Show that If k is odd, then $\Bbb{Q}_{4k}$ is isomorphic to $\Bbb{Z}_k \rtimes_{\alpha} \Bbb{Z}_4$ for some $\alpha$ 
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
 A: Let's be clear on one significant point first:

What confuses me is that we only have one cyclic group in Q, right?

No. Remember that any chosen element $g$ in a group $G$ generates a cyclic group $\langle g\rangle$, so in general a group will have many cyclic subgroups. In fact, since any cyclic group has a generator, any cyclic subgroup of some $G$ will be obtained by generating it from some single element $g\in G$. Moreover, a group $G$ has only one nontrivial cyclic subgroup iff $G$ is itself a cyclic group of prime order.
To prove this, observe that if $G$ is not cyclic of prime order, then either 


*

*$G$ is cyclic of composite order $n$ generated by say $g$, in which case $\langle g^d\rangle$ is a proper nontrivial subgroup of $G$ when $d\mid n$ is a proper divisor and $\ne1$ hence $\langle g^d\rangle$ and $G$ itself are two distinct nontrivial subgroups of $G$ that are cyclic, or

*$G$ is not cyclic, so pick $x\in G\setminus\{e\}$, then pick $y\in G\setminus\langle x\rangle$, in which case $\langle x\rangle$ and $\langle y\rangle$ are two distinct nontrivial cyclic subgroups of $G$.


Now, back to generalized quaternion groups. The group presentation is

$$Q_{4k}=\langle a,b~|~b^{2k}=a^4=1,~b^n=a^2,~a^{-1}ba=b^{-1} \rangle$$

What are some cyclic subgroups? By experimentation, $\langle b\rangle$, $\langle b^2\rangle$, $\langle b^k\rangle$, $\langle a\rangle$, $\langle a^2\rangle$, etc. (At this point, assume $k$ is odd.) What are some subgroups of order $4$? The obvious one is $\langle a\rangle$. What about normal cyclic subgroups of order $k$? Well, $\langle b^2\rangle$ is cyclic of order $k$. We see that $b=a^2(b^2)^{(1-n)/2}\in\langle a,b^2\rangle$, which we can use to prove that $\langle a\rangle\langle b^2\rangle=Q_{4k}$. Furthermore $\langle b^2\rangle\cap\langle a\rangle=\{1\}$. We can check that $a\langle b^2\rangle a^{-1}=\langle b^{-2}\rangle=\langle b^2\rangle$ hence $\langle b^2\rangle$ is normalized by $a$ and $b$ hence normalized by the subgroup generated by $a$ and $b$, which is the full group $Q_{4k}$. So $\langle b^2\rangle$ is normal, cyclic and order $k$.
Thus, $Q_{4k}=\langle b^2\rangle\rtimes\langle a\rangle$. Apply $x\mapsto a^{-1}xa$ to $b^2$ three times to get $ab^2a^{-1}=b^{-2}$, hence we conclude that the homomorphism $\alpha:\langle a\rangle\to{\rm Aut}(\langle b^2\rangle)$ sends $a$ to the inverse map $x\mapsto x^{-1}$ (this is an automorphism of any abelian group), hence sends $a^2$ to the identity and also sends $a^3=a^{-1}$ to the inverse map. In terms of cyclic groups and unit groups,
$$Z_4\to U(k)={\rm Aut}(Z_k):0,2\mapsto 1;~1,3\mapsto -1.$$
However you'll have to judge for yourself from the context of the text and the material what constitutes an explicit bijection for the purposes of this exercise.
