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Let $G=\langle a,b:a^8=b^p=1,a^{-1}ba=b^\alpha \rangle$, where $\alpha$ is a primitive root of $\alpha^4 \equiv 1~\text{mod}(4)$, $4$ divides $p-1$. I wants to compute the commutator of $G$. My attempt is

Let $H=\langle b\rangle$ and $K=\langle a\rangle$. Then the sequence $\{1\}\longrightarrow H\stackrel{i}{\longrightarrow} G\stackrel{\pi}{\longrightarrow} K\longrightarrow \{1\}$, where $i(a)=a$ and $\pi(a^nb^m)=b^m$. Therefore $G/H \simeq K$.

I also thinks that and $[G,G]\simeq H$. One way it is clear $[G,G]\leq H$, but I am stuck in proving of $H\leq [G,G]$.

Please help me and thanks in advance.

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    $\begingroup$ It seems correct. Can you tell the roll of $\alpha$ in this proof. $\endgroup$ – MANI Oct 16 '19 at 11:14
  • $\begingroup$ @MANI Please see the edited question, thanks for your pleased advise. $\endgroup$ – Priya Pandey Oct 16 '19 at 11:21
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Your proof to show that above sequence is short esact is correct, nevertheless define a map $t:K\rightarrow G$, by $t(a)=a$. I have asked similar question (Commutator subgroup of a group of order $8q$, where $q$ is odd prime.) and got the answer, that's why I am giving answer in same pattern:

Now to show that $H\subseteq [G,G]$. Let $g=b^k \in H$. Then consider $[a^{-1},g]=a^{-1}gag^{-1}=b^{-\alpha k}b^{-k}=b^{-k(\alpha+1)}$. Since $p$ is prime therefore if any $n\neq mp,~b^{n}\in [G,G]\Rightarrow b\in [G,G].$

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  • $\begingroup$ why such $n$ will exist? $\endgroup$ – Priya Pandey Oct 16 '19 at 12:54

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