Question

Let G be the group ${\mathit{S}}_{\mathbf{3}}\mathbf{\times }{\mathrm{\mathbb{Z}}}_{\mathbf{4}}$and let $\mathit{a}\mathbf{=}\left(\left(123\right),2\right)$and $\mathit{b}\mathbf{=}\left(\left(12\right),1\right)$.

Show that $\left|a\right|\mathbf{=}\mathbf{6}\mathbf{,}{\mathit{b}}^{\mathbf{2}}\mathbf{=}{\mathit{a}}^{\mathbf{3}}\mathbf{,}$and $\mathit{b}\mathit{a}\mathbf{=}{\mathit{a}}^{\mathbf{-}\mathbf{1}}\mathit{b}$.

Short answer

It is proved that, $\left|a\right|=6,b^2=a^3,$ and$ba=a^{-1}b$ .

Step-by-step solution

Given information

It is given that $G=S_3\times {\mathrm{\mathbb{Z}}}_4$ .

Let $a=\left(\left(123\right),2\right)$and$b=\left(\left(12\right),1\right)$ .

Prove that |a|=6

Computing the powers of a as:

$$\begin{gathered} a=\left(\left(123\right),2\right) \\ {a}^2=\left(\left(132\right),0\right) \\ {a}^3=\left(\left(\right),2\right) \\ {a}^4=\left(\left(123\right),0\right) \end{gathered}$$

Further as:

$$\begin{gathered} a^5=\left(\left(132\right),2\right) \\ {a}^6=\left(\left(\right),0\right) \end{gathered}$$

Hence $\left|a\right|=6$.

Prove that b2=a3

Computing the powers of b as:

$$\begin{gathered} b=\left(\left(12\right),1\right) \\ {b}^2=\left(\left(\right),2\right) \end{gathered}$$

Hence, $b^2=a^3$.

Prove that ba=a-1b

Find ba as:

$\begin{array}{rcl}ba& =& \left(\left(12\right)\left(123\right),2+1\right)\\ & =& \left(\left(13\right),3\right)\end{array}$

Find $a^{-1}b$ as:

$\begin{array}{rcl}{a}^{-1}b& =& \left(\left(132\left(12\right),-2+1\right)\right)\\ & =& \left(\left(23\right),3\right)\end{array}$

Thus, $ba=a^{-1}b$.

Hence, $\left|a\right|=6,b^2=a^3,$and $ba=a^{-1}b$.