1} \\ {\left(ab\right)}^{i+2}=a^{i+2}{b}^{i+2} \end{gathered}$$

Considering ${\left(ab\right)}^{i+1}=a^{i+1}{b}^{i+1}$as:

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

Multiplying by $a^{-i}$ from left on both side of equation as:

$b^i\left(ab\right)=a{b}^{i+1}$

Now, multiplying by $b^{-1}$from right on both side of equation as follows:

$b^{i}a=a{b}^i\mathrm{.........}\left(1\right)$

Now, considering ${\left(ab\right)}^{i+2}=a^{i+2}{b}^{i+2}$as:

$$\begin{gathered} {\left(\mathrm{ab}\right)}^{\mathrm{i}+2}={\mathrm{a}}^{\mathrm{i}+2}{\mathrm{b}}^{\mathrm{i}+2} \\ {\left(\mathrm{ab}\right)}^{\mathrm{i}+1}\left(\mathrm{ab}\right)={\mathrm{a}}^{\mathrm{i}+2}{\mathrm{b}}^{\mathrm{i}+2} \end{gathered}$$

Multiplying by $a^{-i}$ from left on both side of equation as follows:

$b^{i+1}\left(ab\right)=a{b}^{i+2}$

Now, multiplying by $b^{-1}$ from right on both side of equation as:

$b^{i+1}a=a{b}^{i+1}$

$$\begin{gathered} a{b}^{i+1}=b^{i+1}a \\ a{b}^{i+1}=b{b}^{i}a \end{gathered}$$

From equation $\left(1\right)$, we know $b^{i}a=a{b}^i$.

So,$a{b}^{i+1}=ba{b}^i$ .

Now, multiplying by role="math" localid="1654270381111" $b^{-i}$ from right on both side of equation, we get $ab=ba$ this is the property of an abelian group.

Hence, it is proved thatG is abelian.