# Split exact sequence

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 $$H\leq [G,G]$$.

• It seems correct. Can you tell the roll of $\alpha$ in this proof. – MANI Oct 16 '19 at 11:14
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 $$n\neq mp,~b^{n}\in [G,G]\Rightarrow b\in [G,G].$$
• why such $n$ will exist? – Priya Pandey Oct 16 '19 at 12:54