Show that if k is odd, then $\Bbb{Q}_{4k}$ is isomorphic to $\Bbb{Z}_k \rtimes_{\alpha} \Bbb{Z}_4$ for some $\alpha: \Bbb{Z}_4 \rightarrow Aut(\Bbb{Z}_k)$. Calculate $\alpha$ explicitly.
We know that $\Bbb{Q}_{4k} = \{ b^k_{2n}, b^k_{2n}a | 0 \leq k < 2n \}$, and that everything outside of the cyclic group $\langle b \rangle$ is of order 4. What confuses me is that we only have one cyclic group in $\Bbb{Q}$, right? However, in order to write it of the form $\Bbb{Z}_k \rtimes_{\alpha} \Bbb{Z}_4$, we need to find a normal cyclic group of order k in $\Bbb{Q}_{4k}$, and I can't really see that.
For example, in the group $\Bbb{Q}_{12} = \{e, b, b^2, b^3, b^4, b^5, a, ab, ab^2, ab^3, ab^4, ab^5\}$, we only have $\langle b \rangle$ as the cyclic group. I cannot see any normal cyclic group of order 3 here.
For the second part of the problem, suppose we accept the fact that $\Bbb{Q}_{4k} = \Bbb{Z}_k \rtimes_{\alpha} \Bbb{Z}_4$ for some $\alpha: \Bbb{Z}_4 \rightarrow Aut(\Bbb{Z}_k)$.
There is a theorem in the textbook that states:
Corollary Let $\bar{m}$ have exponent k in $\Bbb{Z}^×_n$, and let $α : \Bbb{Z}_k → \Bbb{Z}^×_n$ be the homomorphism that takes the generator to $\bar{m}$. Then writing $\Bbb{Z}_n = \langle b \rangle$, $\Bbb{Z}_k = \langle a \rangle$, and identifying $\Bbb{Z}^×_n$ with $Aut(\Bbb{Z}_n)$, we obtain $$\Bbb{Z}_n \rtimes_{α} \Bbb{Z}_k = \{b^ia^j | 0 ≤ i < n, 0 ≤ j < k\},$$ where b has order n, a has order k, and the multiplication is given by $$b^ia^jb^{i'}a^{j'} = b^{i+m^ji'}a^{j+j'}.$$
Moreover, the nk elements $\{b^ia^j | 0 \leq i < n, 0 \leq j < k\}.$
Since we know that $aba^{-1}=b^{-1}$ in quaternionics, we need to find an $\bar{m}$ such that $b^0aba^{-1}= b^{0+m}a^{-1+1} = b^{-1}$. So $\bar{m} = \bar{-1}$, and $\alpha$ must be send the generator to $\bar{-1}$. I guess this is what the question is asking for when it tells us the calculate $\alpha$ "explicitly". Is that correct?
However, I cannot say that until I prove that $\Bbb{Q}_{4k} \cong \Bbb{Z}_k \rtimes_{\alpha} \Bbb{Z}_{4}$, so I was wondering if anybody could help me with that.
Thank you in advance