Question

Let G be a group generated by elements and b such that $\left|a\right|\mathbf{=}\mathbf{4}\mathbf{,}\mathbf{}\left|b\right|\mathbf{=}\mathbf{2}$, and $\mathit{b}\mathit{a}\mathbf{=}{\mathit{a}}^{\mathbf{3}}\mathit{b}$. Show that G is a group of order 8 and that G is isomorphic to ${\mathit{D}}_{\mathbf{4}}$.

Short answer

It is shown that,G is a group of order 8 and that G is isomorphic to $D_4$.

Step-by-step solution

Determine G is a group of order 8 and that G is isomorphic to D4

Consider that G is the group generated by elements and b with $\left|a\right|=4$and

$\left|b\right|=2$

Suppose that$ba=a^{3}b$ .

Now, let’s consider that, H is the subgroup generated by a.

Then, $H=\left\{e,a,a^2,a^3\right\}$.

Here, b does not belong to H.

As if $b\in H$then, $b=a^2$.

As order of b is 2 and the order of and $a^3$are 4.

It implies from the identity$ba=a^{3}b$ as:

$\begin{array}{rcl}ba& =& a^{3}b\\ & =& a^3{a}^2\\ & =& a^5\\ & =& a\\ & & \end{array}$

This implies,$b=e$ .

This is absurd as order of is 2.

Determine multiplication table

As $b\notin H$and bH are disjoint.

Claim on $H\cup bH$is closed under multiplication.

In order to this, draw the multiplication table,

Further simplification

As G is generated by a,b and $H\cup bH\subset G$and$H\cup bH$ is closed under multiplication, $G=H\cup bH$.

Consequently, G has order 8.

As the group $D_4$is generated by $r_1$and h with $\left|r_1\right|=4,\left|h\right|=2$and $h{r}_1={r_1}^{3}h$, the mapping $\varphi :G\to D_4$defined by, $f\left(a\right)=r_1,f\left(b\right)=h$and $f\left({r_1}^j{h}^i\right)=a^j{b}^i$, for $i=0,1,j=0,1,2,3$.

It turns out to be an isomorphism between G and D.

Hence, it is proved that G is a group of order 8 and that G is isomorphic to $D_4$.