Question

Suppose $\mathit{a}\mathbf{,}\mathit{b}\mathbf{\in }\mathit{G}$ with$\mathbf{\left|}\mathbf{a}\mathbf{\right|}\mathbf{=}\mathbf{5}\mathbf{,}\mathit{b}\mathbf{\ne }\mathit{e}$ , and $\mathit{a}\mathit{b}{\mathit{a}}^{\mathbf{-}\mathbf{1}}\mathbf{=}{\mathit{b}}^{\mathbf{2}}\mathbf{.}$ Find $\mathbf{\left|}\mathbf{b}\mathbf{\right|}$ .

Short answer

$\left|b\right|=31$

Step-by-step solution

Given Information

Given that $\left|a\right|=5$, $b\ne e$ and $ab{a}^{-1}=b^2$.

Then,$\left|a\right|=5\Rightarrow a^5=e$

Finding |b|

Simplify $ab{a}^{-1}=b^2$as:

$$\begin{gathered} a\left(ab{a}^{-1}\right)a^{-1}=a{b}^2{a}^{-1} \\ {a}^{2}b{a}^{-2}=\left(ab{a}^{-1}\right)\left(ab{a}^{-1}\right) \\ {a}^{2}b{a}^{-2}=b^2{b}^2 \\ {a}^{2}b{a}^{-2}=b^4 \end{gathered}$$

Further simplify as:

localid="1659250577137" $$\begin{gathered} a^{3}b{a}^{-3}=a{b}^4{a}^{-1} \\ =\left(ab{a}^{-1}\right)\left(ab{a}^{-1}\right)\left(ab{a}^{-1}\right)\left(ab{a}^{-1}\right) \\ =b^2\cdot b^2\cdot b^2\cdot b^2 \\ =b^8 \end{gathered}$$

By repeating the same we get:

$$\begin{gathered} a^{4}b{a}^{-4}=b^{16} \\ {a}^{5}b{a}^{-5}=b^{32} \end{gathered}$$

Since localid="1654268154520" $a^5=e$, localid="1654268159171" $b=b^{32}$.

As $b\ne e$, this implies, $b^{31}=e$.

Therefore, we get $\left|b\right|=31$.