Question

If $\mathbf{a}\mathbf{,}\mathbf{b}\mathbf{\in }\mathbf{G}\mathbf{,}{\mathbf{b}}^{\mathbf{6}}\mathbf{=}\mathbf{e}\mathbf{,}$and$\mathit{a}\mathit{b}\mathbf{=}{\mathit{b}}^{\mathbf{4}}\mathit{a}\mathbf{,}$ prove that ${\mathit{b}}^{\mathbf{3}}\mathbf{=}\mathit{e}$androle="math" localid="1654266019446" $\mathit{a}\mathit{b}\mathbf{=}\mathit{b}\mathit{a}$ .

Short answer

It is proved that,$b^3=e$ and $ab=ba$.

Step-by-step solution

Proving that b3=  e

Given that $b^6=e$and $ab=b^{4}a$.

Simplify $ab=b^{4}a$as:

$$\begin{gathered} a^{-1}ab=a^{-1}{b}^{4}a \\ b=a^{-1}{b}^{4}a\mathrm{........}\left(1\right) \end{gathered}$$

Therefore, $b\cdot b=b^2$implies $b\cdot b=a^{-1}{b}^{4}a{a}^{-1}{b}^{4}a$ using the value of b from equation (1).

Since $a{a}^{-1}$=1 and simplify as:

$$\begin{gathered} b^2=a^{-1}{b}^4{b}^{4}a \\ =a^{-1}{b}^{8}a \end{gathered}$$

Similarly, find $b^3$as:

$$\begin{gathered} b^3=b^2\cdot b \\ =a^{-1}{b}^{8}a\cdot a^{-1}{b}^{4}a \\ =a^{-1}{b}^{12}a \\ =a^{-1}{\left(b^6\right)}^{2}a \end{gathered}$$

Simplify using $b^6=e$as:

$\Rightarrow b^3=a^{-1}a$

$\Rightarrow b^3=e$

Hence, it is proved that$b^3=e$

Proving that ab = ba

Simplify $ab=b^{4}a$as:

$$\begin{gathered} ab=b^{4}a \\ \Rightarrow ab=b^{3}ba \end{gathered}$$

Using the result of Step 1 i.e.$b^3=e$, we have:

$$\begin{gathered} ab=eba \\ ab=ba \end{gathered}$$

Hence, it is proved ab=ba.

Thus,$b^3=e$ and$ab=ba$ .