# Commutator subgroup of a group of order $8q$, where $q$ is odd prime.

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.

• Please use langle and rangle instead of plain < and > Oct 13 '19 at 12:48
• @ajotatxe, ok sir
– MANI
Oct 13 '19 at 13:03
• @Bernard please take a look on the question.
– MANI
Oct 13 '19 at 13:55
• Since $q$ is odd, $\langle b^2 \rangle = \langle b \rangle$ and $G/\langle b\rangle$ is abelian so $[G,G]= \langle b\rangle$ Oct 13 '19 at 15:24
• @DerekHolt Thanks for giving this hint, but how can we say that $\langle b \rangle$ is the smallest normal subgroup $H$ of $G$ such that $G/H$ is abelian.
– MANI
Oct 13 '19 at 18:59

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 $$m,l\in\mathbb{Z}/8\mathbb{Z}$$.

• thank you sir, can you tell me that what mapping are you using from $G$ to $C_8$?
– MANI
Oct 14 '19 at 16:36
• From the presentation of $G$, you can write any element of $G$ as $b^ua^v$. Then, show that the map $G\to \langle a\rangle$ sending $b^ua^v\mapsto a^v$ is a surjective group homomorphism (with kernel being precisely $\langle b\rangle=H$). Oct 14 '19 at 16:39
• sorry for silly question but this is my last doubt, $b^ua^v$ and $a^vb^u$ are different elements here, how can we say that the map $f$ is well defined and is an homomorphism?
– MANI
Oct 14 '19 at 16:42
• They are different, but you can show that $b^ua^v=a^vb^{\text{something}}$. Oct 14 '19 at 16:43
• Thanks, now I got the key
– MANI
Oct 14 '19 at 16:46