Question

If G is a group of order 8 generated by elements a and b such that $\left|a\right|\mathbf{=}\mathbf{4}\mathbf{,}\mathit{b}\mathbf{\notin }<a>$, and ${\mathit{b}}^{\mathbf{2}}\mathbf{=}{\mathit{a}}^{\mathbf{3}}$, then G is abelian. [This fact is used in the proof of Theorem 9.34, so don’t use Theorem 9.34 to prove it.]

Short answer

It is proved that, G is abelian.

Step-by-step solution

Given information

It is given that G is a group of order 8 generated by elements a and b.

Also, and $b^2=a^3$.

Prove that G is abelian

Find $b^6$ as:

$\begin{array}{rcl}{b}^6& =& {\left(b^2\right)}^3\\ & =& {\left(a^3\right)}^3\\ & =& a^9\end{array}$

Since $\left|a\right|=4,a^9=a$.

Thus, $b^6=a$.

Now, find ab as:

$\begin{array}{rcl}ab& =& b^{6}b\\ & =& b{b}^6\\ & =& ba\end{array}$

Thus, G is abelian.

It can be concluded that, G is abelian.