Let $G=\langle a,b :a^8=b^q=1,a^{-1}b a=b^{-1}\rangle$ be a group of order $8q$, ($q$ is odd prime). Then what will be the commutator subgroup of this group $G$.
What have I done: If $x$ and $y$ be any two elements of $G$, such that $x,y \in \langle a\rangle$ or $x,y \in \langle b\rangle$ then $[x,y]=1$, and if $x \in \langle a\rangle ,~ y \in \langle b\rangle$ then $x=a^{i},~ y=b^j,~\text{for some $i~\text{and}~ j$}$, then the commutator $[x,y]=xyx^{-1}y^{-1}=a^ib^ja^{-i}b^{-j} \in \langle b^2\rangle$. This is similar like commutator operation in Dihedral groups $D_n$ of order $2n$. But if $x=a^ib^j$ and $y=a^lb^m$, for integers $i,j,l ~\text{and}~m$ then what we can do for $[x,y]$. I am stuck here, please help me.
Best Answer
I am writing here to make this thread answered. First, let $H$ be the subgroup generated by $b$. Then, as seen, $H$ is normal and isomorphic to $C_q$, where $C_n$ is the cyclic group of order $n$. We have the splitting exact sequence of groups $$1\to H \to G \to C_8\to 1\,.$$ Thus, $G/H\cong C_8$ is abelian, so the commutator subgroup $K$ of $G$ is contained in $H$. It is easy to see that $K=H$.
Now, from the above paragraph, we see that $G$ is the internal semidirect product $\langle b\rangle\rtimes \langle a\rangle$, with $\langle b\rangle\cong C_q$ and $\langle a\rangle\cong C_8$. Writing each element of $G$ as $b^ua^v$ with $v\in\mathbb{Z}/q\mathbb{Z}$ and $u\in\mathbb{Z}/8\mathbb{Z}$, the multiplication rule of $G=\langle b\rangle\rtimes \langle a\rangle$ is given by $$\left(b^na^m\right)\cdot \left(b^la^k\right)=b^{n+(-1)^mk}a^{m+l}$$ for all $n,k\in\mathbb{Z}/q\mathbb{Z}$ and $m,l\in\mathbb{Z}/8\mathbb{Z}$.