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.

  • $\begingroup$ Please use langle and rangle instead of plain < and > $\endgroup$
    – ajotatxe
    Oct 13 '19 at 12:48
  • $\begingroup$ @ajotatxe, ok sir $\endgroup$
    – MANI
    Oct 13 '19 at 13:03
  • $\begingroup$ @Bernard please take a look on the question. $\endgroup$
    – MANI
    Oct 13 '19 at 13:55
  • 2
    $\begingroup$ Since $q$ is odd, $\langle b^2 \rangle = \langle b \rangle$ and $G/\langle b\rangle$ is abelian so $[G,G]= \langle b\rangle$ $\endgroup$
    – Derek Holt
    Oct 13 '19 at 15:24
  • $\begingroup$ @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. $\endgroup$
    – 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 $m,l\in\mathbb{Z}/8\mathbb{Z}$.

  • $\begingroup$ thank you sir, can you tell me that what mapping are you using from $G$ to $C_8$? $\endgroup$
    – MANI
    Oct 14 '19 at 16:36
  • $\begingroup$ 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 $\langle b\rangle=H$). $\endgroup$ Oct 14 '19 at 16:39
  • $\begingroup$ 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? $\endgroup$
    – MANI
    Oct 14 '19 at 16:42
  • $\begingroup$ They are different, but you can show that $b^ua^v=a^vb^{\text{something}}$. $\endgroup$ Oct 14 '19 at 16:43
  • $\begingroup$ Thanks, now I got the key $\endgroup$
    – MANI
    Oct 14 '19 at 16:46

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